diff --git a/README.md b/README.md index bb07c5c9..9978339f 100644 --- a/README.md +++ b/README.md @@ -1,4 +1,4 @@ -# 1,709 LeetCode solutions in JavaScript +# 1,980 LeetCode solutions in JavaScript [https://leetcodejavascript.com](https://leetcodejavascript.com) @@ -1571,55 +1571,313 @@ 2045|[Second Minimum Time to Reach Destination](./solutions/2045-second-minimum-time-to-reach-destination.js)|Hard| 2047|[Number of Valid Words in a Sentence](./solutions/2047-number-of-valid-words-in-a-sentence.js)|Easy| 2048|[Next Greater Numerically Balanced Number](./solutions/2048-next-greater-numerically-balanced-number.js)|Medium| +2049|[Count Nodes With the Highest Score](./solutions/2049-count-nodes-with-the-highest-score.js)|Medium| +2050|[Parallel Courses III](./solutions/2050-parallel-courses-iii.js)|Hard| 2053|[Kth Distinct String in an Array](./solutions/2053-kth-distinct-string-in-an-array.js)|Medium| +2055|[Plates Between Candles](./solutions/2055-plates-between-candles.js)|Medium| +2056|[Number of Valid Move Combinations On Chessboard](./solutions/2056-number-of-valid-move-combinations-on-chessboard.js)|Hard| +2057|[Smallest Index With Equal Value](./solutions/2057-smallest-index-with-equal-value.js)|Easy| +2058|[Find the Minimum and Maximum Number of Nodes Between Critical Points](./solutions/2058-find-the-minimum-and-maximum-number-of-nodes-between-critical-points.js)|Medium| +2059|[Minimum Operations to Convert Number](./solutions/2059-minimum-operations-to-convert-number.js)|Medium| +2060|[Check if an Original String Exists Given Two Encoded Strings](./solutions/2060-check-if-an-original-string-exists-given-two-encoded-strings.js)|Hard| +2062|[Count Vowel Substrings of a String](./solutions/2062-count-vowel-substrings-of-a-string.js)|Easy| +2063|[Vowels of All Substrings](./solutions/2063-vowels-of-all-substrings.js)|Medium| +2064|[Minimized Maximum of Products Distributed to Any Store](./solutions/2064-minimized-maximum-of-products-distributed-to-any-store.js)|Medium| +2065|[Maximum Path Quality of a Graph](./solutions/2065-maximum-path-quality-of-a-graph.js)|Hard| +2068|[Check Whether Two Strings are Almost Equivalent](./solutions/2068-check-whether-two-strings-are-almost-equivalent.js)|Easy| +2069|[Walking Robot Simulation II](./solutions/2069-walking-robot-simulation-ii.js)|Medium| +2070|[Most Beautiful Item for Each Query](./solutions/2070-most-beautiful-item-for-each-query.js)|Medium| 2071|[Maximum Number of Tasks You Can Assign](./solutions/2071-maximum-number-of-tasks-you-can-assign.js)|Hard| +2074|[Reverse Nodes in Even Length Groups](./solutions/2074-reverse-nodes-in-even-length-groups.js)|Medium| +2075|[Decode the Slanted Ciphertext](./solutions/2075-decode-the-slanted-ciphertext.js)|Medium| +2076|[Process Restricted Friend Requests](./solutions/2076-process-restricted-friend-requests.js)|Hard| +2078|[Two Furthest Houses With Different Colors](./solutions/2078-two-furthest-houses-with-different-colors.js)|Easy| +2079|[Watering Plants](./solutions/2079-watering-plants.js)|Medium| +2080|[Range Frequency Queries](./solutions/2080-range-frequency-queries.js)|Medium| +2081|[Sum of k-Mirror Numbers](./solutions/2081-sum-of-k-mirror-numbers.js)|Hard| 2085|[Count Common Words With One Occurrence](./solutions/2085-count-common-words-with-one-occurrence.js)|Easy| +2086|[Minimum Number of Food Buckets to Feed the Hamsters](./solutions/2086-minimum-number-of-food-buckets-to-feed-the-hamsters.js)|Medium| +2087|[Minimum Cost Homecoming of a Robot in a Grid](./solutions/2087-minimum-cost-homecoming-of-a-robot-in-a-grid.js)|Medium| +2088|[Count Fertile Pyramids in a Land](./solutions/2088-count-fertile-pyramids-in-a-land.js)|Hard| +2089|[Find Target Indices After Sorting Array](./solutions/2089-find-target-indices-after-sorting-array.js)|Easy| +2090|[K Radius Subarray Averages](./solutions/2090-k-radius-subarray-averages.js)|Medium| +2091|[Removing Minimum and Maximum From Array](./solutions/2091-removing-minimum-and-maximum-from-array.js)|Medium| +2092|[Find All People With Secret](./solutions/2092-find-all-people-with-secret.js)|Hard| +2094|[Finding 3-Digit Even Numbers](./solutions/2094-finding-3-digit-even-numbers.js)|Easy| 2095|[Delete the Middle Node of a Linked List](./solutions/2095-delete-the-middle-node-of-a-linked-list.js)|Medium| +2096|[Step-By-Step Directions From a Binary Tree Node to Another](./solutions/2096-step-by-step-directions-from-a-binary-tree-node-to-another.js)|Medium| +2097|[Valid Arrangement of Pairs](./solutions/2097-valid-arrangement-of-pairs.js)|Hard| 2099|[Find Subsequence of Length K With the Largest Sum](./solutions/2099-find-subsequence-of-length-k-with-the-largest-sum.js)|Medium| +2100|[Find Good Days to Rob the Bank](./solutions/2100-find-good-days-to-rob-the-bank.js)|Medium| +2101|[Detonate the Maximum Bombs](./solutions/2101-detonate-the-maximum-bombs.js)|Medium| +2103|[Rings and Rods](./solutions/2103-rings-and-rods.js)|Easy| +2104|[Sum of Subarray Ranges](./solutions/2104-sum-of-subarray-ranges.js)|Medium| +2105|[Watering Plants II](./solutions/2105-watering-plants-ii.js)|Medium| +2106|[Maximum Fruits Harvested After at Most K Steps](./solutions/2106-maximum-fruits-harvested-after-at-most-k-steps.js)|Hard| +2108|[Find First Palindromic String in the Array](./solutions/2108-find-first-palindromic-string-in-the-array.js)|Easy| +2109|[Adding Spaces to a String](./solutions/2109-adding-spaces-to-a-string.js)|Medium| +2110|[Number of Smooth Descent Periods of a Stock](./solutions/2110-number-of-smooth-descent-periods-of-a-stock.js)|Medium| +2111|[Minimum Operations to Make the Array K-Increasing](./solutions/2111-minimum-operations-to-make-the-array-k-increasing.js)|Hard| 2114|[Maximum Number of Words Found in Sentences](./solutions/2114-maximum-number-of-words-found-in-sentences.js)|Easy| 2115|[Find All Possible Recipes from Given Supplies](./solutions/2115-find-all-possible-recipes-from-given-supplies.js)|Medium| 2116|[Check if a Parentheses String Can Be Valid](./solutions/2116-check-if-a-parentheses-string-can-be-valid.js)|Medium| +2117|[Abbreviating the Product of a Range](./solutions/2117-abbreviating-the-product-of-a-range.js)|Hard| +2119|[A Number After a Double Reversal](./solutions/2119-a-number-after-a-double-reversal.js)|Easy| +2120|[Execution of All Suffix Instructions Staying in a Grid](./solutions/2120-execution-of-all-suffix-instructions-staying-in-a-grid.js)|Medium| +2121|[Intervals Between Identical Elements](./solutions/2121-intervals-between-identical-elements.js)|Medium| +2122|[Recover the Original Array](./solutions/2122-recover-the-original-array.js)|Hard| +2124|[Check if All A's Appears Before All B's](./solutions/2124-check-if-all-as-appears-before-all-bs.js)|Easy| +2125|[Number of Laser Beams in a Bank](./solutions/2125-number-of-laser-beams-in-a-bank.js)|Medium| +2126|[Destroying Asteroids](./solutions/2126-destroying-asteroids.js)|Medium| 2127|[Maximum Employees to Be Invited to a Meeting](./solutions/2127-maximum-employees-to-be-invited-to-a-meeting.js)|Hard| 2129|[Capitalize the Title](./solutions/2129-capitalize-the-title.js)|Easy| 2130|[Maximum Twin Sum of a Linked List](./solutions/2130-maximum-twin-sum-of-a-linked-list.js)|Medium| +2131|[Longest Palindrome by Concatenating Two Letter Words](./solutions/2131-longest-palindrome-by-concatenating-two-letter-words.js)|Medium| +2132|[Stamping the Grid](./solutions/2132-stamping-the-grid.js)|Hard| +2133|[Check if Every Row and Column Contains All Numbers](./solutions/2133-check-if-every-row-and-column-contains-all-numbers.js)|Easy| +2134|[Minimum Swaps to Group All 1's Together II](./solutions/2134-minimum-swaps-to-group-all-1s-together-ii.js)|Medium| +2135|[Count Words Obtained After Adding a Letter](./solutions/2135-count-words-obtained-after-adding-a-letter.js)|Medium| +2136|[Earliest Possible Day of Full Bloom](./solutions/2136-earliest-possible-day-of-full-bloom.js)|Hard| +2138|[Divide a String Into Groups of Size k](./solutions/2138-divide-a-string-into-groups-of-size-k.js)|Easy| +2139|[Minimum Moves to Reach Target Score](./solutions/2139-minimum-moves-to-reach-target-score.js)|Medium| 2140|[Solving Questions With Brainpower](./solutions/2140-solving-questions-with-brainpower.js)|Medium| +2141|[Maximum Running Time of N Computers](./solutions/2141-maximum-running-time-of-n-computers.js)|Hard| +2144|[Minimum Cost of Buying Candies With Discount](./solutions/2144-minimum-cost-of-buying-candies-with-discount.js)|Easy| 2145|[Count the Hidden Sequences](./solutions/2145-count-the-hidden-sequences.js)|Medium| +2147|[Number of Ways to Divide a Long Corridor](./solutions/2147-number-of-ways-to-divide-a-long-corridor.js)|Hard| +2148|[Count Elements With Strictly Smaller and Greater Elements](./solutions/2148-count-elements-with-strictly-smaller-and-greater-elements.js)|Easy| +2149|[Rearrange Array Elements by Sign](./solutions/2149-rearrange-array-elements-by-sign.js)|Medium| +2150|[Find All Lonely Numbers in the Array](./solutions/2150-find-all-lonely-numbers-in-the-array.js)|Medium| +2151|[Maximum Good People Based on Statements](./solutions/2151-maximum-good-people-based-on-statements.js)|Hard| 2154|[Keep Multiplying Found Values by Two](./solutions/2154-keep-multiplying-found-values-by-two.js)|Easy| +2155|[All Divisions With the Highest Score of a Binary Array](./solutions/2155-all-divisions-with-the-highest-score-of-a-binary-array.js)|Medium| +2156|[Find Substring With Given Hash Value](./solutions/2156-find-substring-with-given-hash-value.js)|Hard| +2157|[Groups of Strings](./solutions/2157-groups-of-strings.js)|Hard| +2160|[Minimum Sum of Four Digit Number After Splitting Digits](./solutions/2160-minimum-sum-of-four-digit-number-after-splitting-digits.js)|Easy| 2161|[Partition Array According to Given Pivot](./solutions/2161-partition-array-according-to-given-pivot.js)|Medium| +2162|[Minimum Cost to Set Cooking Time](./solutions/2162-minimum-cost-to-set-cooking-time.js)|Medium| +2164|[Sort Even and Odd Indices Independently](./solutions/2164-sort-even-and-odd-indices-independently.js)|Easy| +2165|[Smallest Value of the Rearranged Number](./solutions/2165-smallest-value-of-the-rearranged-number.js)|Medium| +2166|[Design Bitset](./solutions/2166-design-bitset.js)|Medium| +2167|[Minimum Time to Remove All Cars Containing Illegal Goods](./solutions/2167-minimum-time-to-remove-all-cars-containing-illegal-goods.js)|Hard| +2169|[Count Operations to Obtain Zero](./solutions/2169-count-operations-to-obtain-zero.js)|Easy| +2170|[Minimum Operations to Make the Array Alternating](./solutions/2170-minimum-operations-to-make-the-array-alternating.js)|Medium| +2171|[Removing Minimum Number of Magic Beans](./solutions/2171-removing-minimum-number-of-magic-beans.js)|Medium| +2172|[Maximum AND Sum of Array](./solutions/2172-maximum-and-sum-of-array.js)|Hard| 2176|[Count Equal and Divisible Pairs in an Array](./solutions/2176-count-equal-and-divisible-pairs-in-an-array.js)|Easy| +2177|[Find Three Consecutive Integers That Sum to a Given Number](./solutions/2177-find-three-consecutive-integers-that-sum-to-a-given-number.js)|Medium| +2178|[Maximum Split of Positive Even Integers](./solutions/2178-maximum-split-of-positive-even-integers.js)|Medium| 2179|[Count Good Triplets in an Array](./solutions/2179-count-good-triplets-in-an-array.js)|Hard| +2180|[Count Integers With Even Digit Sum](./solutions/2180-count-integers-with-even-digit-sum.js)|Easy| +2181|[Merge Nodes in Between Zeros](./solutions/2181-merge-nodes-in-between-zeros.js)|Medium| +2183|[Count Array Pairs Divisible by K](./solutions/2183-count-array-pairs-divisible-by-k.js)|Hard| 2185|[Counting Words With a Given Prefix](./solutions/2185-counting-words-with-a-given-prefix.js)|Easy| +2186|[Minimum Number of Steps to Make Two Strings Anagram II](./solutions/2186-minimum-number-of-steps-to-make-two-strings-anagram-ii.js)|Medium| +2187|[Minimum Time to Complete Trips](./solutions/2187-minimum-time-to-complete-trips.js)|Medium| +2188|[Minimum Time to Finish the Race](./solutions/2188-minimum-time-to-finish-the-race.js)|Hard| +2190|[Most Frequent Number Following Key In an Array](./solutions/2190-most-frequent-number-following-key-in-an-array.js)|Easy| +2191|[Sort the Jumbled Numbers](./solutions/2191-sort-the-jumbled-numbers.js)|Medium| +2192|[All Ancestors of a Node in a Directed Acyclic Graph](./solutions/2192-all-ancestors-of-a-node-in-a-directed-acyclic-graph.js)|Medium| +2193|[Minimum Number of Moves to Make Palindrome](./solutions/2193-minimum-number-of-moves-to-make-palindrome.js)|Hard| +2194|[Cells in a Range on an Excel Sheet](./solutions/2194-cells-in-a-range-on-an-excel-sheet.js)|Easy| +2195|[Append K Integers With Minimal Sum](./solutions/2195-append-k-integers-with-minimal-sum.js)|Medium| +2196|[Create Binary Tree From Descriptions](./solutions/2196-create-binary-tree-from-descriptions.js)|Medium| +2197|[Replace Non-Coprime Numbers in Array](./solutions/2197-replace-non-coprime-numbers-in-array.js)|Hard| +2200|[Find All K-Distant Indices in an Array](./solutions/2200-find-all-k-distant-indices-in-an-array.js)|Easy| +2201|[Count Artifacts That Can Be Extracted](./solutions/2201-count-artifacts-that-can-be-extracted.js)|Medium| +2202|[Maximize the Topmost Element After K Moves](./solutions/2202-maximize-the-topmost-element-after-k-moves.js)|Medium| +2203|[Minimum Weighted Subgraph With the Required Paths](./solutions/2203-minimum-weighted-subgraph-with-the-required-paths.js)|Hard| 2206|[Divide Array Into Equal Pairs](./solutions/2206-divide-array-into-equal-pairs.js)|Easy| +2207|[Maximize Number of Subsequences in a String](./solutions/2207-maximize-number-of-subsequences-in-a-string.js)|Medium| +2209|[Minimum White Tiles After Covering With Carpets](./solutions/2209-minimum-white-tiles-after-covering-with-carpets.js)|Hard| +2210|[Count Hills and Valleys in an Array](./solutions/2210-count-hills-and-valleys-in-an-array.js)|Easy| +2211|[Count Collisions on a Road](./solutions/2211-count-collisions-on-a-road.js)|Medium| +2212|[Maximum Points in an Archery Competition](./solutions/2212-maximum-points-in-an-archery-competition.js)|Medium| +2213|[Longest Substring of One Repeating Character](./solutions/2213-longest-substring-of-one-repeating-character.js)|Hard| 2215|[Find the Difference of Two Arrays](./solutions/2215-find-the-difference-of-two-arrays.js)|Easy| +2216|[Minimum Deletions to Make Array Beautiful](./solutions/2216-minimum-deletions-to-make-array-beautiful.js)|Medium| +2217|[Find Palindrome With Fixed Length](./solutions/2217-find-palindrome-with-fixed-length.js)|Medium| +2218|[Maximum Value of K Coins From Piles](./solutions/2218-maximum-value-of-k-coins-from-piles.js)|Hard| +2220|[Minimum Bit Flips to Convert Number](./solutions/2220-minimum-bit-flips-to-convert-number.js)|Easy| +2221|[Find Triangular Sum of an Array](./solutions/2221-find-triangular-sum-of-an-array.js)|Medium| +2222|[Number of Ways to Select Buildings](./solutions/2222-number-of-ways-to-select-buildings.js)|Medium| +2223|[Sum of Scores of Built Strings](./solutions/2223-sum-of-scores-of-built-strings.js)|Hard| +2224|[Minimum Number of Operations to Convert Time](./solutions/2224-minimum-number-of-operations-to-convert-time.js)|Easy| +2225|[Find Players With Zero or One Losses](./solutions/2225-find-players-with-zero-or-one-losses.js)|Medium| 2226|[Maximum Candies Allocated to K Children](./solutions/2226-maximum-candies-allocated-to-k-children.js)|Medium| +2227|[Encrypt and Decrypt Strings](./solutions/2227-encrypt-and-decrypt-strings.js)|Hard| +2232|[Minimize Result by Adding Parentheses to Expression](./solutions/2232-minimize-result-by-adding-parentheses-to-expression.js)|Medium| +2234|[Maximum Total Beauty of the Gardens](./solutions/2234-maximum-total-beauty-of-the-gardens.js)|Hard| 2235|[Add Two Integers](./solutions/2235-add-two-integers.js)|Easy| +2236|[Root Equals Sum of Children](./solutions/2236-root-equals-sum-of-children.js)|Easy| +2239|[Find Closest Number to Zero](./solutions/2239-find-closest-number-to-zero.js)|Easy| +2240|[Number of Ways to Buy Pens and Pencils](./solutions/2240-number-of-ways-to-buy-pens-and-pencils.js)|Medium| +2241|[Design an ATM Machine](./solutions/2241-design-an-atm-machine.js)|Medium| +2242|[Maximum Score of a Node Sequence](./solutions/2242-maximum-score-of-a-node-sequence.js)|Hard| +2243|[Calculate Digit Sum of a String](./solutions/2243-calculate-digit-sum-of-a-string.js)|Easy| 2244|[Minimum Rounds to Complete All Tasks](./solutions/2244-minimum-rounds-to-complete-all-tasks.js)|Medium| +2245|[Maximum Trailing Zeros in a Cornered Path](./solutions/2245-maximum-trailing-zeros-in-a-cornered-path.js)|Medium| +2246|[Longest Path With Different Adjacent Characters](./solutions/2246-longest-path-with-different-adjacent-characters.js)|Hard| +2248|[Intersection of Multiple Arrays](./solutions/2248-intersection-of-multiple-arrays.js)|Easy| +2249|[Count Lattice Points Inside a Circle](./solutions/2249-count-lattice-points-inside-a-circle.js)|Medium| +2250|[Count Number of Rectangles Containing Each Point](./solutions/2250-count-number-of-rectangles-containing-each-point.js)|Medium| +2251|[Number of Flowers in Full Bloom](./solutions/2251-number-of-flowers-in-full-bloom.js)|Hard| +2255|[Count Prefixes of a Given String](./solutions/2255-count-prefixes-of-a-given-string.js)|Easy| +2256|[Minimum Average Difference](./solutions/2256-minimum-average-difference.js)|Medium| +2257|[Count Unguarded Cells in the Grid](./solutions/2257-count-unguarded-cells-in-the-grid.js)|Medium| +2258|[Escape the Spreading Fire](./solutions/2258-escape-the-spreading-fire.js)|Hard| +2259|[Remove Digit From Number to Maximize Result](./solutions/2259-remove-digit-from-number-to-maximize-result.js)|Easy| +2260|[Minimum Consecutive Cards to Pick Up](./solutions/2260-minimum-consecutive-cards-to-pick-up.js)|Medium| +2261|[K Divisible Elements Subarrays](./solutions/2261-k-divisible-elements-subarrays.js)|Medium| +2262|[Total Appeal of A String](./solutions/2262-total-appeal-of-a-string.js)|Hard| +2264|[Largest 3-Same-Digit Number in String](./solutions/2264-largest-3-same-digit-number-in-string.js)|Easy| +2265|[Count Nodes Equal to Average of Subtree](./solutions/2265-count-nodes-equal-to-average-of-subtree.js)|Medium| +2266|[Count Number of Texts](./solutions/2266-count-number-of-texts.js)|Medium| +2267|[Check if There Is a Valid Parentheses String Path](./solutions/2267-check-if-there-is-a-valid-parentheses-string-path.js)|Hard| +2269|[Find the K-Beauty of a Number](./solutions/2269-find-the-k-beauty-of-a-number.js)|Easy| 2270|[Number of Ways to Split Array](./solutions/2270-number-of-ways-to-split-array.js)|Medium| +2271|[Maximum White Tiles Covered by a Carpet](./solutions/2271-maximum-white-tiles-covered-by-a-carpet.js)|Medium| +2272|[Substring With Largest Variance](./solutions/2272-substring-with-largest-variance.js)|Hard| +2273|[Find Resultant Array After Removing Anagrams](./solutions/2273-find-resultant-array-after-removing-anagrams.js)|Easy| +2274|[Maximum Consecutive Floors Without Special Floors](./solutions/2274-maximum-consecutive-floors-without-special-floors.js)|Medium| +2275|[Largest Combination With Bitwise AND Greater Than Zero](./solutions/2275-largest-combination-with-bitwise-and-greater-than-zero.js)|Medium| +2278|[Percentage of Letter in String](./solutions/2278-percentage-of-letter-in-string.js)|Easy| +2279|[Maximum Bags With Full Capacity of Rocks](./solutions/2279-maximum-bags-with-full-capacity-of-rocks.js)|Medium| +2280|[Minimum Lines to Represent a Line Chart](./solutions/2280-minimum-lines-to-represent-a-line-chart.js)|Medium| +2283|[Check if Number Has Equal Digit Count and Digit Value](./solutions/2283-check-if-number-has-equal-digit-count-and-digit-value.js)|Easy| +2284|[Sender With Largest Word Count](./solutions/2284-sender-with-largest-word-count.js)|Medium| +2287|[Rearrange Characters to Make Target String](./solutions/2287-rearrange-characters-to-make-target-string.js)|Easy| +2288|[Apply Discount to Prices](./solutions/2288-apply-discount-to-prices.js)|Medium| +2293|[Min Max Game](./solutions/2293-min-max-game.js)|Easy| +2294|[Partition Array Such That Maximum Difference Is K](./solutions/2294-partition-array-such-that-maximum-difference-is-k.js)|Medium| +2295|[Replace Elements in an Array](./solutions/2295-replace-elements-in-an-array.js)|Medium| +2296|[Design a Text Editor](./solutions/2296-design-a-text-editor.js)|Hard| +2299|[Strong Password Checker II](./solutions/2299-strong-password-checker-ii.js)|Easy| 2300|[Successful Pairs of Spells and Potions](./solutions/2300-successful-pairs-of-spells-and-potions.js)|Medium| +2301|[Match Substring After Replacement](./solutions/2301-match-substring-after-replacement.js)|Hard| 2302|[Count Subarrays With Score Less Than K](./solutions/2302-count-subarrays-with-score-less-than-k.js)|Hard| +2303|[Calculate Amount Paid in Taxes](./solutions/2303-calculate-amount-paid-in-taxes.js)|Easy| +2304|[Minimum Path Cost in a Grid](./solutions/2304-minimum-path-cost-in-a-grid.js)|Medium| +2305|[Fair Distribution of Cookies](./solutions/2305-fair-distribution-of-cookies.js)|Medium| +2306|[Naming a Company](./solutions/2306-naming-a-company.js)|Hard| +2309|[Greatest English Letter in Upper and Lower Case](./solutions/2309-greatest-english-letter-in-upper-and-lower-case.js)|Easy| +2312|[Selling Pieces of Wood](./solutions/2312-selling-pieces-of-wood.js)|Hard| +2315|[Count Asterisks](./solutions/2315-count-asterisks.js)|Easy| +2316|[Count Unreachable Pairs of Nodes in an Undirected Graph](./solutions/2316-count-unreachable-pairs-of-nodes-in-an-undirected-graph.js)|Medium| +2317|[Maximum XOR After Operations](./solutions/2317-maximum-xor-after-operations.js)|Medium| +2318|[Number of Distinct Roll Sequences](./solutions/2318-number-of-distinct-roll-sequences.js)|Hard| +2319|[Check if Matrix Is X-Matrix](./solutions/2319-check-if-matrix-is-x-matrix.js)|Easy| +2320|[Count Number of Ways to Place Houses](./solutions/2320-count-number-of-ways-to-place-houses.js)|Medium| +2321|[Maximum Score Of Spliced Array](./solutions/2321-maximum-score-of-spliced-array.js)|Hard| +2322|[Minimum Score After Removals on a Tree](./solutions/2322-minimum-score-after-removals-on-a-tree.js)|Hard| +2325|[Decode the Message](./solutions/2325-decode-the-message.js)|Easy| +2326|[Spiral Matrix IV](./solutions/2326-spiral-matrix-iv.js)|Medium| +2328|[Number of Increasing Paths in a Grid](./solutions/2328-number-of-increasing-paths-in-a-grid.js)|Hard| +2331|[Evaluate Boolean Binary Tree](./solutions/2331-evaluate-boolean-binary-tree.js)|Easy| +2334|[Subarray With Elements Greater Than Varying Threshold](./solutions/2334-subarray-with-elements-greater-than-varying-threshold.js)|Hard| 2336|[Smallest Number in Infinite Set](./solutions/2336-smallest-number-in-infinite-set.js)|Medium| +2337|[Move Pieces to Obtain a String](./solutions/2337-move-pieces-to-obtain-a-string.js)|Medium| 2338|[Count the Number of Ideal Arrays](./solutions/2338-count-the-number-of-ideal-arrays.js)|Hard| +2341|[Maximum Number of Pairs in Array](./solutions/2341-maximum-number-of-pairs-in-array.js)|Easy| 2342|[Max Sum of a Pair With Equal Sum of Digits](./solutions/2342-max-sum-of-a-pair-with-equal-sum-of-digits.js)|Medium| +2347|[Best Poker Hand](./solutions/2347-best-poker-hand.js)|Easy| +2348|[Number of Zero-Filled Subarrays](./solutions/2348-number-of-zero-filled-subarrays.js)|Medium| 2349|[Design a Number Container System](./solutions/2349-design-a-number-container-system.js)|Medium| +2350|[Shortest Impossible Sequence of Rolls](./solutions/2350-shortest-impossible-sequence-of-rolls.js)|Hard| +2351|[First Letter to Appear Twice](./solutions/2351-first-letter-to-appear-twice.js)|Easy| 2352|[Equal Row and Column Pairs](./solutions/2352-equal-row-and-column-pairs.js)|Medium| +2354|[Number of Excellent Pairs](./solutions/2354-number-of-excellent-pairs.js)|Hard| +2358|[Maximum Number of Groups Entering a Competition](./solutions/2358-maximum-number-of-groups-entering-a-competition.js)|Medium| +2359|[Find Closest Node to Given Two Nodes](./solutions/2359-find-closest-node-to-given-two-nodes.js)|Medium| +2360|[Longest Cycle in a Graph](./solutions/2360-longest-cycle-in-a-graph.js)|Hard| +2363|[Merge Similar Items](./solutions/2363-merge-similar-items.js)|Easy| 2364|[Count Number of Bad Pairs](./solutions/2364-count-number-of-bad-pairs.js)|Medium| +2365|[Task Scheduler II](./solutions/2365-task-scheduler-ii.js)|Medium| +2366|[Minimum Replacements to Sort the Array](./solutions/2366-minimum-replacements-to-sort-the-array.js)|Hard| +2367|[Number of Arithmetic Triplets](./solutions/2367-number-of-arithmetic-triplets.js)|Easy| +2368|[Reachable Nodes With Restrictions](./solutions/2368-reachable-nodes-with-restrictions.js)|Medium| +2369|[Check if There is a Valid Partition For The Array](./solutions/2369-check-if-there-is-a-valid-partition-for-the-array.js)|Medium| +2370|[Longest Ideal Subsequence](./solutions/2370-longest-ideal-subsequence.js)|Medium| +2373|[Largest Local Values in a Matrix](./solutions/2373-largest-local-values-in-a-matrix.js)|Easy| +2374|[Node With Highest Edge Score](./solutions/2374-node-with-highest-edge-score.js)|Medium| 2375|[Construct Smallest Number From DI String](./solutions/2375-construct-smallest-number-from-di-string.js)|Medium| 2379|[Minimum Recolors to Get K Consecutive Black Blocks](./solutions/2379-minimum-recolors-to-get-k-consecutive-black-blocks.js)|Easy| +2380|[Time Needed to Rearrange a Binary String](./solutions/2380-time-needed-to-rearrange-a-binary-string.js)|Medium| 2381|[Shifting Letters II](./solutions/2381-shifting-letters-ii.js)|Medium| +2382|[Maximum Segment Sum After Removals](./solutions/2382-maximum-segment-sum-after-removals.js)|Hard| +2383|[Minimum Hours of Training to Win a Competition](./solutions/2383-minimum-hours-of-training-to-win-a-competition.js)|Easy| +2385|[Amount of Time for Binary Tree to Be Infected](./solutions/2385-amount-of-time-for-binary-tree-to-be-infected.js)|Medium| +2389|[Longest Subsequence With Limited Sum](./solutions/2389-longest-subsequence-with-limited-sum.js)|Easy| 2390|[Removing Stars From a String](./solutions/2390-removing-stars-from-a-string.js)|Medium| +2391|[Minimum Amount of Time to Collect Garbage](./solutions/2391-minimum-amount-of-time-to-collect-garbage.js)|Medium| +2392|[Build a Matrix With Conditions](./solutions/2392-build-a-matrix-with-conditions.js)|Hard| +2395|[Find Subarrays With Equal Sum](./solutions/2395-find-subarrays-with-equal-sum.js)|Easy| 2396|[Strictly Palindromic Number](./solutions/2396-strictly-palindromic-number.js)|Medium| +2397|[Maximum Rows Covered by Columns](./solutions/2397-maximum-rows-covered-by-columns.js)|Medium| +2399|[Check Distances Between Same Letters](./solutions/2399-check-distances-between-same-letters.js)|Easy| 2401|[Longest Nice Subarray](./solutions/2401-longest-nice-subarray.js)|Medium| +2404|[Most Frequent Even Element](./solutions/2404-most-frequent-even-element.js)|Easy| +2405|[Optimal Partition of String](./solutions/2405-optimal-partition-of-string.js)|Medium| +2409|[Count Days Spent Together](./solutions/2409-count-days-spent-together.js)|Easy| +2410|[Maximum Matching of Players With Trainers](./solutions/2410-maximum-matching-of-players-with-trainers.js)|Medium| +2411|[Smallest Subarrays With Maximum Bitwise OR](./solutions/2411-smallest-subarrays-with-maximum-bitwise-or.js)|Medium| +2412|[Minimum Money Required Before Transactions](./solutions/2412-minimum-money-required-before-transactions.js)|Hard| 2413|[Smallest Even Multiple](./solutions/2413-smallest-even-multiple.js)|Easy| +2414|[Length of the Longest Alphabetical Continuous Substring](./solutions/2414-length-of-the-longest-alphabetical-continuous-substring.js)|Medium| +2415|[Reverse Odd Levels of Binary Tree](./solutions/2415-reverse-odd-levels-of-binary-tree.js)|Medium| +2416|[Sum of Prefix Scores of Strings](./solutions/2416-sum-of-prefix-scores-of-strings.js)|Hard| +2418|[Sort the People](./solutions/2418-sort-the-people.js)|Easy| +2419|[Longest Subarray With Maximum Bitwise AND](./solutions/2419-longest-subarray-with-maximum-bitwise-and.js)|Medium| +2421|[Number of Good Paths](./solutions/2421-number-of-good-paths.js)|Hard| 2425|[Bitwise XOR of All Pairings](./solutions/2425-bitwise-xor-of-all-pairings.js)|Medium| +2426|[Number of Pairs Satisfying Inequality](./solutions/2426-number-of-pairs-satisfying-inequality.js)|Hard| 2427|[Number of Common Factors](./solutions/2427-number-of-common-factors.js)|Easy| +2428|[Maximum Sum of an Hourglass](./solutions/2428-maximum-sum-of-an-hourglass.js)|Medium| 2429|[Minimize XOR](./solutions/2429-minimize-xor.js)|Medium| +2432|[The Employee That Worked on the Longest Task](./solutions/2432-the-employee-that-worked-on-the-longest-task.js)|Easy| +2433|[Find The Original Array of Prefix Xor](./solutions/2433-find-the-original-array-of-prefix-xor.js)|Medium| +2434|[Using a Robot to Print the Lexicographically Smallest String](./solutions/2434-using-a-robot-to-print-the-lexicographically-smallest-string.js)|Medium| +2435|[Paths in Matrix Whose Sum Is Divisible by K](./solutions/2435-paths-in-matrix-whose-sum-is-divisible-by-k.js)|Hard| +2437|[Number of Valid Clock Times](./solutions/2437-number-of-valid-clock-times.js)|Easy| +2438|[Range Product Queries of Powers](./solutions/2438-range-product-queries-of-powers.js)|Medium| +2439|[Minimize Maximum of Array](./solutions/2439-minimize-maximum-of-array.js)|Medium| +2440|[Create Components With Same Value](./solutions/2440-create-components-with-same-value.js)|Hard| +2441|[Largest Positive Integer That Exists With Its Negative](./solutions/2441-largest-positive-integer-that-exists-with-its-negative.js)|Easy| +2442|[Count Number of Distinct Integers After Reverse Operations](./solutions/2442-count-number-of-distinct-integers-after-reverse-operations.js)|Medium| +2443|[Sum of Number and Its Reverse](./solutions/2443-sum-of-number-and-its-reverse.js)|Medium| 2444|[Count Subarrays With Fixed Bounds](./solutions/2444-count-subarrays-with-fixed-bounds.js)|Hard| +2446|[Determine if Two Events Have Conflict](./solutions/2446-determine-if-two-events-have-conflict.js)|Easy| +2447|[Number of Subarrays With GCD Equal to K](./solutions/2447-number-of-subarrays-with-gcd-equal-to-k.js)|Medium| +2448|[Minimum Cost to Make Array Equal](./solutions/2448-minimum-cost-to-make-array-equal.js)|Hard| +2449|[Minimum Number of Operations to Make Arrays Similar](./solutions/2449-minimum-number-of-operations-to-make-arrays-similar.js)|Hard| +2451|[Odd String Difference](./solutions/2451-odd-string-difference.js)|Easy| +2452|[Words Within Two Edits of Dictionary](./solutions/2452-words-within-two-edits-of-dictionary.js)|Medium| +2453|[Destroy Sequential Targets](./solutions/2453-destroy-sequential-targets.js)|Medium| +2455|[Average Value of Even Numbers That Are Divisible by Three](./solutions/2455-average-value-of-even-numbers-that-are-divisible-by-three.js)|Easy| +2458|[Height of Binary Tree After Subtree Removal Queries](./solutions/2458-height-of-binary-tree-after-subtree-removal-queries.js)|Hard| 2460|[Apply Operations to an Array](./solutions/2460-apply-operations-to-an-array.js)|Easy| +2461|[Maximum Sum of Distinct Subarrays With Length K](./solutions/2461-maximum-sum-of-distinct-subarrays-with-length-k.js)|Medium| 2462|[Total Cost to Hire K Workers](./solutions/2462-total-cost-to-hire-k-workers.js)|Medium| +2463|[Minimum Total Distance Traveled](./solutions/2463-minimum-total-distance-traveled.js)|Hard| +2465|[Number of Distinct Averages](./solutions/2465-number-of-distinct-averages.js)|Easy| +2466|[Count Ways To Build Good Strings](./solutions/2466-count-ways-to-build-good-strings.js)|Medium| 2467|[Most Profitable Path in a Tree](./solutions/2467-most-profitable-path-in-a-tree.js)|Medium| +2468|[Split Message Based on Limit](./solutions/2468-split-message-based-on-limit.js)|Hard| 2469|[Convert the Temperature](./solutions/2469-convert-the-temperature.js)|Easy| +2471|[Minimum Number of Operations to Sort a Binary Tree by Level](./solutions/2471-minimum-number-of-operations-to-sort-a-binary-tree-by-level.js)|Medium| +2472|[Maximum Number of Non-overlapping Palindrome Substrings](./solutions/2472-maximum-number-of-non-overlapping-palindrome-substrings.js)|Hard| +2475|[Number of Unequal Triplets in Array](./solutions/2475-number-of-unequal-triplets-in-array.js)|Easy| +2476|[Closest Nodes Queries in a Binary Search Tree](./solutions/2476-closest-nodes-queries-in-a-binary-search-tree.js)|Medium| +2477|[Minimum Fuel Cost to Report to the Capital](./solutions/2477-minimum-fuel-cost-to-report-to-the-capital.js)|Medium| +2481|[Minimum Cuts to Divide a Circle](./solutions/2481-minimum-cuts-to-divide-a-circle.js)|Easy| 2482|[Difference Between Ones and Zeros in Row and Column](./solutions/2482-difference-between-ones-and-zeros-in-row-and-column.js)|Medium| +2483|[Minimum Penalty for a Shop](./solutions/2483-minimum-penalty-for-a-shop.js)|Medium| +2485|[Find the Pivot Integer](./solutions/2485-find-the-pivot-integer.js)|Easy| +2486|[Append Characters to String to Make Subsequence](./solutions/2486-append-characters-to-string-to-make-subsequence.js)|Medium| +2487|[Remove Nodes From Linked List](./solutions/2487-remove-nodes-from-linked-list.js)|Medium| +2488|[Count Subarrays With Median K](./solutions/2488-count-subarrays-with-median-k.js)|Hard| 2490|[Circular Sentence](./solutions/2490-circular-sentence.js)|Easy| +2491|[Divide Players Into Teams of Equal Skill](./solutions/2491-divide-players-into-teams-of-equal-skill.js)|Medium| 2493|[Divide Nodes Into the Maximum Number of Groups](./solutions/2493-divide-nodes-into-the-maximum-number-of-groups.js)|Hard| 2503|[Maximum Number of Points From Grid Queries](./solutions/2503-maximum-number-of-points-from-grid-queries.js)|Hard| 2523|[Closest Prime Numbers in Range](./solutions/2523-closest-prime-numbers-in-range.js)|Medium| @@ -1634,6 +1892,7 @@ 2570|[Merge Two 2D Arrays by Summing Values](./solutions/2570-merge-two-2d-arrays-by-summing-values.js)|Easy| 2579|[Count Total Number of Colored Cells](./solutions/2579-count-total-number-of-colored-cells.js)|Medium| 2594|[Minimum Time to Repair Cars](./solutions/2594-minimum-time-to-repair-cars.js)|Medium| +2615|[Sum of Distances](./solutions/2615-sum-of-distances.js)|Medium| 2618|[Check if Object Instance of Class](./solutions/2618-check-if-object-instance-of-class.js)|Medium| 2619|[Array Prototype Last](./solutions/2619-array-prototype-last.js)|Easy| 2620|[Counter](./solutions/2620-counter.js)|Easy| @@ -1684,12 +1943,18 @@ 2845|[Count of Interesting Subarrays](./solutions/2845-count-of-interesting-subarrays.js)|Medium| 2873|[Maximum Value of an Ordered Triplet I](./solutions/2873-maximum-value-of-an-ordered-triplet-i.js)|Easy| 2874|[Maximum Value of an Ordered Triplet II](./solutions/2874-maximum-value-of-an-ordered-triplet-ii.js)|Medium| +2900|[Longest Unequal Adjacent Groups Subsequence I](./solutions/2900-longest-unequal-adjacent-groups-subsequence-i.js)|Easy| +2901|[Longest Unequal Adjacent Groups Subsequence II](./solutions/2901-longest-unequal-adjacent-groups-subsequence-ii.js)|Medium| +2918|[Minimum Equal Sum of Two Arrays After Replacing Zeros](./solutions/2918-minimum-equal-sum-of-two-arrays-after-replacing-zeros.js)|Medium| +2942|[Find Words Containing Character](./solutions/2942-find-words-containing-character.js)|Easy| 2948|[Make Lexicographically Smallest Array by Swapping Elements](./solutions/2948-make-lexicographically-smallest-array-by-swapping-elements.js)|Medium| 2962|[Count Subarrays Where Max Element Appears at Least K Times](./solutions/2962-count-subarrays-where-max-element-appears-at-least-k-times.js)|Medium| 2965|[Find Missing and Repeated Values](./solutions/2965-find-missing-and-repeated-values.js)|Easy| 2999|[Count the Number of Powerful Integers](./solutions/2999-count-the-number-of-powerful-integers.js)|Hard| +3024|[Type of Triangle](./solutions/3024-type-of-triangle.js)|Easy| 3042|[Count Prefix and Suffix Pairs I](./solutions/3042-count-prefix-and-suffix-pairs-i.js)|Easy| 3066|[Minimum Operations to Exceed Threshold Value II](./solutions/3066-minimum-operations-to-exceed-threshold-value-ii.js)|Medium| +3068|[Find the Maximum Sum of Node Values](./solutions/3068-find-the-maximum-sum-of-node-values.js)|Hard| 3105|[Longest Strictly Increasing or Strictly Decreasing Subarray](./solutions/3105-longest-strictly-increasing-or-strictly-decreasing-subarray.js)|Easy| 3108|[Minimum Cost Walk in Weighted Graph](./solutions/3108-minimum-cost-walk-in-weighted-graph.js)|Hard| 3110|[Score of a String](./solutions/3110-score-of-a-string.js)|Easy| @@ -1702,8 +1967,14 @@ 3223|[Minimum Length of String After Operations](./solutions/3223-minimum-length-of-string-after-operations.js)|Medium| 3272|[Find the Count of Good Integers](./solutions/3272-find-the-count-of-good-integers.js)|Hard| 3306|[Count of Substrings Containing Every Vowel and K Consonants II](./solutions/3306-count-of-substrings-containing-every-vowel-and-k-consonants-ii.js)|Medium| +3335|[Total Characters in String After Transformations I](./solutions/3335-total-characters-in-string-after-transformations-i.js)|Medium| +3337|[Total Characters in String After Transformations II](./solutions/3337-total-characters-in-string-after-transformations-ii.js)|Hard| +3341|[Find Minimum Time to Reach Last Room I](./solutions/3341-find-minimum-time-to-reach-last-room-i.js)|Medium| 3342|[Find Minimum Time to Reach Last Room II](./solutions/3342-find-minimum-time-to-reach-last-room-ii.js)|Medium| +3343|[Count Number of Balanced Permutations](./solutions/3343-count-number-of-balanced-permutations.js)|Hard| +3355|[Zero Array Transformation I](./solutions/3355-zero-array-transformation-i.js)|Medium| 3356|[Zero Array Transformation II](./solutions/3356-zero-array-transformation-ii.js)|Medium| +3362|[Zero Array Transformation III](./solutions/3362-zero-array-transformation-iii.js)|Medium| 3375|[Minimum Operations to Make Array Values Equal to K](./solutions/3375-minimum-operations-to-make-array-values-equal-to-k.js)|Easy| 3392|[Count Subarrays of Length Three With a Condition](./solutions/3392-count-subarrays-of-length-three-with-a-condition.js)|Easy| 3394|[Check if Grid can be Cut into Sections](./solutions/3394-check-if-grid-can-be-cut-into-sections.js)|Medium| diff --git a/solutions/2049-count-nodes-with-the-highest-score.js b/solutions/2049-count-nodes-with-the-highest-score.js new file mode 100644 index 00000000..1f13961e --- /dev/null +++ b/solutions/2049-count-nodes-with-the-highest-score.js @@ -0,0 +1,61 @@ +/** + * 2049. Count Nodes With the Highest Score + * https://leetcode.com/problems/count-nodes-with-the-highest-score/ + * Difficulty: Medium + * + * There is a binary tree rooted at 0 consisting of n nodes. The nodes are labeled from 0 to n - 1. + * You are given a 0-indexed integer array parents representing the tree, where parents[i] is the + * parent of node i. Since node 0 is the root, parents[0] == -1. + * + * Each node has a score. To find the score of a node, consider if the node and the edges connected + * to it were removed. The tree would become one or more non-empty subtrees. The size of a subtree + * is the number of the nodes in it. The score of the node is the product of the sizes of all + * those subtrees. + * + * Return the number of nodes that have the highest score. + */ + +/** + * @param {number[]} parents + * @return {number} + */ +var countHighestScoreNodes = function(parents) { + const n = parents.length; + const children = Array.from({ length: n }, () => []); + for (let i = 1; i < n; i++) { + children[parents[i]].push(i); + } + + const sizes = new Array(n).fill(0); + let maxScore = 0n; + let maxScoreCount = 0; + + calculateSize(0); + + return maxScoreCount; + + function calculateSize(node) { + let leftSize = 0; + let rightSize = 0; + + if (children[node].length > 0) leftSize = calculateSize(children[node][0]); + if (children[node].length > 1) rightSize = calculateSize(children[node][1]); + + sizes[node] = leftSize + rightSize + 1; + + let score = 1n; + if (leftSize > 0) score *= BigInt(leftSize); + if (rightSize > 0) score *= BigInt(rightSize); + const parentSize = n - sizes[node]; + if (parentSize > 0) score *= BigInt(parentSize); + + if (score > maxScore) { + maxScore = score; + maxScoreCount = 1; + } else if (score === maxScore) { + maxScoreCount++; + } + + return sizes[node]; + } +}; diff --git a/solutions/2050-parallel-courses-iii.js b/solutions/2050-parallel-courses-iii.js new file mode 100644 index 00000000..62cdec65 --- /dev/null +++ b/solutions/2050-parallel-courses-iii.js @@ -0,0 +1,61 @@ +/** + * 2050. Parallel Courses III + * https://leetcode.com/problems/parallel-courses-iii/ + * Difficulty: Hard + * + * You are given an integer n, which indicates that there are n courses labeled from 1 to n. + * You are also given a 2D integer array relations where relations[j] = [prevCoursej, nextCoursej] + * denotes that course prevCoursej has to be completed before course nextCoursej (prerequisite + * relationship). Furthermore, you are given a 0-indexed integer array time where time[i] denotes + * how many months it takes to complete the (i+1)th course. + * + * You must find the minimum number of months needed to complete all the courses following these + * rules: + * - You may start taking a course at any time if the prerequisites are met. + * - Any number of courses can be taken at the same time. + * + * Return the minimum number of months needed to complete all the courses. + * + * Note: The test cases are generated such that it is possible to complete every course (i.e., the + * graph is a directed acyclic graph). + */ + +/** + * @param {number} n + * @param {number[][]} relations + * @param {number[]} time + * @return {number} + */ +var minimumTime = function(n, relations, time) { + const adjacencyList = Array.from({ length: n + 1 }, () => []); + const inDegree = new Array(n + 1).fill(0); + const completionTime = new Array(n + 1).fill(0); + + for (const [prev, next] of relations) { + adjacencyList[prev].push(next); + inDegree[next]++; + } + + const queue = []; + for (let i = 1; i <= n; i++) { + if (inDegree[i] === 0) { + queue.push(i); + completionTime[i] = time[i - 1]; + } + } + + while (queue.length) { + const current = queue.shift(); + for (const next of adjacencyList[current]) { + completionTime[next] = Math.max( + completionTime[next], + completionTime[current] + time[next - 1] + ); + if (--inDegree[next] === 0) { + queue.push(next); + } + } + } + + return Math.max(...completionTime); +}; diff --git a/solutions/2055-plates-between-candles.js b/solutions/2055-plates-between-candles.js new file mode 100644 index 00000000..347ed4ae --- /dev/null +++ b/solutions/2055-plates-between-candles.js @@ -0,0 +1,55 @@ +/** + * 2055. Plates Between Candles + * https://leetcode.com/problems/plates-between-candles/ + * Difficulty: Medium + * + * There is a long table with a line of plates and candles arranged on top of it. You are given + * a 0-indexed string s consisting of characters '*' and '|' only, where a '*' represents a plate + * and a '|' represents a candle. + * + * You are also given a 0-indexed 2D integer array queries where queries[i] = [lefti, righti] + * denotes the substring s[lefti...righti] (inclusive). For each query, you need to find the + * number of plates between candles that are in the substring. A plate is considered between + * candles if there is at least one candle to its left and at least one candle to its right + * in the substring. + * - For example, s = "||**||**|*", and a query [3, 8] denotes the substring "*||**|". The number + * of plates between candles in this substring is 2, as each of the two plates has at least one + * candle in the substring to its left and right. + * + * Return an integer array answer where answer[i] is the answer to the ith query. + */ + +/** + * @param {string} s + * @param {number[][]} queries + * @return {number[]} + */ +var platesBetweenCandles = function(s, queries) { + const n = s.length; + const prefixPlates = new Array(n + 1).fill(0); + const leftCandle = new Array(n).fill(-1); + const rightCandle = new Array(n).fill(-1); + + for (let i = 0, candle = -1; i < n; i++) { + prefixPlates[i + 1] = prefixPlates[i] + (s[i] === '*' ? 1 : 0); + if (s[i] === '|') candle = i; + leftCandle[i] = candle; + } + + for (let i = n - 1, candle = -1; i >= 0; i--) { + if (s[i] === '|') candle = i; + rightCandle[i] = candle; + } + + const result = new Array(queries.length).fill(0); + for (let i = 0; i < queries.length; i++) { + const [start, end] = queries[i]; + const left = rightCandle[start]; + const right = leftCandle[end]; + if (left !== -1 && right !== -1 && left < right) { + result[i] = prefixPlates[right] - prefixPlates[left]; + } + } + + return result; +}; diff --git a/solutions/2056-number-of-valid-move-combinations-on-chessboard.js b/solutions/2056-number-of-valid-move-combinations-on-chessboard.js new file mode 100644 index 00000000..94716f2b --- /dev/null +++ b/solutions/2056-number-of-valid-move-combinations-on-chessboard.js @@ -0,0 +1,108 @@ +/** + * 2056. Number of Valid Move Combinations On Chessboard + * https://leetcode.com/problems/number-of-valid-move-combinations-on-chessboard/ + * Difficulty: Hard + * + * There is an 8 x 8 chessboard containing n pieces (rooks, queens, or bishops). You are given a + * string array pieces of length n, where pieces[i] describes the type (rook, queen, or bishop) + * of the ith piece. In addition, you are given a 2D integer array positions also of length n, + * where positions[i] = [ri, ci] indicates that the ith piece is currently at the 1-based + * coordinate (ri, ci) on the chessboard. + * + * When making a move for a piece, you choose a destination square that the piece will travel + * toward and stop on. + * - A rook can only travel horizontally or vertically from (r, c) to the direction of (r+1, c), + * (r-1, c), (r, c+1), or (r, c-1). + * - A queen can only travel horizontally, vertically, or diagonally from (r, c) to the direction + * of (r+1, c), (r-1, c), (r, c+1), (r, c-1), (r+1, c+1), (r+1, c-1), (r-1, c+1), (r-1, c-1). + * - A bishop can only travel diagonally from (r, c) to the direction of (r+1, c+1), (r+1, c-1), + * (r-1, c+1), (r-1, c-1). + * + * You must make a move for every piece on the board simultaneously. A move combination consists of + * all the moves performed on all the given pieces. Every second, each piece will instantaneously + * travel one square towards their destination if they are not already at it. All pieces start + * traveling at the 0th second. A move combination is invalid if, at a given time, two or more + * pieces occupy the same square. + * + * Return the number of valid move combinations. + * + * Notes: + * - No two pieces will start in the same square. + * - You may choose the square a piece is already on as its destination. + * - If two pieces are directly adjacent to each other, it is valid for them to move past each other + * and swap positions in one second. + */ + +/** + * @param {string[]} pieces + * @param {number[][]} positions + * @return {number} + */ +var countCombinations = function(pieces, positions) { + const n = pieces.length; + const directions = { + rook: [[0, 1], [0, -1], [1, 0], [-1, 0]], + bishop: [[1, 1], [1, -1], [-1, 1], [-1, -1]], + queen: [[0, 1], [0, -1], [1, 0], [-1, 0], [1, 1], [1, -1], [-1, 1], [-1, -1]] + }; + + return generateCombinations(0, [], []); + + function isValidPosition(row, col) { + return row >= 1 && row <= 8 && col >= 1 && col <= 8; + } + + function simulateMoves(moves, steps) { + const current = positions.map(([r, c]) => [r, c]); + const maxSteps = Math.max(...steps); + + for (let t = 0; t <= maxSteps; t++) { + const positionsAtTime = new Set(); + for (let i = 0; i < n; i++) { + const [dr, dc] = moves[i]; + const step = Math.min(t, steps[i]); + const newRow = current[i][0] + dr * step; + const newCol = current[i][1] + dc * step; + + if (t > steps[i] && steps[i] > 0 && !isValidPosition(newRow, newCol)) continue; + + const posKey = `${newRow},${newCol}`; + if (positionsAtTime.has(posKey)) return false; + positionsAtTime.add(posKey); + } + } + + return true; + } + + function generateCombinations(index, selectedMoves, selectedSteps) { + if (index === n) { + return simulateMoves(selectedMoves, selectedSteps) ? 1 : 0; + } + + let total = 0; + const [startRow, startCol] = positions[index]; + const pieceDirections = directions[pieces[index]]; + + total += generateCombinations(index + 1, [...selectedMoves, [0, 0]], [...selectedSteps, 0]); + + for (const [dr, dc] of pieceDirections) { + let row = startRow; + let col = startCol; + let step = 0; + + while (isValidPosition(row + dr, col + dc)) { + row += dr; + col += dc; + step++; + total += generateCombinations( + index + 1, + [...selectedMoves, [dr, dc]], + [...selectedSteps, step] + ); + } + } + + return total; + } +}; diff --git a/solutions/2057-smallest-index-with-equal-value.js b/solutions/2057-smallest-index-with-equal-value.js new file mode 100644 index 00000000..3fe1c453 --- /dev/null +++ b/solutions/2057-smallest-index-with-equal-value.js @@ -0,0 +1,23 @@ +/** + * 2057. Smallest Index With Equal Value + * https://leetcode.com/problems/smallest-index-with-equal-value/ + * Difficulty: Easy + * + * Given a 0-indexed integer array nums, return the smallest index i of nums such that i mod + * 10 == nums[i], or -1 if such index does not exist. + * + * x mod y denotes the remainder when x is divided by y. + */ + +/** + * @param {number[]} nums + * @return {number} + */ +var smallestEqual = function(nums) { + for (let i = 0; i < nums.length; i++) { + if (i % 10 === nums[i]) { + return i; + } + } + return -1; +}; diff --git a/solutions/2058-find-the-minimum-and-maximum-number-of-nodes-between-critical-points.js b/solutions/2058-find-the-minimum-and-maximum-number-of-nodes-between-critical-points.js new file mode 100644 index 00000000..da92ff90 --- /dev/null +++ b/solutions/2058-find-the-minimum-and-maximum-number-of-nodes-between-critical-points.js @@ -0,0 +1,63 @@ +/** + * 2058. Find the Minimum and Maximum Number of Nodes Between Critical Points + * https://leetcode.com/problems/find-the-minimum-and-maximum-number-of-nodes-between-critical-points/ + * Difficulty: Medium + * + * A critical point in a linked list is defined as either a local maxima or a local minima. + * + * A node is a local maxima if the current node has a value strictly greater than the previous node + * and the next node. + * + * A node is a local minima if the current node has a value strictly smaller than the previous node + * and the next node. + * + * Note that a node can only be a local maxima/minima if there exists both a previous node and a + * next node. + * + * Given a linked list head, return an array of length 2 containing [minDistance, maxDistance] + * where minDistance is the minimum distance between any two distinct critical points and + * maxDistance is the maximum distance between any two distinct critical points. If there are + * fewer than two critical points, return [-1, -1]. + */ + +/** + * Definition for singly-linked list. + * function ListNode(val, next) { + * this.val = (val===undefined ? 0 : val) + * this.next = (next===undefined ? null : next) + * } + */ +/** + * @param {ListNode} head + * @return {number[]} + */ +var nodesBetweenCriticalPoints = function(head) { + const criticalPositions = []; + let position = 1; + let prev = head; + let current = head.next; + + while (current.next) { + const next = current.next; + if ((current.val > prev.val && current.val > next.val) || + (current.val < prev.val && current.val < next.val)) { + criticalPositions.push(position); + } + prev = current; + current = next; + position++; + } + + if (criticalPositions.length < 2) { + return [-1, -1]; + } + + let minDistance = Infinity; + for (let i = 1; i < criticalPositions.length; i++) { + minDistance = Math.min(minDistance, criticalPositions[i] - criticalPositions[i - 1]); + } + + const maxDistance = criticalPositions[criticalPositions.length - 1] - criticalPositions[0]; + + return [minDistance, maxDistance]; +}; diff --git a/solutions/2059-minimum-operations-to-convert-number.js b/solutions/2059-minimum-operations-to-convert-number.js new file mode 100644 index 00000000..677e5de9 --- /dev/null +++ b/solutions/2059-minimum-operations-to-convert-number.js @@ -0,0 +1,57 @@ +/** + * 2059. Minimum Operations to Convert Number + * https://leetcode.com/problems/minimum-operations-to-convert-number/ + * Difficulty: Medium + * + * You are given a 0-indexed integer array nums containing distinct numbers, an integer start, + * and an integer goal. There is an integer x that is initially set to start, and you want to + * perform operations on x such that it is converted to goal. You can perform the following + * operation repeatedly on the number x: + * + * If 0 <= x <= 1000, then for any index i in the array (0 <= i < nums.length), you can set x + * to any of the following: + * - x + nums[i] + * - x - nums[i] + * - x ^ nums[i] (bitwise-XOR) + * + * Note that you can use each nums[i] any number of times in any order. Operations that set x + * to be out of the range 0 <= x <= 1000 are valid, but no more operations can be done afterward. + * + * Return the minimum number of operations needed to convert x = start into goal, and -1 if it + * is not possible. + */ + +/** + * @param {number[]} nums + * @param {number} start + * @param {number} goal + * @return {number} + */ +var minimumOperations = function(nums, start, goal) { + const queue = [[start, 0]]; + const visited = new Set([start]); + + while (queue.length) { + const [current, steps] = queue.shift(); + + for (const num of nums) { + const candidates = [ + current + num, + current - num, + current ^ num + ]; + + for (const next of candidates) { + if (next === goal) { + return steps + 1; + } + if (next >= 0 && next <= 1000 && !visited.has(next)) { + visited.add(next); + queue.push([next, steps + 1]); + } + } + } + } + + return -1; +}; diff --git a/solutions/2060-check-if-an-original-string-exists-given-two-encoded-strings.js b/solutions/2060-check-if-an-original-string-exists-given-two-encoded-strings.js new file mode 100644 index 00000000..520e531e --- /dev/null +++ b/solutions/2060-check-if-an-original-string-exists-given-two-encoded-strings.js @@ -0,0 +1,73 @@ +/** + * 2060. Check if an Original String Exists Given Two Encoded Strings + * https://leetcode.com/problems/check-if-an-original-string-exists-given-two-encoded-strings/ + * Difficulty: Hard + * + * An original string, consisting of lowercase English letters, can be encoded by the following + * steps: + * - Arbitrarily split it into a sequence of some number of non-empty substrings. + * - Arbitrarily choose some elements (possibly none) of the sequence, and replace each with its + * length (as a numeric string). + * - Concatenate the sequence as the encoded string. + * + * For example, one way to encode an original string "abcdefghijklmnop" might be: + * - Split it as a sequence: ["ab", "cdefghijklmn", "o", "p"]. + * - Choose the second and third elements to be replaced by their lengths, respectively. The + * sequence becomes ["ab", "12", "1", "p"]. + * - Concatenate the elements of the sequence to get the encoded string: "ab121p". + * + * Given two encoded strings s1 and s2, consisting of lowercase English letters and digits 1-9 + * (inclusive), return true if there exists an original string that could be encoded as both s1 + * and s2. Otherwise, return false. + * + * Note: The test cases are generated such that the number of consecutive digits in s1 and s2 + * does not exceed 3. + */ + +/** + * @param {string} s1 + * @param {string} s2 + * @return {boolean} + */ +var possiblyEquals = function(s1, s2) { + const memo = new Array(s1.length + 1).fill().map(() => new Array(s2.length + 1).fill() + .map(() => ({}))); + + function match(pos1, pos2, diff) { + if (pos1 === s1.length && pos2 === s2.length) return diff === 0; + if (memo[pos1][pos2][diff] !== undefined) return memo[pos1][pos2][diff]; + + const char1 = s1[pos1]; + const char2 = s2[pos2]; + + if (pos1 < s1.length && pos2 < s2.length && char1 === char2 && diff === 0) { + if (match(pos1 + 1, pos2 + 1, 0)) return true; + } + + if (pos1 < s1.length && isNaN(char1) && diff < 0) { + if (match(pos1 + 1, pos2, diff + 1)) return true; + } + + if (pos2 < s2.length && isNaN(char2) && diff > 0) { + if (match(pos1, pos2 + 1, diff - 1)) return true; + } + + let num = 0; + for (let i = 0; i < 3 && pos1 + i < s1.length; i++) { + if (isNaN(s1[pos1 + i])) break; + num = num * 10 + parseInt(s1[pos1 + i]); + if (match(pos1 + i + 1, pos2, diff + num)) return true; + } + + num = 0; + for (let i = 0; i < 3 && pos2 + i < s2.length; i++) { + if (isNaN(s2[pos2 + i])) break; + num = num * 10 + parseInt(s2[pos2 + i]); + if (match(pos1, pos2 + i + 1, diff - num)) return true; + } + + return memo[pos1][pos2][diff] = false; + } + + return match(0, 0, 0); +}; diff --git a/solutions/2062-count-vowel-substrings-of-a-string.js b/solutions/2062-count-vowel-substrings-of-a-string.js new file mode 100644 index 00000000..8d427ac8 --- /dev/null +++ b/solutions/2062-count-vowel-substrings-of-a-string.js @@ -0,0 +1,33 @@ +/** + * 2062. Count Vowel Substrings of a String + * https://leetcode.com/problems/count-vowel-substrings-of-a-string/ + * Difficulty: Easy + * + * A substring is a contiguous (non-empty) sequence of characters within a string. + * + * A vowel substring is a substring that only consists of vowels ('a', 'e', 'i', 'o', and 'u') + * and has all five vowels present in it. + * + * Given a string word, return the number of vowel substrings in word. + */ + +/** + * @param {string} word + * @return {number} + */ +var countVowelSubstrings = function(word) { + const vowels = new Set(['a', 'e', 'i', 'o', 'u']); + let total = 0; + + for (let start = 0; start < word.length; start++) { + const vowelCount = new Map(); + for (let end = start; end < word.length && vowels.has(word[end]); end++) { + vowelCount.set(word[end], (vowelCount.get(word[end]) || 0) + 1); + if (vowelCount.size === 5) { + total++; + } + } + } + + return total; +}; diff --git a/solutions/2063-vowels-of-all-substrings.js b/solutions/2063-vowels-of-all-substrings.js new file mode 100644 index 00000000..bde705e8 --- /dev/null +++ b/solutions/2063-vowels-of-all-substrings.js @@ -0,0 +1,30 @@ +/** + * 2063. Vowels of All Substrings + * https://leetcode.com/problems/vowels-of-all-substrings/ + * Difficulty: Medium + * + * Given a string word, return the sum of the number of vowels ('a', 'e', 'i', 'o', and 'u') in + * every substring of word. + * + * A substring is a contiguous (non-empty) sequence of characters within a string. + * + * Note: Due to the large constraints, the answer may not fit in a signed 32-bit integer. Please + * be careful during the calculations. + */ + +/** + * @param {string} word + * @return {number} + */ +var countVowels = function(word) { + const vowels = new Set(['a', 'e', 'i', 'o', 'u']); + let total = 0; + + for (let i = 0; i < word.length; i++) { + if (vowels.has(word[i])) { + total += (i + 1) * (word.length - i); + } + } + + return total; +}; diff --git a/solutions/2064-minimized-maximum-of-products-distributed-to-any-store.js b/solutions/2064-minimized-maximum-of-products-distributed-to-any-store.js new file mode 100644 index 00000000..b1781081 --- /dev/null +++ b/solutions/2064-minimized-maximum-of-products-distributed-to-any-store.js @@ -0,0 +1,45 @@ +/** + * 2064. Minimized Maximum of Products Distributed to Any Store + * https://leetcode.com/problems/minimized-maximum-of-products-distributed-to-any-store/ + * Difficulty: Medium + * + * You are given an integer n indicating there are n specialty retail stores. There are m product + * types of varying amounts, which are given as a 0-indexed integer array quantities, where + * quantities[i] represents the number of products of the ith product type. + * + * You need to distribute all products to the retail stores following these rules: + * - A store can only be given at most one product type but can be given any amount of it. + * - After distribution, each store will have been given some number of products (possibly 0). + * Let x represent the maximum number of products given to any store. You want x to be as small + * as possible, i.e., you want to minimize the maximum number of products that are given to any + * store. + * + * Return the minimum possible x. + */ + +/** + * @param {number} n + * @param {number[]} quantities + * @return {number} + */ +var minimizedMaximum = function(n, quantities) { + let left = 1; + let right = Math.max(...quantities); + + while (left < right) { + const mid = Math.floor((left + right) / 2); + let storesNeeded = 0; + + for (const quantity of quantities) { + storesNeeded += Math.ceil(quantity / mid); + } + + if (storesNeeded <= n) { + right = mid; + } else { + left = mid + 1; + } + } + + return left; +}; diff --git a/solutions/2065-maximum-path-quality-of-a-graph.js b/solutions/2065-maximum-path-quality-of-a-graph.js new file mode 100644 index 00000000..9d07048c --- /dev/null +++ b/solutions/2065-maximum-path-quality-of-a-graph.js @@ -0,0 +1,59 @@ +/** + * 2065. Maximum Path Quality of a Graph + * https://leetcode.com/problems/maximum-path-quality-of-a-graph/ + * Difficulty: Hard + * + * There is an undirected graph with n nodes numbered from 0 to n - 1 (inclusive). You are given a + * 0-indexed integer array values where values[i] is the value of the ith node. You are also given + * a 0-indexed 2D integer array edges, where each edges[j] = [uj, vj, timej] indicates that there + * is an undirected edge between the nodes uj and vj, and it takes timej seconds to travel between + * the two nodes. Finally, you are given an integer maxTime. + * + * A valid path in the graph is any path that starts at node 0, ends at node 0, and takes at most + * maxTime seconds to complete. You may visit the same node multiple times. The quality of a valid + * path is the sum of the values of the unique nodes visited in the path (each node's value is + * added at most once to the sum). + * + * Return the maximum quality of a valid path. + * + * Note: There are at most four edges connected to each node. + */ + +/** + * @param {number[]} values + * @param {number[][]} edges + * @param {number} maxTime + * @return {number} + */ +var maximalPathQuality = function(values, edges, maxTime) { + const n = values.length; + const adjList = Array.from({ length: n }, () => []); + + for (const [u, v, time] of edges) { + adjList[u].push([v, time]); + adjList[v].push([u, time]); + } + + let result = 0; + const initialVisited = new Set([0]); + + dfs(0, maxTime, values[0], initialVisited); + + return result; + + function dfs(node, timeLeft, quality, visited) { + if (timeLeft < 0) return 0; + if (node === 0) { + result = Math.max(result, quality); + } + + for (const [nextNode, travelTime] of adjList[node]) { + const newVisited = new Set(visited); + const newQuality = newVisited.has(nextNode) ? quality : quality + values[nextNode]; + newVisited.add(nextNode); + dfs(nextNode, timeLeft - travelTime, newQuality, newVisited); + } + + return result; + } +}; diff --git a/solutions/2068-check-whether-two-strings-are-almost-equivalent.js b/solutions/2068-check-whether-two-strings-are-almost-equivalent.js new file mode 100644 index 00000000..210aeb88 --- /dev/null +++ b/solutions/2068-check-whether-two-strings-are-almost-equivalent.js @@ -0,0 +1,29 @@ +/** + * 2068. Check Whether Two Strings are Almost Equivalent + * https://leetcode.com/problems/check-whether-two-strings-are-almost-equivalent/ + * Difficulty: Easy + * + * Two strings word1 and word2 are considered almost equivalent if the differences between the + * frequencies of each letter from 'a' to 'z' between word1 and word2 is at most 3. + * + * Given two strings word1 and word2, each of length n, return true if word1 and word2 are almost + * equivalent, or false otherwise. + * + * The frequency of a letter x is the number of times it occurs in the string. + */ + +/** + * @param {string} word1 + * @param {string} word2 + * @return {boolean} + */ +var checkAlmostEquivalent = function(word1, word2) { + const map = new Array(26).fill(0); + + for (let i = 0; i < word1.length; i++) { + map[word1.charCodeAt(i) - 97]++; + map[word2.charCodeAt(i) - 97]--; + } + + return map.every(diff => Math.abs(diff) <= 3); +}; diff --git a/solutions/2069-walking-robot-simulation-ii.js b/solutions/2069-walking-robot-simulation-ii.js new file mode 100644 index 00000000..e4896040 --- /dev/null +++ b/solutions/2069-walking-robot-simulation-ii.js @@ -0,0 +1,84 @@ +/** + * 2069. Walking Robot Simulation II + * https://leetcode.com/problems/walking-robot-simulation-ii/ + * Difficulty: Medium + * + * A width x height grid is on an XY-plane with the bottom-left cell at (0, 0) and the + * top-right cell at (width - 1, height - 1). The grid is aligned with the four cardinal + * directions ("North", "East", "South", and "West"). A robot is initially at cell (0, 0) + * facing direction "East". + * + * The robot can be instructed to move for a specific number of steps. For each step, it does + * the following. + * 1. Attempts to move forward one cell in the direction it is facing. + * 2. If the cell the robot is moving to is out of bounds, the robot instead turns 90 degrees + * counterclockwise and retries the step. + * + * After the robot finishes moving the number of steps required, it stops and awaits the next + * instruction. + * + * Implement the Robot class: + * - Robot(int width, int height) Initializes the width x height grid with the robot at (0, 0) + * facing "East". + * - void step(int num) Instructs the robot to move forward num steps. + * - int[] getPos() Returns the current cell the robot is at, as an array of length 2, [x, y]. + * - String getDir() Returns the current direction of the robot, "North", "East", "South", or + * "West". + */ + +/** + * @param {number} width + * @param {number} height + */ +var Robot = function(width, height) { + this.width = width; + this.height = height; + this.position = [0, 0]; + this.directionIndex = 0; + this.directions = ['East', 'North', 'West', 'South']; + this.moves = [[1, 0], [0, 1], [-1, 0], [0, -1]]; + this.perimeter = 2 * (width + height - 2); +}; + +/** + * @param {number} num + * @return {void} + */ +Robot.prototype.step = function(num) { + num %= this.perimeter; + if (num === 0) num = this.perimeter; + + while (num > 0) { + const [dx, dy] = this.moves[this.directionIndex]; + const [x, y] = this.position; + let steps; + + if (this.directionIndex === 0) steps = this.width - 1 - x; + else if (this.directionIndex === 1) steps = this.height - 1 - y; + else if (this.directionIndex === 2) steps = x; + else steps = y; + + if (steps >= num) { + this.position = [x + dx * num, y + dy * num]; + num = 0; + } else { + this.position = [x + dx * steps, y + dy * steps]; + this.directionIndex = (this.directionIndex + 1) % 4; + num -= steps; + } + } +}; + +/** + * @return {number[]} + */ +Robot.prototype.getPos = function() { + return this.position; +}; + +/** + * @return {string} + */ +Robot.prototype.getDir = function() { + return this.directions[this.directionIndex]; +}; diff --git a/solutions/2070-most-beautiful-item-for-each-query.js b/solutions/2070-most-beautiful-item-for-each-query.js new file mode 100644 index 00000000..1931e875 --- /dev/null +++ b/solutions/2070-most-beautiful-item-for-each-query.js @@ -0,0 +1,54 @@ +/** + * 2070. Most Beautiful Item for Each Query + * https://leetcode.com/problems/most-beautiful-item-for-each-query/ + * Difficulty: Medium + * + * You are given a 2D integer array items where items[i] = [pricei, beautyi] denotes the price and + * beauty of an item respectively. + * + * You are also given a 0-indexed integer array queries. For each queries[j], you want to determine + * the maximum beauty of an item whose price is less than or equal to queries[j]. If no such item + * exists, then the answer to this query is 0. + * + * Return an array answer of the same length as queries where answer[j] is the answer to the jth + * query. + */ + +/** + * @param {number[][]} items + * @param {number[]} queries + * @return {number[]} + */ +var maximumBeauty = function(items, queries) { + items.sort((a, b) => a[0] - b[0]); + const maxBeauty = []; + let currentMax = 0; + + for (const [, beauty] of items) { + currentMax = Math.max(currentMax, beauty); + maxBeauty.push(currentMax); + } + + const result = new Array(queries.length).fill(0); + + for (let i = 0; i < queries.length; i++) { + const price = queries[i]; + let left = 0; + let right = items.length - 1; + + while (left <= right) { + const mid = Math.floor((left + right) / 2); + if (items[mid][0] <= price) { + left = mid + 1; + } else { + right = mid - 1; + } + } + + if (right >= 0) { + result[i] = maxBeauty[right]; + } + } + + return result; +}; diff --git a/solutions/2074-reverse-nodes-in-even-length-groups.js b/solutions/2074-reverse-nodes-in-even-length-groups.js new file mode 100644 index 00000000..2608277e --- /dev/null +++ b/solutions/2074-reverse-nodes-in-even-length-groups.js @@ -0,0 +1,72 @@ +/** + * 2074. Reverse Nodes in Even Length Groups + * https://leetcode.com/problems/reverse-nodes-in-even-length-groups/ + * Difficulty: Medium + * + * You are given the head of a linked list. + * + * The nodes in the linked list are sequentially assigned to non-empty groups whose lengths form + * the sequence of the natural numbers (1, 2, 3, 4, ...). The length of a group is the number + * of nodes assigned to it. In other words: + * - The 1st node is assigned to the first group. + * - The 2nd and the 3rd nodes are assigned to the second group. + * - The 4th, 5th, and 6th nodes are assigned to the third group, and so on. + * + * Note that the length of the last group may be less than or equal to 1 + the length of the second + * to last group. + * + * Reverse the nodes in each group with an even length, and return the head of the modified linked + * list. + */ + +/** + * Definition for singly-linked list. + * function ListNode(val, next) { + * this.val = (val===undefined ? 0 : val) + * this.next = (next===undefined ? null : next) + * } + */ +/** + * @param {ListNode} head + * @return {ListNode} + */ +var reverseEvenLengthGroups = function(head) { + let prevGroup = head; + let groupLength = 2; + + while (prevGroup.next) { + let count = 0; + let current = prevGroup.next; + const groupStart = current; + + while (current && count < groupLength) { + current = current.next; + count++; + } + + if (count % 2 === 0) { + prevGroup.next = reverseGroup(groupStart, count); + } + + for (let i = 0; i < count; i++) { + prevGroup = prevGroup.next; + } + + groupLength++; + } + + return head; + + function reverseGroup(start, length) { + let prev = null; + let current = start; + for (let i = 0; i < length; i++) { + const next = current.next; + current.next = prev; + prev = current; + current = next; + } + start.next = current; + return prev; + } +}; diff --git a/solutions/2075-decode-the-slanted-ciphertext.js b/solutions/2075-decode-the-slanted-ciphertext.js new file mode 100644 index 00000000..63f5a41d --- /dev/null +++ b/solutions/2075-decode-the-slanted-ciphertext.js @@ -0,0 +1,51 @@ +/** + * 2075. Decode the Slanted Ciphertext + * https://leetcode.com/problems/decode-the-slanted-ciphertext/ + * Difficulty: Medium + * + * A string originalText is encoded using a slanted transposition cipher to a string encodedText + * with the help of a matrix having a fixed number of rows rows. + * + * originalText is placed first in a top-left to bottom-right manner. + * + * The blue cells are filled first, followed by the red cells, then the yellow cells, and so on, + * until we reach the end of originalText. The arrow indicates the order in which the cells are + * filled. All empty cells are filled with ' '. The number of columns is chosen such that the + * rightmost column will not be empty after filling in originalText. + * + * encodedText is then formed by appending all characters of the matrix in a row-wise fashion. + * + * The characters in the blue cells are appended first to encodedText, then the red cells, and so + * on, and finally the yellow cells. The arrow indicates the order in which the cells are accessed. + * + * For example, if originalText = "cipher" and rows = 3, then we encode it in the following manner. + * + * The blue arrows depict how originalText is placed in the matrix, and the red arrows denote the + * order in which encodedText is formed. In the above example, encodedText = "ch ie pr". + * + * Given the encoded string encodedText and number of rows rows, return the original string + * originalText. + * + * Note: originalText does not have any trailing spaces ' '. The test cases are generated such that + * there is only one possible originalText. + */ + +/** + * @param {string} encodedText + * @param {number} rows + * @return {string} + */ +var decodeCiphertext = function(encodedText, rows) { + if (rows === 1) return encodedText; + + const count = encodedText.length / rows; + let result = ''; + + for (let i = 0; i < count; i++) { + for (let row = 0; row < rows && row + i < count; row++) { + result += encodedText[row * count + i + row]; + } + } + + return result.trimEnd(); +}; diff --git a/solutions/2076-process-restricted-friend-requests.js b/solutions/2076-process-restricted-friend-requests.js new file mode 100644 index 00000000..dff94d09 --- /dev/null +++ b/solutions/2076-process-restricted-friend-requests.js @@ -0,0 +1,70 @@ +/** + * 2076. Process Restricted Friend Requests + * https://leetcode.com/problems/process-restricted-friend-requests/ + * Difficulty: Hard + * + * You are given an integer n indicating the number of people in a network. Each person is + * labeled from 0 to n - 1. + * + * You are also given a 0-indexed 2D integer array restrictions, where restrictions[i] = [xi, yi] + * means that person xi and person yi cannot become friends, either directly or indirectly through + * other people. + * + * Initially, no one is friends with each other. You are given a list of friend requests as a + * 0-indexed 2D integer array requests, where requests[j] = [uj, vj] is a friend request between + * person uj and person vj. + * + * A friend request is successful if uj and vj can be friends. Each friend request is processed + * in the given order (i.e., requests[j] occurs before requests[j + 1]), and upon a successful + * request, uj and vj become direct friends for all future friend requests. + * + * Return a boolean array result, where each result[j] is true if the jth friend request is + * successful or false if it is not. + * + * Note: If uj and vj are already direct friends, the request is still successful. + */ + +/** + * @param {number} n + * @param {number[][]} restrictions + * @param {number[][]} requests + * @return {boolean[]} + */ +var friendRequests = function(n, restrictions, requests) { + const parent = Array.from({ length: n }, (_, i) => i); + const result = new Array(requests.length).fill(true); + + for (let j = 0; j < requests.length; j++) { + const [u, v] = requests[j]; + const pu = find(u); + const pv = find(v); + + let canBeFriends = true; + for (const [x, y] of restrictions) { + const px = find(x); + const py = find(y); + if ((pu === px && pv === py) || (pu === py && pv === px)) { + canBeFriends = false; + break; + } + } + + result[j] = canBeFriends; + if (canBeFriends) { + union(u, v); + } + } + + return result; + + function find(x) { + if (parent[x] !== x) { + parent[x] = find(parent[x]); + } + return parent[x]; + } + + function union(x, y) { + parent[find(x)] = find(y); + } +}; diff --git a/solutions/2078-two-furthest-houses-with-different-colors.js b/solutions/2078-two-furthest-houses-with-different-colors.js new file mode 100644 index 00000000..9a42b696 --- /dev/null +++ b/solutions/2078-two-furthest-houses-with-different-colors.js @@ -0,0 +1,32 @@ +/** + * 2078. Two Furthest Houses With Different Colors + * https://leetcode.com/problems/two-furthest-houses-with-different-colors/ + * Difficulty: Easy + * + * There are n houses evenly lined up on the street, and each house is beautifully painted. + * You are given a 0-indexed integer array colors of length n, where colors[i] represents + * the color of the ith house. + * + * Return the maximum distance between two houses with different colors. + * + * The distance between the ith and jth houses is abs(i - j), where abs(x) is the absolute + * value of x. + */ + +/** + * @param {number[]} colors + * @return {number} + */ +var maxDistance = function(colors) { + let result = 0; + + for (let i = 0; i < colors.length; i++) { + for (let j = i + 1; j < colors.length; j++) { + if (colors[i] !== colors[j]) { + result = Math.max(result, j - i); + } + } + } + + return result; +}; diff --git a/solutions/2079-watering-plants.js b/solutions/2079-watering-plants.js new file mode 100644 index 00000000..ba77b302 --- /dev/null +++ b/solutions/2079-watering-plants.js @@ -0,0 +1,42 @@ +/** + * 2079. Watering Plants + * https://leetcode.com/problems/watering-plants/ + * Difficulty: Medium + * + * You want to water n plants in your garden with a watering can. The plants are arranged in + * a row and are labeled from 0 to n - 1 from left to right where the ith plant is located at + * x = i. There is a river at x = -1 that you can refill your watering can at. + * + * Each plant needs a specific amount of water. You will water the plants in the following way: + * - Water the plants in order from left to right. + * - After watering the current plant, if you do not have enough water to completely water the + * next plant, return to the river to fully refill the watering can. + * - You cannot refill the watering can early. + * + * You are initially at the river (i.e., x = -1). It takes one step to move one unit on the x-axis. + * + * Given a 0-indexed integer array plants of n integers, where plants[i] is the amount of water + * the ith plant needs, and an integer capacity representing the watering can capacity, return + * the number of steps needed to water all the plants. + */ + +/** + * @param {number[]} plants + * @param {number} capacity + * @return {number} + */ +var wateringPlants = function(plants, capacity) { + let water = capacity; + let result = 0; + + for (let i = 0; i < plants.length; i++) { + if (water < plants[i]) { + result += 2 * i; + water = capacity; + } + water -= plants[i]; + result++; + } + + return result; +}; diff --git a/solutions/2080-range-frequency-queries.js b/solutions/2080-range-frequency-queries.js new file mode 100644 index 00000000..32b1bb27 --- /dev/null +++ b/solutions/2080-range-frequency-queries.js @@ -0,0 +1,72 @@ +/** + * 2080. Range Frequency Queries + * https://leetcode.com/problems/range-frequency-queries/ + * Difficulty: Medium + * + * Design a data structure to find the frequency of a given value in a given subarray. + * + * The frequency of a value in a subarray is the number of occurrences of that value in + * the subarray. + * + * Implement the RangeFreqQuery class: + * - RangeFreqQuery(int[] arr) Constructs an instance of the class with the given 0-indexed + * integer array arr. + * - int query(int left, int right, int value) Returns the frequency of value in the subarray + * arr[left...right]. + * - A subarray is a contiguous sequence of elements within an array. arr[left...right] denotes + * the subarray that contains the elements of nums between indices left and right (inclusive). + */ + +/** + * @param {number[]} arr + */ +var RangeFreqQuery = function(arr) { + this.frequencyMap = new Map(); + + for (let i = 0; i < arr.length; i++) { + if (!this.frequencyMap.has(arr[i])) { + this.frequencyMap.set(arr[i], []); + } + this.frequencyMap.get(arr[i]).push(i); + } +}; + +/** + * @param {number} left + * @param {number} right + * @param {number} value + * @return {number} + */ +RangeFreqQuery.prototype.query = function(left, right, value) { + if (!this.frequencyMap.has(value)) return 0; + + const indices = this.frequencyMap.get(value); + let start = 0; + let end = indices.length - 1; + let leftBound = -1; + let rightBound = -1; + + while (start <= end) { + const mid = Math.floor((start + end) / 2); + if (indices[mid] >= left) { + leftBound = mid; + end = mid - 1; + } else { + start = mid + 1; + } + } + + start = 0; + end = indices.length - 1; + while (start <= end) { + const mid = Math.floor((start + end) / 2); + if (indices[mid] <= right) { + rightBound = mid; + start = mid + 1; + } else { + end = mid - 1; + } + } + + return leftBound === -1 || rightBound === -1 ? 0 : rightBound - leftBound + 1; +}; diff --git a/solutions/2081-sum-of-k-mirror-numbers.js b/solutions/2081-sum-of-k-mirror-numbers.js new file mode 100644 index 00000000..600d62d8 --- /dev/null +++ b/solutions/2081-sum-of-k-mirror-numbers.js @@ -0,0 +1,105 @@ +/** + * 2081. Sum of k-Mirror Numbers + * https://leetcode.com/problems/sum-of-k-mirror-numbers/ + * Difficulty: Hard + * + * A k-mirror number is a positive integer without leading zeros that reads the same both forward + * and backward in base-10 as well as in base-k. + * - For example, 9 is a 2-mirror number. The representation of 9 in base-10 and base-2 are 9 and + * 1001 respectively, which read the same both forward and backward. + * - On the contrary, 4 is not a 2-mirror number. The representation of 4 in base-2 is 100, which + * does not read the same both forward and backward. + * + * Given the base k and the number n, return the sum of the n smallest k-mirror numbers. + */ + +/** + * @param {number} k + * @param {number} n + * @return {number} + */ +var kMirror = function(k, n) { + const kMirrorNumbers = []; + let sum = 0; + let length = 1; + + while (kMirrorNumbers.length < n) { + generatePalindromes(length).forEach(num => { + if (kMirrorNumbers.length < n && isKMirror(num, k)) { + kMirrorNumbers.push(num); + sum += num; + } + }); + length++; + } + + return sum; +}; + +function toBaseK(num, k) { + if (num === 0) return '0'; + + let result = ''; + while (num > 0) { + result = (num % k) + result; + num = Math.floor(num / k); + } + + return result; +} + +function isKMirror(num, k) { + const baseKRepresentation = toBaseK(num, k); + return isPalindrome(baseKRepresentation); +} + +function isPalindrome(str) { + let left = 0; + let right = str.length - 1; + + while (left < right) { + if (str[left] !== str[right]) return false; + left++; + right--; + } + + return true; +} + +function generatePalindromes(length) { + const palindromes = []; + + if (length === 1) { + for (let i = 1; i <= 9; i++) { + palindromes.push(i); + } + return palindromes; + } + + if (length % 2 === 0) { + const half = length / 2; + const start = Math.pow(10, half - 1); + const end = Math.pow(10, half) - 1; + + for (let i = start; i <= end; i++) { + const firstHalf = i.toString(); + const secondHalf = firstHalf.split('').reverse().join(''); + palindromes.push(parseInt(firstHalf + secondHalf)); + } + } else { + const half = Math.floor(length / 2); + const start = Math.pow(10, half); + const end = Math.pow(10, half + 1) - 1; + + for (let i = start; i <= end; i++) { + const withoutMiddle = Math.floor(i / 10); + const middle = i % 10; + const firstHalf = withoutMiddle.toString(); + const secondHalf = firstHalf.split('').reverse().join(''); + palindromes.push(parseInt(firstHalf + middle + secondHalf)); + } + } + + return palindromes; +} + diff --git a/solutions/2086-minimum-number-of-food-buckets-to-feed-the-hamsters.js b/solutions/2086-minimum-number-of-food-buckets-to-feed-the-hamsters.js new file mode 100644 index 00000000..5cda9132 --- /dev/null +++ b/solutions/2086-minimum-number-of-food-buckets-to-feed-the-hamsters.js @@ -0,0 +1,43 @@ +/** + * 2086. Minimum Number of Food Buckets to Feed the Hamsters + * https://leetcode.com/problems/minimum-number-of-food-buckets-to-feed-the-hamsters/ + * Difficulty: Medium + * + * You are given a 0-indexed string hamsters where hamsters[i] is either: + * - 'H' indicating that there is a hamster at index i, or + * - '.' indicating that index i is empty. + * + * You will add some number of food buckets at the empty indices in order to feed the hamsters. + * A hamster can be fed if there is at least one food bucket to its left or to its right. More + * formally, a hamster at index i can be fed if you place a food bucket at index i - 1 and/or + * at index i + 1. + * + * Return the minimum number of food buckets you should place at empty indices to feed all the + * hamsters or -1 if it is impossible to feed all of them. + */ + +/** + * @param {string} hamsters + * @return {number} + */ +var minimumBuckets = function(hamsters) { + const n = hamsters.length; + let result = 0; + + for (let i = 0; i < n; i++) { + if (hamsters[i] === 'H') { + if (i > 0 && hamsters[i - 1] === 'B') continue; + if (i < n - 1 && hamsters[i + 1] === '.') { + result++; + hamsters = hamsters.slice(0, i + 1) + 'B' + hamsters.slice(i + 2); + } else if (i > 0 && hamsters[i - 1] === '.') { + result++; + hamsters = hamsters.slice(0, i - 1) + 'B' + hamsters.slice(i); + } else { + return -1; + } + } + } + + return result; +}; diff --git a/solutions/2087-minimum-cost-homecoming-of-a-robot-in-a-grid.js b/solutions/2087-minimum-cost-homecoming-of-a-robot-in-a-grid.js new file mode 100644 index 00000000..8fde0697 --- /dev/null +++ b/solutions/2087-minimum-cost-homecoming-of-a-robot-in-a-grid.js @@ -0,0 +1,45 @@ +/** + * 2087. Minimum Cost Homecoming of a Robot in a Grid + * https://leetcode.com/problems/minimum-cost-homecoming-of-a-robot-in-a-grid/ + * Difficulty: Medium + * + * There is an m x n grid, where (0, 0) is the top-left cell and (m - 1, n - 1) is the bottom-right + * cell. You are given an integer array startPos where startPos = [startrow, startcol] indicates + * that initially, a robot is at the cell (startrow, startcol). You are also given an integer array + * homePos where homePos = [homerow, homecol] indicates that its home is at the cell (homerow, + * homecol). + * + * The robot needs to go to its home. It can move one cell in four directions: left, right, up, + * or down, and it can not move outside the boundary. Every move incurs some cost. You are further + * given two 0-indexed integer arrays: rowCosts of length m and colCosts of length n. + * - If the robot moves up or down into a cell whose row is r, then this move costs rowCosts[r]. + * - If the robot moves left or right into a cell whose column is c, then this move costs + * colCosts[c]. + * + * Return the minimum total cost for this robot to return home. + */ + +/** + * @param {number[]} startPos + * @param {number[]} homePos + * @param {number[]} rowCosts + * @param {number[]} colCosts + * @return {number} + */ +var minCost = function(startPos, homePos, rowCosts, colCosts) { + let result = 0; + const [startRow, startCol] = startPos; + const [homeRow, homeCol] = homePos; + + const rowStep = startRow < homeRow ? 1 : -1; + for (let row = startRow + rowStep; row !== homeRow + rowStep; row += rowStep) { + result += rowCosts[row]; + } + + const colStep = startCol < homeCol ? 1 : -1; + for (let col = startCol + colStep; col !== homeCol + colStep; col += colStep) { + result += colCosts[col]; + } + + return result; +}; diff --git a/solutions/2088-count-fertile-pyramids-in-a-land.js b/solutions/2088-count-fertile-pyramids-in-a-land.js new file mode 100644 index 00000000..0780710b --- /dev/null +++ b/solutions/2088-count-fertile-pyramids-in-a-land.js @@ -0,0 +1,71 @@ +/** + * 2088. Count Fertile Pyramids in a Land + * https://leetcode.com/problems/count-fertile-pyramids-in-a-land/ + * Difficulty: Hard + * + * A farmer has a rectangular grid of land with m rows and n columns that can be divided into + * unit cells. Each cell is either fertile (represented by a 1) or barren (represented by a 0). + * All cells outside the grid are considered barren. + * + * A pyramidal plot of land can be defined as a set of cells with the following criteria: + * 1. The number of cells in the set has to be greater than 1 and all cells must be fertile. + * 2. The apex of a pyramid is the topmost cell of the pyramid. The height of a pyramid is the + * number of rows it covers. Let (r, c) be the apex of the pyramid, and its height be h. + * Then, the plot comprises of cells (i, j) where r <= i <= r + h - 1 and + * c - (i - r) <= j <= c + (i - r). + * + * An inverse pyramidal plot of land can be defined as a set of cells with similar criteria: + * 1. The number of cells in the set has to be greater than 1 and all cells must be fertile. + * 2. The apex of an inverse pyramid is the bottommost cell of the inverse pyramid. The height + * of an inverse pyramid is the number of rows it covers. Let (r, c) be the apex of the + * pyramid, and its height be h. Then, the plot comprises of cells (i, j) where + * r - h + 1 <= i <= r and c - (r - i) <= j <= c + (r - i). + * + * Some examples of valid and invalid pyramidal (and inverse pyramidal) plots are shown below. + * Black cells indicate fertile cells. + * + * Given a 0-indexed m x n binary matrix grid representing the farmland, return the total number + * of pyramidal and inverse pyramidal plots that can be found in grid. + */ + +/** + * @param {number[][]} grid + * @return {number} + */ +var countPyramids = function(grid) { + const rows = grid.length; + const cols = grid[0].length; + let result = 0; + + for (let i = 0; i < rows; i++) { + for (let j = 0; j < cols; j++) { + if (grid[i][j] === 1) { + result += countPyramidAt(i, j, false); + result += countPyramidAt(i, j, true); + } + } + } + + return result; + + function countPyramidAt(row, col, isInverse) { + let height = 0; + for (let h = 1; ; h++) { + const r = isInverse ? row - h : row + h; + if (r < 0 || r >= rows) break; + const left = col - h; + const right = col + h; + if (left < 0 || right >= cols) break; + let valid = true; + for (let j = left; j <= right; j++) { + if (grid[r][j] === 0) { + valid = false; + break; + } + } + if (!valid) break; + height++; + } + return height; + } +}; diff --git a/solutions/2089-find-target-indices-after-sorting-array.js b/solutions/2089-find-target-indices-after-sorting-array.js new file mode 100644 index 00000000..c7235faa --- /dev/null +++ b/solutions/2089-find-target-indices-after-sorting-array.js @@ -0,0 +1,32 @@ +/** + * 2089. Find Target Indices After Sorting Array + * https://leetcode.com/problems/find-target-indices-after-sorting-array/ + * Difficulty: Easy + * + * You are given a 0-indexed integer array nums and a target element target. + * + * A target index is an index i such that nums[i] == target. + * + * Return a list of the target indices of nums after sorting nums in non-decreasing order. + * If there are no target indices, return an empty list. The returned list must be sorted + * in increasing order. + */ + +/** + * @param {number[]} nums + * @param {number} target + * @return {number[]} + */ +var targetIndices = function(nums, target) { + const result = []; + + nums.sort((a, b) => a - b); + + for (let i = 0; i < nums.length; i++) { + if (nums[i] === target) { + result.push(i); + } + } + + return result; +}; diff --git a/solutions/2090-k-radius-subarray-averages.js b/solutions/2090-k-radius-subarray-averages.js new file mode 100644 index 00000000..3808077d --- /dev/null +++ b/solutions/2090-k-radius-subarray-averages.js @@ -0,0 +1,46 @@ +/** + * 2090. K Radius Subarray Averages + * https://leetcode.com/problems/k-radius-subarray-averages/ + * Difficulty: Medium + * + * You are given a 0-indexed array nums of n integers, and an integer k. + * + * The k-radius average for a subarray of nums centered at some index i with the radius k is + * the average of all elements in nums between the indices i - k and i + k (inclusive). If + * there are less than k elements before or after the index i, then the k-radius average is -1. + * + * Build and return an array avgs of length n where avgs[i] is the k-radius average for the + * subarray centered at index i. + * + * The average of x elements is the sum of the x elements divided by x, using integer division. + * The integer division truncates toward zero, which means losing its fractional part. + * - For example, the average of four elements 2, 3, 1, and + * 5 is (2 + 3 + 1 + 5) / 4 = 11 / 4 = 2.75, which truncates to 2. + */ + +/** + * @param {number[]} nums + * @param {number} k + * @return {number[]} + */ +var getAverages = function(nums, k) { + const n = nums.length; + const windowSize = 2 * k + 1; + const result = new Array(n).fill(-1); + + if (windowSize > n) return result; + + let sum = 0; + for (let i = 0; i < windowSize; i++) { + sum += nums[i]; + } + + result[k] = Math.floor(sum / windowSize); + + for (let i = k + 1; i < n - k; i++) { + sum = sum - nums[i - k - 1] + nums[i + k]; + result[i] = Math.floor(sum / windowSize); + } + + return result; +}; diff --git a/solutions/2091-removing-minimum-and-maximum-from-array.js b/solutions/2091-removing-minimum-and-maximum-from-array.js new file mode 100644 index 00000000..b62e781e --- /dev/null +++ b/solutions/2091-removing-minimum-and-maximum-from-array.js @@ -0,0 +1,41 @@ +/** + * 2091. Removing Minimum and Maximum From Array + * https://leetcode.com/problems/removing-minimum-and-maximum-from-array/ + * Difficulty: Medium + * + * You are given a 0-indexed array of distinct integers nums. + * + * There is an element in nums that has the lowest value and an element that has the highest + * value. We call them the minimum and maximum respectively. Your goal is to remove both these + * elements from the array. + * + * A deletion is defined as either removing an element from the front of the array or removing + * an element from the back of the array. + * + * Return the minimum number of deletions it would take to remove both the minimum and maximum + * element from the array. + */ + +/** + * @param {number[]} nums + * @return {number} + */ +var minimumDeletions = function(nums) { + const n = nums.length; + let minIndex = 0; + let maxIndex = 0; + + for (let i = 1; i < n; i++) { + if (nums[i] < nums[minIndex]) minIndex = i; + if (nums[i] > nums[maxIndex]) maxIndex = i; + } + + const left = Math.min(minIndex, maxIndex); + const right = Math.max(minIndex, maxIndex); + + return Math.min( + right + 1, + n - left, + left + 1 + n - right + ); +}; diff --git a/solutions/2092-find-all-people-with-secret.js b/solutions/2092-find-all-people-with-secret.js new file mode 100644 index 00000000..f5fecbdc --- /dev/null +++ b/solutions/2092-find-all-people-with-secret.js @@ -0,0 +1,73 @@ +/** + * 2092. Find All People With Secret + * https://leetcode.com/problems/find-all-people-with-secret/ + * Difficulty: Hard + * + * You are given an integer n indicating there are n people numbered from 0 to n - 1. You are also + * given a 0-indexed 2D integer array meetings where meetings[i] = [xi, yi, timei] indicates that + * person xi and person yi have a meeting at timei. A person may attend multiple meetings at the + * same time. Finally, you are given an integer firstPerson. + * + * Person 0 has a secret and initially shares the secret with a person firstPerson at time 0. This + * secret is then shared every time a meeting takes place with a person that has the secret. More + * formally, for every meeting, if a person xi has the secret at timei, then they will share the + * secret with person yi, and vice versa. + * + * The secrets are shared instantaneously. That is, a person may receive the secret and share it + * with people in other meetings within the same time frame. + * + * Return a list of all the people that have the secret after all the meetings have taken place. + * You may return the answer in any order. + */ + +/** + * @param {number} n + * @param {number[][]} meetings + * @param {number} firstPerson + * @return {number[]} + */ +var findAllPeople = function(n, meetings, firstPerson) { + const parent = Array.from({ length: n }, (_, i) => i); + + union(0, firstPerson); + meetings.sort((a, b) => a[2] - b[2]); + + let i = 0; + while (i < meetings.length) { + const currentTime = meetings[i][2]; + const group = []; + + while (i < meetings.length && meetings[i][2] === currentTime) { + const [x, y] = meetings[i]; + group.push(x, y); + union(x, y); + i++; + } + + for (const person of group) { + if (find(person) !== find(0)) { + parent[person] = person; + } + } + } + + const result = []; + for (let j = 0; j < n; j++) { + if (find(j) === find(0)) { + result.push(j); + } + } + + return result; + + function find(x) { + if (parent[x] !== x) { + parent[x] = find(parent[x]); + } + return parent[x]; + } + + function union(x, y) { + parent[find(x)] = find(y); + } +}; diff --git a/solutions/2094-finding-3-digit-even-numbers.js b/solutions/2094-finding-3-digit-even-numbers.js new file mode 100644 index 00000000..f021bdff --- /dev/null +++ b/solutions/2094-finding-3-digit-even-numbers.js @@ -0,0 +1,51 @@ +/** + * 2094. Finding 3-Digit Even Numbers + * https://leetcode.com/problems/finding-3-digit-even-numbers/ + * Difficulty: Easy + * + * You are given an integer array digits, where each element is a digit. The array may contain + * duplicates. + * + * You need to find all the unique integers that follow the given requirements: + * - The integer consists of the concatenation of three elements from digits in any arbitrary order. + * - The integer does not have leading zeros. + * - The integer is even. + * + * For example, if the given digits were [1, 2, 3], integers 132 and 312 follow the requirements. + * + * Return a sorted array of the unique integers. + */ + +/** + * @param {number[]} digits + * @return {number[]} + */ +var findEvenNumbers = function(digits) { + const frequency = new Array(10).fill(0); + const uniqueNumbers = new Set(); + + for (const digit of digits) { + frequency[digit]++; + } + + for (let hundreds = 1; hundreds <= 9; hundreds++) { + if (frequency[hundreds] === 0) continue; + frequency[hundreds]--; + + for (let tens = 0; tens <= 9; tens++) { + if (frequency[tens] === 0) continue; + frequency[tens]--; + + for (let ones = 0; ones <= 8; ones += 2) { + if (frequency[ones] === 0) continue; + uniqueNumbers.add(hundreds * 100 + tens * 10 + ones); + } + + frequency[tens]++; + } + + frequency[hundreds]++; + } + + return Array.from(uniqueNumbers).sort((a, b) => a - b); +}; diff --git a/solutions/2096-step-by-step-directions-from-a-binary-tree-node-to-another.js b/solutions/2096-step-by-step-directions-from-a-binary-tree-node-to-another.js new file mode 100644 index 00000000..4eb7d4e9 --- /dev/null +++ b/solutions/2096-step-by-step-directions-from-a-binary-tree-node-to-another.js @@ -0,0 +1,65 @@ +/** + * 2096. Step-By-Step Directions From a Binary Tree Node to Another + * https://leetcode.com/problems/step-by-step-directions-from-a-binary-tree-node-to-another/ + * Difficulty: Medium + * + * You are given the root of a binary tree with n nodes. Each node is uniquely assigned a value + * from 1 to n. You are also given an integer startValue representing the value of the start node + * s, and a different integer destValue representing the value of the destination node t. + * + * Find the shortest path starting from node s and ending at node t. Generate step-by-step + * directions of such path as a string consisting of only the uppercase letters 'L', 'R', and + * 'U'. Each letter indicates a specific direction: + * - 'L' means to go from a node to its left child node. + * - 'R' means to go from a node to its right child node. + * - 'U' means to go from a node to its parent node. + * + * Return the step-by-step directions of the shortest path from node s to node t. + */ + +/** + * Definition for a binary tree node. + * function TreeNode(val, left, right) { + * this.val = (val===undefined ? 0 : val) + * this.left = (left===undefined ? null : left) + * this.right = (right===undefined ? null : right) + * } + */ +/** + * @param {TreeNode} root + * @param {number} startValue + * @param {number} destValue + * @return {string} + */ +var getDirections = function(root, startValue, destValue) { + const startPath = []; + const destPath = []; + + findPath(root, startValue, startPath); + findPath(root, destValue, destPath); + + let i = 0; + while (i < startPath.length && i < destPath.length && startPath[i] === destPath[i]) { + i++; + } + + const upMoves = 'U'.repeat(startPath.length - i); + const downMoves = destPath.slice(i).join(''); + + return upMoves + downMoves; + + function findPath(node, value, path) { + if (!node) return false; + if (node.val === value) return true; + + path.push('L'); + if (findPath(node.left, value, path)) return true; + path.pop(); + + path.push('R'); + if (findPath(node.right, value, path)) return true; + path.pop(); + + return false; + } +}; diff --git a/solutions/2097-valid-arrangement-of-pairs.js b/solutions/2097-valid-arrangement-of-pairs.js new file mode 100644 index 00000000..d3c5ee1c --- /dev/null +++ b/solutions/2097-valid-arrangement-of-pairs.js @@ -0,0 +1,50 @@ +/** + * 2097. Valid Arrangement of Pairs + * https://leetcode.com/problems/valid-arrangement-of-pairs/ + * Difficulty: Hard + * + * You are given a 0-indexed 2D integer array pairs where pairs[i] = [starti, endi]. An arrangement + * of pairs is valid if for every index i where 1 <= i < pairs.length, we have endi-1 == starti. + * + * Return any valid arrangement of pairs. + * + * Note: The inputs will be generated such that there exists a valid arrangement of pairs. + */ + +/** + * @param {number[][]} pairs + * @return {number[][]} + */ +var validArrangement = function(pairs) { + const graph = new Map(); + const degree = new Map(); + + for (const [start, end] of pairs) { + if (!graph.has(start)) graph.set(start, []); + graph.get(start).push(end); + + degree.set(start, (degree.get(start) || 0) + 1); + degree.set(end, (degree.get(end) || 0) - 1); + } + + let startNode = pairs[0][0]; + for (const [node, deg] of degree) { + if (deg > 0) { + startNode = node; + break; + } + } + + const result = []; + helper(startNode); + + return result.reverse(); + + function helper(node) { + while (graph.get(node)?.length) { + const next = graph.get(node).pop(); + helper(next); + result.push([node, next]); + } + } +}; diff --git a/solutions/2100-find-good-days-to-rob-the-bank.js b/solutions/2100-find-good-days-to-rob-the-bank.js new file mode 100644 index 00000000..1dc59611 --- /dev/null +++ b/solutions/2100-find-good-days-to-rob-the-bank.js @@ -0,0 +1,50 @@ +/** + * 2100. Find Good Days to Rob the Bank + * https://leetcode.com/problems/find-good-days-to-rob-the-bank/ + * Difficulty: Medium + * + * You and a gang of thieves are planning on robbing a bank. You are given a 0-indexed integer + * array security, where security[i] is the number of guards on duty on the ith day. The days + * are numbered starting from 0. You are also given an integer time. + * + * The ith day is a good day to rob the bank if: + * - There are at least time days before and after the ith day, + * - The number of guards at the bank for the time days before i are non-increasing, and + * - The number of guards at the bank for the time days after i are non-decreasing. + * + * More formally, this means day i is a good day to rob the bank if and only if + * security[i - time] >= security[i - time + 1] >= ... >= security[i] + * <= ... <= security[i + time - 1] <= security[i + time]. + * + * Return a list of all days (0-indexed) that are good days to rob the bank. The order that the + * days are returned in does not matter. + */ + +/** + * @param {number[]} security + * @param {number} time + * @return {number[]} + */ +var goodDaysToRobBank = function(security, time) { + const n = security.length; + const nonIncreasing = new Array(n).fill(0); + const nonDecreasing = new Array(n).fill(0); + const result = []; + + for (let i = 1; i < n; i++) { + if (security[i] <= security[i - 1]) { + nonIncreasing[i] = nonIncreasing[i - 1] + 1; + } + if (security[n - i - 1] <= security[n - i]) { + nonDecreasing[n - i - 1] = nonDecreasing[n - i] + 1; + } + } + + for (let i = time; i < n - time; i++) { + if (nonIncreasing[i] >= time && nonDecreasing[i] >= time) { + result.push(i); + } + } + + return result; +}; diff --git a/solutions/2101-detonate-the-maximum-bombs.js b/solutions/2101-detonate-the-maximum-bombs.js new file mode 100644 index 00000000..11069857 --- /dev/null +++ b/solutions/2101-detonate-the-maximum-bombs.js @@ -0,0 +1,57 @@ +/** + * 2101. Detonate the Maximum Bombs + * https://leetcode.com/problems/detonate-the-maximum-bombs/ + * Difficulty: Medium + * + * You are given a list of bombs. The range of a bomb is defined as the area where its effect + * can be felt. This area is in the shape of a circle with the center as the location of the bomb. + * + * The bombs are represented by a 0-indexed 2D integer array bombs where bombs[i] = [xi, yi, ri]. + * xi and yi denote the X-coordinate and Y-coordinate of the location of the ith bomb, whereas ri + * denotes the radius of its range. + * + * You may choose to detonate a single bomb. When a bomb is detonated, it will detonate all bombs + * that lie in its range. These bombs will further detonate the bombs that lie in their ranges. + * + * Given the list of bombs, return the maximum number of bombs that can be detonated if you are + * allowed to detonate only one bomb. + */ + +/** + * @param {number[][]} bombs + * @return {number} + */ +var maximumDetonation = function(bombs) { + const n = bombs.length; + const graph = Array.from({ length: n }, () => []); + + for (let i = 0; i < n; i++) { + const [x1, y1, r1] = bombs[i]; + for (let j = 0; j < n; j++) { + if (i === j) continue; + const [x2, y2] = bombs[j]; + const distance = Math.sqrt((x2 - x1) ** 2 + (y2 - y1) ** 2); + if (distance <= r1) { + graph[i].push(j); + } + } + } + + let result = 0; + for (let i = 0; i < n; i++) { + result = Math.max(result, dfs(i, new Set())); + } + + return result; + + function dfs(node, visited) { + visited.add(node); + let count = 1; + for (const neighbor of graph[node]) { + if (!visited.has(neighbor)) { + count += dfs(neighbor, visited); + } + } + return count; + } +}; diff --git a/solutions/2103-rings-and-rods.js b/solutions/2103-rings-and-rods.js new file mode 100644 index 00000000..b7cc219e --- /dev/null +++ b/solutions/2103-rings-and-rods.js @@ -0,0 +1,42 @@ +/** + * 2103. Rings and Rods + * https://leetcode.com/problems/rings-and-rods/ + * Difficulty: Easy + * + * There are n rings and each ring is either red, green, or blue. The rings are distributed across + * ten rods labeled from 0 to 9. + * + * You are given a string rings of length 2n that describes the n rings that are placed onto the + * rods. Every two characters in rings forms a color-position pair that is used to describe each + * ring where: + * - The first character of the ith pair denotes the ith ring's color ('R', 'G', 'B'). + * - The second character of the ith pair denotes the rod that the ith ring is placed on + * ('0' to '9'). + * + * For example, "R3G2B1" describes n == 3 rings: a red ring placed onto the rod labeled 3, a green + * ring placed onto the rod labeled 2, and a blue ring placed onto the rod labeled 1. + * + * Return the number of rods that have all three colors of rings on them. + */ + +/** + * @param {string} rings + * @return {number} + */ +var countPoints = function(rings) { + const map = new Map(); + + for (let i = 0; i < rings.length; i += 2) { + const color = rings[i]; + const rod = rings[i + 1]; + if (!map.has(rod)) map.set(rod, new Set()); + map.get(rod).add(color); + } + + let result = 0; + for (const colors of map.values()) { + if (colors.size === 3) result++; + } + + return result; +}; diff --git a/solutions/2104-sum-of-subarray-ranges.js b/solutions/2104-sum-of-subarray-ranges.js new file mode 100644 index 00000000..0efaf1d7 --- /dev/null +++ b/solutions/2104-sum-of-subarray-ranges.js @@ -0,0 +1,33 @@ +/** + * 2104. Sum of Subarray Ranges + * https://leetcode.com/problems/sum-of-subarray-ranges/ + * Difficulty: Medium + * + * You are given an integer array nums. The range of a subarray of nums is the difference between + * the largest and smallest element in the subarray. + * + * Return the sum of all subarray ranges of nums. + * + * A subarray is a contiguous non-empty sequence of elements within an array. + */ + +/** + * @param {number[]} nums + * @return {number} + */ +var subArrayRanges = function(nums) { + let result = 0; + + for (let i = 0; i < nums.length; i++) { + let min = nums[i]; + let max = nums[i]; + + for (let j = i; j < nums.length; j++) { + min = Math.min(min, nums[j]); + max = Math.max(max, nums[j]); + result += max - min; + } + } + + return result; +}; diff --git a/solutions/2105-watering-plants-ii.js b/solutions/2105-watering-plants-ii.js new file mode 100644 index 00000000..dcdaf994 --- /dev/null +++ b/solutions/2105-watering-plants-ii.js @@ -0,0 +1,61 @@ +/** + * 2105. Watering Plants II + * https://leetcode.com/problems/watering-plants-ii/ + * Difficulty: Medium + * + * Alice and Bob want to water n plants in their garden. The plants are arranged in a row and are + * labeled from 0 to n - 1 from left to right where the ith plant is located at x = i. + * + * Each plant needs a specific amount of water. Alice and Bob have a watering can each, initially + * full. They water the plants in the following way: + * - Alice waters the plants in order from left to right, starting from the 0th plant. Bob waters + * the plants in order from right to left, starting from the (n - 1)th plant. They begin watering + * the plants simultaneously. + * - It takes the same amount of time to water each plant regardless of how much water it needs. + * - Alice/Bob must water the plant if they have enough in their can to fully water it. Otherwise, + * they first refill their can (instantaneously) then water the plant. + * - In case both Alice and Bob reach the same plant, the one with more water currently in his/her + * watering can should water this plant. If they have the same amount of water, then Alice should + * water this plant. + * + * Given a 0-indexed integer array plants of n integers, where plants[i] is the amount of water the + * ith plant needs, and two integers capacityA and capacityB representing the capacities of Alice's + * and Bob's watering cans respectively, return the number of times they have to refill to water all + * the plants. + */ + +/** + * @param {number[]} plants + * @param {number} capacityA + * @param {number} capacityB + * @return {number} + */ +var minimumRefill = function(plants, capacityA, capacityB) { + let result = 0; + let left = 0; + let right = plants.length - 1; + let waterA = capacityA; + let waterB = capacityB; + + while (left <= right) { + if (left === right) { + if (waterA >= waterB && waterA < plants[left]) result++; + if (waterB > waterA && waterB < plants[right]) result++; + break; + } + + if (waterA < plants[left]) { + waterA = capacityA; + result++; + } + waterA -= plants[left++]; + + if (waterB < plants[right]) { + waterB = capacityB; + result++; + } + waterB -= plants[right--]; + } + + return result; +}; diff --git a/solutions/2106-maximum-fruits-harvested-after-at-most-k-steps.js b/solutions/2106-maximum-fruits-harvested-after-at-most-k-steps.js new file mode 100644 index 00000000..bd81e17e --- /dev/null +++ b/solutions/2106-maximum-fruits-harvested-after-at-most-k-steps.js @@ -0,0 +1,52 @@ +/** + * 2106. Maximum Fruits Harvested After at Most K Steps + * https://leetcode.com/problems/maximum-fruits-harvested-after-at-most-k-steps/ + * Difficulty: Hard + * + * Fruits are available at some positions on an infinite x-axis. You are given a 2D integer array + * fruits where fruits[i] = [positioni, amounti] depicts amounti fruits at the position positioni. + * fruits is already sorted by positioni in ascending order, and each positioni is unique. + * + * You are also given an integer startPos and an integer k. Initially, you are at the position + * startPos. From any position, you can either walk to the left or right. It takes one step to + * move one unit on the x-axis, and you can walk at most k steps in total. For every position + * you reach, you harvest all the fruits at that position, and the fruits will disappear from + * that position. + * + * Return the maximum total number of fruits you can harvest. + */ + +/** + * @param {number[][]} fruits + * @param {number} startPos + * @param {number} k + * @return {number} + */ +var maxTotalFruits = function(fruits, startPos, k) { + let result = 0; + let left = 0; + let currentSum = 0; + + for (let right = 0; right < fruits.length; right++) { + currentSum += fruits[right][1]; + + while (left <= right) { + const minPos = fruits[left][0]; + const maxPos = fruits[right][0]; + const steps = Math.min( + Math.abs(startPos - minPos) + (maxPos - minPos), + Math.abs(startPos - maxPos) + (maxPos - minPos) + ); + + if (steps <= k) break; + currentSum -= fruits[left][1]; + left++; + } + + if (left <= right) { + result = Math.max(result, currentSum); + } + } + + return result; +}; diff --git a/solutions/2108-find-first-palindromic-string-in-the-array.js b/solutions/2108-find-first-palindromic-string-in-the-array.js new file mode 100644 index 00000000..a72a02b0 --- /dev/null +++ b/solutions/2108-find-first-palindromic-string-in-the-array.js @@ -0,0 +1,23 @@ +/** + * 2108. Find First Palindromic String in the Array + * https://leetcode.com/problems/find-first-palindromic-string-in-the-array/ + * Difficulty: Easy + * + * Given an array of strings words, return the first palindromic string in the array. If there is + * no such string, return an empty string "". + * + * A string is palindromic if it reads the same forward and backward. + */ + +/** + * @param {string[]} words + * @return {string} + */ +var firstPalindrome = function(words) { + for (const word of words) { + if (word === word.split('').reverse().join('')) { + return word; + } + } + return ''; +}; diff --git a/solutions/2109-adding-spaces-to-a-string.js b/solutions/2109-adding-spaces-to-a-string.js new file mode 100644 index 00000000..7f4a85ba --- /dev/null +++ b/solutions/2109-adding-spaces-to-a-string.js @@ -0,0 +1,33 @@ +/** + * 2109. Adding Spaces to a String + * https://leetcode.com/problems/adding-spaces-to-a-string/ + * Difficulty: Medium + * + * You are given a 0-indexed string s and a 0-indexed integer array spaces that describes the + * indices in the original string where spaces will be added. Each space should be inserted + * before the character at the given index. + * - For example, given s = "EnjoyYourCoffee" and spaces = [5, 9], we place spaces before 'Y' + * and 'C', which are at indices 5 and 9 respectively. Thus, we obtain "Enjoy Your Coffee". + * + * Return the modified string after the spaces have been added. + */ + +/** + * @param {string} s + * @param {number[]} spaces + * @return {string} + */ +var addSpaces = function(s, spaces) { + let result = ''; + let spaceIndex = 0; + + for (let i = 0; i < s.length; i++) { + if (spaceIndex < spaces.length && i === spaces[spaceIndex]) { + result += ' '; + spaceIndex++; + } + result += s[i]; + } + + return result; +}; diff --git a/solutions/2110-number-of-smooth-descent-periods-of-a-stock.js b/solutions/2110-number-of-smooth-descent-periods-of-a-stock.js new file mode 100644 index 00000000..09a08e95 --- /dev/null +++ b/solutions/2110-number-of-smooth-descent-periods-of-a-stock.js @@ -0,0 +1,34 @@ +/** + * 2110. Number of Smooth Descent Periods of a Stock + * https://leetcode.com/problems/number-of-smooth-descent-periods-of-a-stock/ + * Difficulty: Medium + * + * You are given an integer array prices representing the daily price history of a stock, + * where prices[i] is the stock price on the ith day. + * + * A smooth descent period of a stock consists of one or more contiguous days such that the + * price on each day is lower than the price on the preceding day by exactly 1. The first + * day of the period is exempted from this rule. + * + * Return the number of smooth descent periods. + */ + +/** + * @param {number[]} prices + * @return {number} + */ +var getDescentPeriods = function(prices) { + let result = 1; + let currentLength = 1; + + for (let i = 1; i < prices.length; i++) { + if (prices[i] === prices[i - 1] - 1) { + currentLength++; + } else { + currentLength = 1; + } + result += currentLength; + } + + return result; +}; diff --git a/solutions/2111-minimum-operations-to-make-the-array-k-increasing.js b/solutions/2111-minimum-operations-to-make-the-array-k-increasing.js new file mode 100644 index 00000000..8f861b84 --- /dev/null +++ b/solutions/2111-minimum-operations-to-make-the-array-k-increasing.js @@ -0,0 +1,58 @@ +/** + * 2111. Minimum Operations to Make the Array K-Increasing + * https://leetcode.com/problems/minimum-operations-to-make-the-array-k-increasing/ + * Difficulty: Hard + * + * You are given a 0-indexed array arr consisting of n positive integers, and a positive integer k. + * + * The array arr is called K-increasing if arr[i-k] <= arr[i] holds for every index i, where + * k <= i <= n-1. + * + * - For example, arr = [4, 1, 5, 2, 6, 2] is K-increasing for k = 2 because: + * - arr[0] <= arr[2] (4 <= 5) + * - arr[1] <= arr[3] (1 <= 2) + * - arr[2] <= arr[4] (5 <= 6) + * - arr[3] <= arr[5] (2 <= 2) + * - However, the same arr is not K-increasing for k = 1 (because arr[0] > arr[1]) or k = 3 (because + * arr[0] > arr[3]). + * + * In one operation, you can choose an index i and change arr[i] into any positive integer. + * + * Return the minimum number of operations required to make the array K-increasing for the given k. + */ + +/** + * @param {number[]} arr + * @param {number} k + * @return {number} + */ +var kIncreasing = function(arr, k) { + let result = 0; + for (let i = 0; i < k; i++) { + const subsequence = []; + for (let j = i; j < arr.length; j += k) { + subsequence.push(arr[j]); + } + result += longestNonDecreasingSubsequence(subsequence); + } + + return result; + + function longestNonDecreasingSubsequence(nums) { + const tails = []; + for (const num of nums) { + let left = 0; + let right = tails.length; + while (left < right) { + const mid = Math.floor((left + right) / 2); + if (tails[mid] <= num) { + left = mid + 1; + } else { + right = mid; + } + } + tails[left] = num; + } + return nums.length - tails.length; + } +}; diff --git a/solutions/2117-abbreviating-the-product-of-a-range.js b/solutions/2117-abbreviating-the-product-of-a-range.js new file mode 100644 index 00000000..4334b25b --- /dev/null +++ b/solutions/2117-abbreviating-the-product-of-a-range.js @@ -0,0 +1,91 @@ +/** + * 2117. Abbreviating the Product of a Range + * https://leetcode.com/problems/abbreviating-the-product-of-a-range/ + * Difficulty: Hard + * + * You are given two positive integers left and right with left <= right. Calculate the product of + * all integers in the inclusive range [left, right]. + * + * Since the product may be very large, you will abbreviate it following these steps: + * 1. Count all trailing zeros in the product and remove them. Let us denote this count as C. + * - For example, there are 3 trailing zeros in 1000, and there are 0 trailing zeros in 546. + * 2. Denote the remaining number of digits in the product as d. If d > 10, then express the product + * as 
... where 
 denotes the first 5 digits of the product, and  denotes the
+ *    last 5 digits of the product after removing all trailing zeros. If d <= 10, we keep it
+ *    unchanged.
+ *    - For example, we express 1234567654321 as 12345...54321, but 1234567 is represented as
+ *      1234567.
+ * 3. Finally, represent the product as a string "
...eC".
+ *    - For example, 12345678987600000 will be represented as "12345...89876e5".
+ *
+ * Return a string denoting the abbreviated product of all integers in the inclusive range [left,
+ * right].
+ */
+
+/**
+ * @param {number} left
+ * @param {number} right
+ * @return {string}
+ */
+var abbreviateProduct = function(left, right) {
+  let zeros = 0;
+  let count2 = 0;
+  let count5 = 0;
+
+  for (let i = left; i <= right; i++) {
+    let n = i;
+    while (n % 2 === 0) {
+      count2++;
+      n = Math.floor(n / 2);
+    }
+    n = i;
+    while (n % 5 === 0) {
+      count5++;
+      n = Math.floor(n / 5);
+    }
+  }
+  zeros = Math.min(count2, count5);
+
+  let digits = 0;
+  for (let i = left; i <= right; i++) {
+    digits += Math.log10(i);
+  }
+  digits = Math.floor(digits) + 1;
+
+  if (digits - zeros <= 10) {
+    let product = 1n;
+    for (let i = left; i <= right; i++) {
+      product *= BigInt(i);
+    }
+    for (let i = 0; i < zeros; i++) {
+      product /= 10n;
+    }
+    return product.toString() + 'e' + zeros;
+  }
+
+  let prefix = 1;
+  for (let i = left; i <= right; i++) {
+    prefix *= i;
+    while (prefix >= 1e10) {
+      prefix /= 10;
+    }
+  }
+  prefix = prefix.toString().slice(0, 5);
+
+  let suffix = 1n;
+  for (let i = right; i >= left; i--) {
+    suffix = (suffix * BigInt(i));
+    while (suffix % 10n === 0n) {
+      suffix /= 10n;
+    }
+    suffix = suffix % (10n ** 15n);
+  }
+
+  suffix = suffix.toString();
+  while (suffix.length < 5) {
+    suffix = '0' + suffix;
+  }
+  suffix = suffix.slice(-5);
+
+  return prefix + '...' + suffix + 'e' + zeros;
+};
diff --git a/solutions/2119-a-number-after-a-double-reversal.js b/solutions/2119-a-number-after-a-double-reversal.js
new file mode 100644
index 00000000..efc1d321
--- /dev/null
+++ b/solutions/2119-a-number-after-a-double-reversal.js
@@ -0,0 +1,20 @@
+/**
+ * 2119. A Number After a Double Reversal
+ * https://leetcode.com/problems/a-number-after-a-double-reversal/
+ * Difficulty: Easy
+ *
+ * Reversing an integer means to reverse all its digits.
+ * - For example, reversing 2021 gives 1202. Reversing 12300 gives 321 as the leading zeros
+ *   are not retained.
+ *
+ * Given an integer num, reverse num to get reversed1, then reverse reversed1 to get reversed2.
+ * Return true if reversed2 equals num. Otherwise return false.
+ */
+
+/**
+ * @param {number} num
+ * @return {boolean}
+ */
+var isSameAfterReversals = function(num) {
+  return num === 0 || num % 10 !== 0;
+};
diff --git a/solutions/2120-execution-of-all-suffix-instructions-staying-in-a-grid.js b/solutions/2120-execution-of-all-suffix-instructions-staying-in-a-grid.js
new file mode 100644
index 00000000..940ddc72
--- /dev/null
+++ b/solutions/2120-execution-of-all-suffix-instructions-staying-in-a-grid.js
@@ -0,0 +1,51 @@
+/**
+ * 2120. Execution of All Suffix Instructions Staying in a Grid
+ * https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/
+ * Difficulty: Medium
+ *
+ * There is an n x n grid, with the top-left cell at (0, 0) and the bottom-right cell at
+ * (n - 1, n - 1). You are given the integer n and an integer array startPos where
+ * startPos = [startrow, startcol] indicates that a robot is initially at cell (startrow, startcol).
+ *
+ * You are also given a 0-indexed string s of length m where s[i] is the ith instruction for the
+ * robot: 'L' (move left), 'R' (move right), 'U' (move up), and 'D' (move down).
+ *
+ * The robot can begin executing from any ith instruction in s. It executes the instructions one
+ * by one towards the end of s but it stops if either of these conditions is met:
+ * - The next instruction will move the robot off the grid.
+ * - There are no more instructions left to execute.
+ *
+ * Return an array answer of length m where answer[i] is the number of instructions the robot can
+ * execute if the robot begins executing from the ith instruction in s.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[]} startPos
+ * @param {string} s
+ * @return {number[]}
+ */
+var executeInstructions = function(n, startPos, s) {
+  const m = s.length;
+  const result = new Array(m).fill(0);
+
+  for (let i = 0; i < m; i++) {
+    let row = startPos[0];
+    let col = startPos[1];
+    let steps = 0;
+
+    for (let j = i; j < m; j++) {
+      if (s[j] === 'L') col--;
+      else if (s[j] === 'R') col++;
+      else if (s[j] === 'U') row--;
+      else row++;
+
+      if (row < 0 || row >= n || col < 0 || col >= n) break;
+      steps++;
+    }
+
+    result[i] = steps;
+  }
+
+  return result;
+};
diff --git a/solutions/2121-intervals-between-identical-elements.js b/solutions/2121-intervals-between-identical-elements.js
new file mode 100644
index 00000000..3686726b
--- /dev/null
+++ b/solutions/2121-intervals-between-identical-elements.js
@@ -0,0 +1,48 @@
+/**
+ * 2121. Intervals Between Identical Elements
+ * https://leetcode.com/problems/intervals-between-identical-elements/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed array of n integers arr.
+ *
+ * The interval between two elements in arr is defined as the absolute difference between their
+ * indices. More formally, the interval between arr[i] and arr[j] is |i - j|.
+ *
+ * Return an array intervals of length n where intervals[i] is the sum of intervals between arr[i]
+ * and each element in arr with the same value as arr[i].
+ *
+ * Note: |x| is the absolute value of x.
+ */
+
+/**
+ * @param {number[]} arr
+ * @return {number[]}
+ */
+var getDistances = function(arr) {
+  const valueIndices = new Map();
+  const result = new Array(arr.length).fill(0);
+
+  for (let i = 0; i < arr.length; i++) {
+    if (!valueIndices.has(arr[i])) {
+      valueIndices.set(arr[i], []);
+    }
+    valueIndices.get(arr[i]).push(i);
+  }
+
+  for (const indices of valueIndices.values()) {
+    let prefixSum = 0;
+    for (let i = 1; i < indices.length; i++) {
+      prefixSum += indices[i] - indices[0];
+    }
+
+    result[indices[0]] = prefixSum;
+
+    for (let i = 1; i < indices.length; i++) {
+      const diff = indices[i] - indices[i - 1];
+      prefixSum += diff * (i - (indices.length - i));
+      result[indices[i]] = prefixSum;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2122-recover-the-original-array.js b/solutions/2122-recover-the-original-array.js
new file mode 100644
index 00000000..7b6030fb
--- /dev/null
+++ b/solutions/2122-recover-the-original-array.js
@@ -0,0 +1,55 @@
+/**
+ * 2122. Recover the Original Array
+ * https://leetcode.com/problems/recover-the-original-array/
+ * Difficulty: Hard
+ *
+ * Alice had a 0-indexed array arr consisting of n positive integers. She chose an arbitrary
+ * positive integer k and created two new 0-indexed integer arrays lower and higher in the
+ * following manner:
+ * 1. lower[i] = arr[i] - k, for every index i where 0 <= i < n
+ * 2. higher[i] = arr[i] + k, for every index i where 0 <= i < n
+ *
+ * Unfortunately, Alice lost all three arrays. However, she remembers the integers that were
+ * present in the arrays lower and higher, but not the array each integer belonged to. Help
+ * Alice and recover the original array.
+ *
+ * Given an array nums consisting of 2n integers, where exactly n of the integers were present
+ * in lower and the remaining in higher, return the original array arr. In case the answer is
+ * not unique, return any valid array.
+ *
+ * Note: The test cases are generated such that there exists at least one valid array arr.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var recoverArray = function(nums) {
+  const n = nums.length / 2;
+  nums.sort((a, b) => a - b);
+
+  for (let i = 1; i < 2 * n; i++) {
+    const k = nums[i] - nums[0];
+    if (k <= 0 || k % 2 !== 0) continue;
+
+    const original = [];
+    const used = new Array(2 * n).fill(false);
+    let count = 0;
+
+    for (let left = 0, right = i; right < 2 * n && count < n; right++) {
+      while (left < right && used[left]) left++;
+      if (left >= right) continue;
+
+      if (nums[right] - nums[left] === k) {
+        original.push(nums[left] + k / 2);
+        used[left] = used[right] = true;
+        count++;
+        left++;
+      }
+    }
+
+    if (count === n) return original;
+  }
+
+  return [];
+};
diff --git a/solutions/2124-check-if-all-as-appears-before-all-bs.js b/solutions/2124-check-if-all-as-appears-before-all-bs.js
new file mode 100644
index 00000000..57f06772
--- /dev/null
+++ b/solutions/2124-check-if-all-as-appears-before-all-bs.js
@@ -0,0 +1,23 @@
+/**
+ * 2124. Check if All A's Appears Before All B's
+ * https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/
+ * Difficulty: Easy
+ *
+ * Given a string s consisting of only the characters 'a' and 'b', return true if every 'a' appears
+ * before every 'b' in the string. Otherwise, return false.
+ */
+
+/**
+ * @param {string} s
+ * @return {boolean}
+ */
+var checkString = function(s) {
+  let seen = false;
+
+  for (const char of s) {
+    if (char === 'b') seen = true;
+    else if (seen) return false;
+  }
+
+  return true;
+};
diff --git a/solutions/2125-number-of-laser-beams-in-a-bank.js b/solutions/2125-number-of-laser-beams-in-a-bank.js
new file mode 100644
index 00000000..58a8afb7
--- /dev/null
+++ b/solutions/2125-number-of-laser-beams-in-a-bank.js
@@ -0,0 +1,40 @@
+/**
+ * 2125. Number of Laser Beams in a Bank
+ * https://leetcode.com/problems/number-of-laser-beams-in-a-bank/
+ * Difficulty: Medium
+ *
+ * Anti-theft security devices are activated inside a bank. You are given a 0-indexed binary string
+ * array bank representing the floor plan of the bank, which is an m x n 2D matrix. bank[i]
+ * represents the ith row, consisting of '0's and '1's. '0' means the cell is empty, while'1'
+ * means the cell has a security device.
+ *
+ * There is one laser beam between any two security devices if both conditions are met:
+ * - The two devices are located on two different rows: r1 and r2, where r1 < r2.
+ * - For each row i where r1 < i < r2, there are no security devices in the ith row.
+ *
+ * Laser beams are independent, i.e., one beam does not interfere nor join with another.
+ *
+ * Return the total number of laser beams in the bank.
+ */
+
+/**
+ * @param {string[]} bank
+ * @return {number}
+ */
+var numberOfBeams = function(bank) {
+  let result = 0;
+  let prevDevices = 0;
+
+  for (const row of bank) {
+    let currentDevices = 0;
+    for (const cell of row) {
+      if (cell === '1') currentDevices++;
+    }
+    if (currentDevices > 0) {
+      result += prevDevices * currentDevices;
+      prevDevices = currentDevices;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2126-destroying-asteroids.js b/solutions/2126-destroying-asteroids.js
new file mode 100644
index 00000000..8c0ff17e
--- /dev/null
+++ b/solutions/2126-destroying-asteroids.js
@@ -0,0 +1,31 @@
+/**
+ * 2126. Destroying Asteroids
+ * https://leetcode.com/problems/destroying-asteroids/
+ * Difficulty: Medium
+ *
+ * You are given an integer mass, which represents the original mass of a planet. You are further
+ * given an integer array asteroids, where asteroids[i] is the mass of the ith asteroid.
+ *
+ * You can arrange for the planet to collide with the asteroids in any arbitrary order. If the mass
+ * of the planet is greater than or equal to the mass of the asteroid, the asteroid is destroyed
+ * and the planet gains the mass of the asteroid. Otherwise, the planet is destroyed.
+ *
+ * Return true if all asteroids can be destroyed. Otherwise, return false.
+ */
+
+/**
+ * @param {number} mass
+ * @param {number[]} asteroids
+ * @return {boolean}
+ */
+var asteroidsDestroyed = function(mass, asteroids) {
+  asteroids.sort((a, b) => a - b);
+
+  let planetMass = BigInt(mass);
+  for (const asteroid of asteroids) {
+    if (planetMass < BigInt(asteroid)) return false;
+    planetMass += BigInt(asteroid);
+  }
+
+  return true;
+};
diff --git a/solutions/2131-longest-palindrome-by-concatenating-two-letter-words.js b/solutions/2131-longest-palindrome-by-concatenating-two-letter-words.js
new file mode 100644
index 00000000..296b3bd0
--- /dev/null
+++ b/solutions/2131-longest-palindrome-by-concatenating-two-letter-words.js
@@ -0,0 +1,47 @@
+/**
+ * 2131. Longest Palindrome by Concatenating Two Letter Words
+ * https://leetcode.com/problems/longest-palindrome-by-concatenating-two-letter-words/
+ * Difficulty: Medium
+ *
+ * You are given an array of strings words. Each element of words consists of two lowercase
+ * English letters.
+ *
+ * Create the longest possible palindrome by selecting some elements from words and concatenating
+ * them in any order. Each element can be selected at most once.
+ *
+ * Return the length of the longest palindrome that you can create. If it is impossible to create
+ * any palindrome, return 0.
+ *
+ * A palindrome is a string that reads the same forward and backward.
+ */
+
+/**
+ * @param {string[]} words
+ * @return {number}
+ */
+var longestPalindrome = function(words) {
+  const map = new Map();
+  let length = 0;
+  let hasCenter = false;
+
+  for (const word of words) {
+    map.set(word, (map.get(word) || 0) + 1);
+  }
+
+  for (const word of map.keys()) {
+    const reverse = word[1] + word[0];
+
+    if (word === reverse) {
+      const count = map.get(word);
+      length += Math.floor(count / 2) * 4;
+      if (count % 2 === 1) hasCenter = true;
+    } else if (map.has(reverse)) {
+      const pairs = Math.min(map.get(word), map.get(reverse));
+      length += pairs * 4;
+      map.set(word, 0);
+      map.set(reverse, 0);
+    }
+  }
+
+  return hasCenter ? length + 2 : length;
+};
diff --git a/solutions/2132-stamping-the-grid.js b/solutions/2132-stamping-the-grid.js
new file mode 100644
index 00000000..3143e1e3
--- /dev/null
+++ b/solutions/2132-stamping-the-grid.js
@@ -0,0 +1,64 @@
+/**
+ * 2132. Stamping the Grid
+ * https://leetcode.com/problems/stamping-the-grid/
+ * Difficulty: Hard
+ *
+ * You are given an m x n binary matrix grid where each cell is either 0 (empty) or 1 (occupied).
+ *
+ * You are then given stamps of size stampHeight x stampWidth. We want to fit the stamps such that
+ * they follow the given restrictions and requirements:
+ * 1. Cover all the empty cells.
+ * 2. Do not cover any of the occupied cells.
+ * 3. We can put as many stamps as we want.
+ * 4. Stamps can overlap with each other.
+ * 5. Stamps are not allowed to be rotated.
+ * 6. Stamps must stay completely inside the grid.
+ *
+ * Return true if it is possible to fit the stamps while following the given restrictions and
+ * requirements. Otherwise, return false.
+ */
+
+/**
+ * @param {number[][]} grid
+ * @param {number} stampHeight
+ * @param {number} stampWidth
+ * @return {boolean}
+ */
+var possibleToStamp = function(grid, stampHeight, stampWidth) {
+  const rows = grid.length;
+  const cols = grid[0].length;
+  const prefixSum = new Array(rows + 1).fill().map(() => new Array(cols + 1).fill(0));
+  const diff = new Array(rows + 1).fill().map(() => new Array(cols + 1).fill(0));
+
+  for (let i = 0; i < rows; i++) {
+    for (let j = 0; j < cols; j++) {
+      prefixSum[i + 1][j + 1] = prefixSum[i + 1][j] + prefixSum[i][j + 1]
+        - prefixSum[i][j] + grid[i][j];
+    }
+  }
+
+  for (let i = 0; i <= rows - stampHeight; i++) {
+    for (let j = 0; j <= cols - stampWidth; j++) {
+      const x = i + stampHeight;
+      const y = j + stampWidth;
+      if (prefixSum[x][y] - prefixSum[x][j] - prefixSum[i][y] + prefixSum[i][j] === 0) {
+        diff[i][j]++;
+        diff[i][y]--;
+        diff[x][j]--;
+        diff[x][y]++;
+      }
+    }
+  }
+
+  const covered = new Array(rows).fill().map(() => new Array(cols).fill(0));
+  for (let i = 0; i < rows; i++) {
+    for (let j = 0; j < cols; j++) {
+      covered[i][j] = (i > 0 ? covered[i - 1][j] : 0)
+        + (j > 0 ? covered[i][j - 1] : 0)
+        - (i > 0 && j > 0 ? covered[i - 1][j - 1] : 0) + diff[i][j];
+      if (grid[i][j] === 0 && covered[i][j] === 0) return false;
+    }
+  }
+
+  return true;
+};
diff --git a/solutions/2133-check-if-every-row-and-column-contains-all-numbers.js b/solutions/2133-check-if-every-row-and-column-contains-all-numbers.js
new file mode 100644
index 00000000..e8d6990a
--- /dev/null
+++ b/solutions/2133-check-if-every-row-and-column-contains-all-numbers.js
@@ -0,0 +1,33 @@
+/**
+ * 2133. Check if Every Row and Column Contains All Numbers
+ * https://leetcode.com/problems/check-if-every-row-and-column-contains-all-numbers/
+ * Difficulty: Easy
+ *
+ * An n x n matrix is valid if every row and every column contains all the integers from 1
+ * to n (inclusive).
+ *
+ * Given an n x n integer matrix matrix, return true if the matrix is valid. Otherwise,
+ * return false.
+ */
+
+/**
+ * @param {number[][]} matrix
+ * @return {boolean}
+ */
+var checkValid = function(matrix) {
+  const n = matrix.length;
+
+  for (let i = 0; i < n; i++) {
+    const rowSet = new Set();
+    const colSet = new Set();
+
+    for (let j = 0; j < n; j++) {
+      rowSet.add(matrix[i][j]);
+      colSet.add(matrix[j][i]);
+    }
+
+    if (rowSet.size !== n || colSet.size !== n) return false;
+  }
+
+  return true;
+};
diff --git a/solutions/2134-minimum-swaps-to-group-all-1s-together-ii.js b/solutions/2134-minimum-swaps-to-group-all-1s-together-ii.js
new file mode 100644
index 00000000..1105dced
--- /dev/null
+++ b/solutions/2134-minimum-swaps-to-group-all-1s-together-ii.js
@@ -0,0 +1,38 @@
+/**
+ * 2134. Minimum Swaps to Group All 1's Together II
+ * https://leetcode.com/problems/minimum-swaps-to-group-all-1s-together-ii/
+ * Difficulty: Medium
+ *
+ * A swap is defined as taking two distinct positions in an array and swapping the values in them.
+ *
+ * A circular array is defined as an array where we consider the first element and the last element
+ * to be adjacent.
+ *
+ * Given a binary circular array nums, return the minimum number of swaps required to group all
+ * 1's present in the array together at any location.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var minSwaps = function(nums) {
+  const onesCount = nums.reduce((sum, num) => sum + num, 0);
+  const n = nums.length;
+  let result = Infinity;
+  let currentZeros = 0;
+
+  for (let i = 0; i < onesCount; i++) {
+    if (nums[i] === 0) currentZeros++;
+  }
+
+  result = currentZeros;
+
+  for (let i = 1; i < n; i++) {
+    if (nums[(i - 1) % n] === 0) currentZeros--;
+    if (nums[(i + onesCount - 1) % n] === 0) currentZeros++;
+    result = Math.min(result, currentZeros);
+  }
+
+  return result;
+};
diff --git a/solutions/2135-count-words-obtained-after-adding-a-letter.js b/solutions/2135-count-words-obtained-after-adding-a-letter.js
new file mode 100644
index 00000000..c7b08943
--- /dev/null
+++ b/solutions/2135-count-words-obtained-after-adding-a-letter.js
@@ -0,0 +1,49 @@
+/**
+ * 2135. Count Words Obtained After Adding a Letter
+ * https://leetcode.com/problems/count-words-obtained-after-adding-a-letter/
+ * Difficulty: Medium
+ *
+ * You are given two 0-indexed arrays of strings startWords and targetWords. Each string consists
+ * of lowercase English letters only.
+ *
+ * For each string in targetWords, check if it is possible to choose a string from startWords and
+ * perform a conversion operation on it to be equal to that from targetWords.
+ *
+ * The conversion operation is described in the following two steps:
+ * 1. Append any lowercase letter that is not present in the string to its end.
+ *    - For example, if the string is "abc", the letters 'd', 'e', or 'y' can be added to it, but
+ *      not 'a'. If 'd' is added, the resulting string will be "abcd".
+ * 2. Rearrange the letters of the new string in any arbitrary order.
+ *    - For example, "abcd" can be rearranged to "acbd", "bacd", "cbda", and so on. Note that it
+ *      can also be rearranged to "abcd" itself.
+ *
+ * Return the number of strings in targetWords that can be obtained by performing the operations
+ * on any string of startWords.
+ *
+ * Note that you will only be verifying if the string in targetWords can be obtained from a string
+ * in startWords by performing the operations. The strings in startWords do not actually change
+ * during this process.
+ */
+
+/**
+ * @param {string[]} startWords
+ * @param {string[]} targetWords
+ * @return {number}
+ */
+var wordCount = function(startWords, targetWords) {
+  const sortedStartWords = new Set(startWords.map(word => [...word].sort().join('')));
+  let result = 0;
+
+  for (const target of targetWords) {
+    const sortedTarget = [...target].sort().join('');
+    for (let i = 0; i < target.length; i++) {
+      const candidate = sortedTarget.slice(0, i) + sortedTarget.slice(i + 1);
+      if (sortedStartWords.has(candidate)) {
+        result++;
+        break;
+      }
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2136-earliest-possible-day-of-full-bloom.js b/solutions/2136-earliest-possible-day-of-full-bloom.js
new file mode 100644
index 00000000..e4a07b6c
--- /dev/null
+++ b/solutions/2136-earliest-possible-day-of-full-bloom.js
@@ -0,0 +1,39 @@
+/**
+ * 2136. Earliest Possible Day of Full Bloom
+ * https://leetcode.com/problems/earliest-possible-day-of-full-bloom/
+ * Difficulty: Hard
+ *
+ * You have n flower seeds. Every seed must be planted first before it can begin to grow, then
+ * bloom. Planting a seed takes time and so does the growth of a seed. You are given two 0-indexed
+ * integer arrays plantTime and growTime, of length n each:
+ * - plantTime[i] is the number of full days it takes you to plant the ith seed. Every day, you can
+ *   work on planting exactly one seed. You do not have to work on planting the same seed on
+ *   consecutive days, but the planting of a seed is not complete until you have worked plantTime[i]
+ *   days on planting it in total.
+ * - growTime[i] is the number of full days it takes the ith seed to grow after being completely
+ *   planted. After the last day of its growth, the flower blooms and stays bloomed forever.
+ *
+ * From the beginning of day 0, you can plant the seeds in any order.
+ *
+ * Return the earliest possible day where all seeds are blooming.
+ */
+
+/**
+ * @param {number[]} plantTime
+ * @param {number[]} growTime
+ * @return {number}
+ */
+var earliestFullBloom = function(plantTime, growTime) {
+  const seeds = plantTime.map((plant, index) => ({ plant, grow: growTime[index] }));
+  seeds.sort((a, b) => b.grow - a.grow);
+
+  let plantingDays = 0;
+  let result = 0;
+
+  for (const { plant, grow } of seeds) {
+    plantingDays += plant;
+    result = Math.max(result, plantingDays + grow);
+  }
+
+  return result;
+};
diff --git a/solutions/2138-divide-a-string-into-groups-of-size-k.js b/solutions/2138-divide-a-string-into-groups-of-size-k.js
new file mode 100644
index 00000000..24d0197f
--- /dev/null
+++ b/solutions/2138-divide-a-string-into-groups-of-size-k.js
@@ -0,0 +1,35 @@
+/**
+ * 2138. Divide a String Into Groups of Size k
+ * https://leetcode.com/problems/divide-a-string-into-groups-of-size-k/
+ * Difficulty: Easy
+ *
+ * A string s can be partitioned into groups of size k using the following procedure:
+ * - The first group consists of the first k characters of the string, the second group consists
+ *   of the next k characters of the string, and so on. Each element can be a part of exactly one
+ *   group.
+ * - For the last group, if the string does not have k characters remaining, a character fill is
+ *   used to complete the group.
+ *
+ * Note that the partition is done so that after removing the fill character from the last group
+ * (if it exists) and concatenating all the groups in order, the resultant string should be s.
+ *
+ * Given the string s, the size of each group k and the character fill, return a string array
+ * denoting the composition of every group s has been divided into, using the above procedure.
+ */
+
+/**
+ * @param {string} s
+ * @param {number} k
+ * @param {character} fill
+ * @return {string[]}
+ */
+var divideString = function(s, k, fill) {
+  const paddedString = s + fill.repeat((k - s.length % k) % k);
+  const result = [];
+
+  for (let i = 0; i < paddedString.length; i += k) {
+    result.push(paddedString.slice(i, i + k));
+  }
+
+  return result;
+};
diff --git a/solutions/2139-minimum-moves-to-reach-target-score.js b/solutions/2139-minimum-moves-to-reach-target-score.js
new file mode 100644
index 00000000..a074595f
--- /dev/null
+++ b/solutions/2139-minimum-moves-to-reach-target-score.js
@@ -0,0 +1,40 @@
+/**
+ * 2139. Minimum Moves to Reach Target Score
+ * https://leetcode.com/problems/minimum-moves-to-reach-target-score/
+ * Difficulty: Medium
+ *
+ * You are playing a game with integers. You start with the integer 1 and you want to reach
+ * the integer target.
+ *
+ * In one move, you can either:
+ * - Increment the current integer by one (i.e., x = x + 1).
+ * - Double the current integer (i.e., x = 2 * x).
+ *
+ * You can use the increment operation any number of times, however, you can only use the double
+ * operation at most maxDoubles times.
+ *
+ * Given the two integers target and maxDoubles, return the minimum number of moves needed to
+ * reach target starting with 1.
+ */
+
+/**
+ * @param {number} target
+ * @param {number} maxDoubles
+ * @return {number}
+ */
+var minMoves = function(target, maxDoubles) {
+  let moves = 0;
+  let current = target;
+
+  while (current > 1 && maxDoubles > 0) {
+    if (current % 2 === 0) {
+      current /= 2;
+      maxDoubles--;
+    } else {
+      current--;
+    }
+    moves++;
+  }
+
+  return moves + (current - 1);
+};
diff --git a/solutions/2141-maximum-running-time-of-n-computers.js b/solutions/2141-maximum-running-time-of-n-computers.js
new file mode 100644
index 00000000..b5f7b090
--- /dev/null
+++ b/solutions/2141-maximum-running-time-of-n-computers.js
@@ -0,0 +1,41 @@
+/**
+ * 2141. Maximum Running Time of N Computers
+ * https://leetcode.com/problems/maximum-running-time-of-n-computers/
+ * Difficulty: Hard
+ *
+ * You have n computers. You are given the integer n and a 0-indexed integer array batteries
+ * where the ith battery can run a computer for batteries[i] minutes. You are interested in
+ * running all n computers simultaneously using the given batteries.
+ *
+ * Initially, you can insert at most one battery into each computer. After that and at any
+ * integer time moment, you can remove a battery from a computer and insert another battery
+ * any number of times. The inserted battery can be a totally new battery or a battery from
+ * another computer. You may assume that the removing and inserting processes take no time.
+ *
+ * Note that the batteries cannot be recharged.
+ *
+ * Return the maximum number of minutes you can run all the n computers simultaneously.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[]} batteries
+ * @return {number}
+ */
+var maxRunTime = function(n, batteries) {
+  let left = 1;
+  let right = Math.floor(batteries.reduce((sum, battery) => sum + battery, 0) / n);
+
+  while (left < right) {
+    const mid = Math.floor((left + right + 1) / 2);
+    const total = batteries.reduce((sum, battery) => sum + Math.min(battery, mid), 0);
+
+    if (total >= n * mid) {
+      left = mid;
+    } else {
+      right = mid - 1;
+    }
+  }
+
+  return left;
+};
diff --git a/solutions/2144-minimum-cost-of-buying-candies-with-discount.js b/solutions/2144-minimum-cost-of-buying-candies-with-discount.js
new file mode 100644
index 00000000..86eca1e2
--- /dev/null
+++ b/solutions/2144-minimum-cost-of-buying-candies-with-discount.js
@@ -0,0 +1,33 @@
+/**
+ * 2144. Minimum Cost of Buying Candies With Discount
+ * https://leetcode.com/problems/minimum-cost-of-buying-candies-with-discount/
+ * Difficulty: Easy
+ *
+ * A shop is selling candies at a discount. For every two candies sold, the shop gives a
+ * third candy for free.
+ *
+ * The customer can choose any candy to take away for free as long as the cost of the
+ * chosen candy is less than or equal to the minimum cost of the two candies bought.
+ *
+ * - For example, if there are 4 candies with costs 1, 2, 3, and 4, and the customer buys
+ *   candies with costs 2 and 3, they can take the candy with cost 1 for free, but not the
+ *   candy with cost 4.
+ *
+ * Given a 0-indexed integer array cost, where cost[i] denotes the cost of the ith candy,
+ * return the minimum cost of buying all the candies.
+ */
+
+/**
+ * @param {number[]} cost
+ * @return {number}
+ */
+var minimumCost = function(cost) {
+  cost.sort((a, b) => b - a);
+  let result = 0;
+
+  for (let i = 0; i < cost.length; i += 3) {
+    result += cost[i] + (cost[i + 1] || 0);
+  }
+
+  return result;
+};
diff --git a/solutions/2147-number-of-ways-to-divide-a-long-corridor.js b/solutions/2147-number-of-ways-to-divide-a-long-corridor.js
new file mode 100644
index 00000000..e504b765
--- /dev/null
+++ b/solutions/2147-number-of-ways-to-divide-a-long-corridor.js
@@ -0,0 +1,58 @@
+/**
+ * 2147. Number of Ways to Divide a Long Corridor
+ * https://leetcode.com/problems/number-of-ways-to-divide-a-long-corridor/
+ * Difficulty: Hard
+ *
+ * Along a long library corridor, there is a line of seats and decorative plants. You are given
+ * a 0-indexed string corridor of length n consisting of letters 'S' and 'P' where each 'S'
+ * represents a seat and each 'P' represents a plant.
+ *
+ * One room divider has already been installed to the left of index 0, and another to the right
+ * of index n - 1. Additional room dividers can be installed. For each position between indices
+ * i - 1 and i (1 <= i <= n - 1), at most one divider can be installed.
+ *
+ * Divide the corridor into non-overlapping sections, where each section has exactly two seats
+ * with any number of plants. There may be multiple ways to perform the division. Two ways are
+ * different if there is a position with a room divider installed in the first way but not in
+ * the second way.
+ *
+ * Return the number of ways to divide the corridor. Since the answer may be very large, return
+ * it modulo 109 + 7. If there is no way, return 0.
+ */
+
+/**
+ * @param {string} corridor
+ * @return {number}
+ */
+var numberOfWays = function(corridor) {
+  const MOD = 1e9 + 7;
+  let seatCount = 0;
+  let result = 1;
+  let lastPairEnd = -1;
+
+  for (let i = 0; i < corridor.length; i++) {
+    if (corridor[i] === 'S') {
+      seatCount++;
+    }
+  }
+  if (seatCount === 0 || seatCount % 2 !== 0) {
+    return 0;
+  }
+
+  seatCount = 0;
+
+  for (let i = 0; i < corridor.length; i++) {
+    if (corridor[i] === 'S') {
+      seatCount++;
+
+      if (seatCount % 2 === 0) {
+        lastPairEnd = i;
+      } else if (seatCount > 1) {
+        const plantsCount = i - lastPairEnd - 1;
+        result = (result * (plantsCount + 1)) % MOD;
+      }
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2148-count-elements-with-strictly-smaller-and-greater-elements.js b/solutions/2148-count-elements-with-strictly-smaller-and-greater-elements.js
new file mode 100644
index 00000000..194880af
--- /dev/null
+++ b/solutions/2148-count-elements-with-strictly-smaller-and-greater-elements.js
@@ -0,0 +1,25 @@
+/**
+ * 2148. Count Elements With Strictly Smaller and Greater Elements
+ * https://leetcode.com/problems/count-elements-with-strictly-smaller-and-greater-elements/
+ * Difficulty: Easy
+ *
+ * Given an integer array nums, return the number of elements that have both a strictly
+ * smaller and a strictly greater element appear in nums.
+ */
+
+/**
+* @param {number[]} nums
+* @return {number}
+*/
+var countElements = function(nums) {
+  let result = 0;
+
+  if (nums.length <= 2) return 0;
+  for (const num of nums) {
+    if (num > Math.min(...nums) && num < Math.max(...nums)) {
+      result++;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2149-rearrange-array-elements-by-sign.js b/solutions/2149-rearrange-array-elements-by-sign.js
new file mode 100644
index 00000000..7979e5ef
--- /dev/null
+++ b/solutions/2149-rearrange-array-elements-by-sign.js
@@ -0,0 +1,33 @@
+/**
+ * 2149. Rearrange Array Elements by Sign
+ * https://leetcode.com/problems/rearrange-array-elements-by-sign/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed integer array nums of even length consisting of an equal number
+ * of positive and negative integers.
+ *
+ * You should return the array of nums such that the the array follows the given conditions:
+ * 1. Every consecutive pair of integers have opposite signs.
+ * 2. For all integers with the same sign, the order in which they were present in nums is
+ *    preserved.
+ * 3. The rearranged array begins with a positive integer.
+ *
+ * Return the modified array after rearranging the elements to satisfy the aforementioned
+ * conditions.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var rearrangeArray = function(nums) {
+  const positives = nums.filter(num => num > 0);
+  const negatives = nums.filter(num => num < 0);
+  const result = [];
+
+  for (let i = 0; i < positives.length; i++) {
+    result.push(positives[i], negatives[i]);
+  }
+
+  return result;
+};
diff --git a/solutions/2150-find-all-lonely-numbers-in-the-array.js b/solutions/2150-find-all-lonely-numbers-in-the-array.js
new file mode 100644
index 00000000..676badc6
--- /dev/null
+++ b/solutions/2150-find-all-lonely-numbers-in-the-array.js
@@ -0,0 +1,31 @@
+/**
+ * 2150. Find All Lonely Numbers in the Array
+ * https://leetcode.com/problems/find-all-lonely-numbers-in-the-array/
+ * Difficulty: Medium
+ *
+ * You are given an integer array nums. A number x is lonely when it appears only once, and
+ * no adjacent numbers (i.e. x + 1 and x - 1) appear in the array.
+ *
+ * Return all lonely numbers in nums. You may return the answer in any order.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var findLonely = function(nums) {
+  const map = new Map();
+
+  for (const num of nums) {
+    map.set(num, (map.get(num) || 0) + 1);
+  }
+
+  const result = [];
+  for (const [num, count] of map) {
+    if (count === 1 && !map.has(num - 1) && !map.has(num + 1)) {
+      result.push(num);
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2151-maximum-good-people-based-on-statements.js b/solutions/2151-maximum-good-people-based-on-statements.js
new file mode 100644
index 00000000..9a5d71cd
--- /dev/null
+++ b/solutions/2151-maximum-good-people-based-on-statements.js
@@ -0,0 +1,59 @@
+/**
+ * 2151. Maximum Good People Based on Statements
+ * https://leetcode.com/problems/maximum-good-people-based-on-statements/
+ * Difficulty: Hard
+ *
+ * There are two types of persons:
+ * - The good person: The person who always tells the truth.
+ * - The bad person: The person who might tell the truth and might lie.
+ *
+ * You are given a 0-indexed 2D integer array statements of size n x n that represents the
+ * statements made by n people about each other. More specifically, statements[i][j] could
+ * be one of the following:
+ * - 0 which represents a statement made by person i that person j is a bad person.
+ * - 1 which represents a statement made by person i that person j is a good person.
+ * - 2 represents that no statement is made by person i about person j.
+ *
+ * Additionally, no person ever makes a statement about themselves. Formally, we have that
+ * statements[i][i] = 2 for all 0 <= i < n.
+ *
+ * Return the maximum number of people who can be good based on the statements made by the
+ * n people.
+ */
+
+/**
+ * @param {number[][]} statements
+ * @return {number}
+ */
+var maximumGood = function(statements) {
+  const n = statements.length;
+  let result = 0;
+
+  backtrack(0, new Array(n).fill(false), 0);
+  return result;
+
+  function isValid(goodPeople) {
+    for (let i = 0; i < n; i++) {
+      if (!goodPeople[i]) continue;
+      for (let j = 0; j < n; j++) {
+        if (statements[i][j] === 0 && goodPeople[j]) return false;
+        if (statements[i][j] === 1 && !goodPeople[j]) return false;
+      }
+    }
+    return true;
+  }
+
+  function backtrack(index, goodPeople, goodCount) {
+    if (index === n) {
+      if (isValid(goodPeople)) {
+        result = Math.max(result, goodCount);
+      }
+      return;
+    }
+
+    goodPeople[index] = true;
+    backtrack(index + 1, goodPeople, goodCount + 1);
+    goodPeople[index] = false;
+    backtrack(index + 1, goodPeople, goodCount);
+  }
+};
diff --git a/solutions/2155-all-divisions-with-the-highest-score-of-a-binary-array.js b/solutions/2155-all-divisions-with-the-highest-score-of-a-binary-array.js
new file mode 100644
index 00000000..66db5b2f
--- /dev/null
+++ b/solutions/2155-all-divisions-with-the-highest-score-of-a-binary-array.js
@@ -0,0 +1,45 @@
+/**
+ * 2155. All Divisions With the Highest Score of a Binary Array
+ * https://leetcode.com/problems/all-divisions-with-the-highest-score-of-a-binary-array/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed binary array nums of length n. nums can be divided at index i
+ * (where 0 <= i <= n) into two arrays (possibly empty) numsleft and numsright:
+ * - numsleft has all the elements of nums between index 0 and i - 1 (inclusive), while numsright
+ *   has all the elements of nums between index i and n - 1 (inclusive).
+ * - If i == 0, numsleft is empty, while numsright has all the elements of nums.
+ * - If i == n, numsleft has all the elements of nums, while numsright is empty.
+ *
+ * The division score of an index i is the sum of the number of 0's in numsleft and the number
+ * of 1's in numsright.
+ *
+ * Return all distinct indices that have the highest possible division score. You may return
+ * the answer in any order.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var maxScoreIndices = function(nums) {
+  const result = [0];
+  let leftZeros = 0;
+  let rightOnes = nums.reduce((sum, num) => sum + num, 0);
+  let maxScore = rightOnes;
+
+  for (let i = 0; i < nums.length; i++) {
+    leftZeros += nums[i] === 0 ? 1 : 0;
+    rightOnes -= nums[i];
+    const score = leftZeros + rightOnes;
+
+    if (score > maxScore) {
+      maxScore = score;
+      result.length = 0;
+      result.push(i + 1);
+    } else if (score === maxScore) {
+      result.push(i + 1);
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2156-find-substring-with-given-hash-value.js b/solutions/2156-find-substring-with-given-hash-value.js
new file mode 100644
index 00000000..a43f4a10
--- /dev/null
+++ b/solutions/2156-find-substring-with-given-hash-value.js
@@ -0,0 +1,51 @@
+/**
+ * 2156. Find Substring With Given Hash Value
+ * https://leetcode.com/problems/find-substring-with-given-hash-value/
+ * Difficulty: Hard
+ *
+ * The hash of a 0-indexed string s of length k, given integers p and m, is computed using
+ * the following function:
+ * - hash(s, p, m) = (val(s[0]) * p0 + val(s[1]) * p1 + ... + val(s[k-1]) * pk-1) mod m.
+ *
+ * Where val(s[i]) represents the index of s[i] in the alphabet from val('a') = 1 to val('z') = 26.
+ *
+ * You are given a string s and the integers power, modulo, k, and hashValue. Return sub, the
+ * first substring of s of length k such that hash(sub, power, modulo) == hashValue.
+ *
+ * The test cases will be generated such that an answer always exists.
+ *
+ * A substring is a contiguous non-empty sequence of characters within a string.
+ */
+
+/**
+* @param {string} s
+* @param {number} power
+* @param {number} modulo
+* @param {number} k
+* @param {number} hashValue
+* @return {string}
+*/
+var subStrHash = function(s, power, modulo, k, hashValue) {
+  power = BigInt(power);
+  modulo = BigInt(modulo);
+  hashValue = BigInt(hashValue);
+
+  const powers = new Array(k);
+  powers[0] = 1n;
+  for (let i = 1; i < k; i++) {
+    powers[i] = (powers[i - 1] * power) % modulo;
+  }
+
+  for (let start = 0; start <= s.length - k; start++) {
+    let hash = 0n;
+    for (let i = 0; i < k; i++) {
+      const charVal = BigInt(s.charCodeAt(start + i) - 'a'.charCodeAt(0) + 1);
+      hash = (hash + charVal * powers[i]) % modulo;
+    }
+    if (hash === hashValue) {
+      return s.substring(start, start + k);
+    }
+  }
+
+  return '';
+};
diff --git a/solutions/2157-groups-of-strings.js b/solutions/2157-groups-of-strings.js
new file mode 100644
index 00000000..8af51c12
--- /dev/null
+++ b/solutions/2157-groups-of-strings.js
@@ -0,0 +1,140 @@
+/**
+ * 2157. Groups of Strings
+ * https://leetcode.com/problems/groups-of-strings/
+ * Difficulty: Hard
+ *
+ * You are given a 0-indexed array of strings words. Each string consists of lowercase English
+ * letters only. No letter occurs more than once in any string of words.
+ *
+ * Two strings s1 and s2 are said to be connected if the set of letters of s2 can be obtained
+ * from the set of letters of s1 by any one of the following operations:
+ * - Adding exactly one letter to the set of the letters of s1.
+ * - Deleting exactly one letter from the set of the letters of s1.
+ * - Replacing exactly one letter from the set of the letters of s1 with any letter, including
+ *   itself.
+ *
+ * The array words can be divided into one or more non-intersecting groups. A string belongs
+ * to a group if any one of the following is true:
+ * - It is connected to at least one other string of the group.
+ * - It is the only string present in the group.
+ *
+ * Note that the strings in words should be grouped in such a manner that a string belonging
+ * to a group cannot be connected to a string present in any other group. It can be proved
+ * that such an arrangement is always unique.
+ *
+ * Return an array ans of size 2 where:
+ * - ans[0] is the maximum number of groups words can be divided into, and
+ * - ans[1] is the size of the largest group.
+ */
+
+/**
+ * @param {string[]} words
+ * @return {number[]}
+ */
+var groupStrings = function(words) {
+  const n = words.length;
+  const masks = words.map(word => {
+    let mask = 0;
+    for (let i = 0; i < word.length; i++) {
+      mask |= (1 << (word.charCodeAt(i) - 'a'.charCodeAt(0)));
+    }
+    return mask;
+  });
+  const uf = new UnionFind(n);
+  const maskToIndex = new Map();
+  for (let i = 0; i < n; i++) {
+    if (maskToIndex.has(masks[i])) {
+      uf.union(maskToIndex.get(masks[i]), i);
+    } else {
+      maskToIndex.set(masks[i], i);
+    }
+  }
+  const processed = new Set();
+
+  for (let i = 0; i < n; i++) {
+    const mask = masks[i];
+    if (processed.has(mask)) continue;
+    processed.add(mask);
+
+    for (let bit = 0; bit < 26; bit++) {
+      if ((mask & (1 << bit)) === 0) {
+        const newMask = mask | (1 << bit);
+        if (maskToIndex.has(newMask)) {
+          uf.union(i, maskToIndex.get(newMask));
+        }
+      }
+    }
+
+    for (let bit = 0; bit < 26; bit++) {
+      if ((mask & (1 << bit)) !== 0) {
+        const newMask = mask & ~(1 << bit);
+        if (maskToIndex.has(newMask)) {
+          uf.union(i, maskToIndex.get(newMask));
+        }
+      }
+    }
+
+    for (let remove = 0; remove < 26; remove++) {
+      if ((mask & (1 << remove)) !== 0) {
+        for (let add = 0; add < 26; add++) {
+          if ((mask & (1 << add)) === 0) {
+            const newMask = (mask & ~(1 << remove)) | (1 << add);
+            if (maskToIndex.has(newMask)) {
+              uf.union(i, maskToIndex.get(newMask));
+            }
+          }
+        }
+      }
+    }
+  }
+
+  return [uf.count, uf.getMaxSize()];
+};
+
+class UnionFind {
+  constructor(n) {
+    this.parent = Array(n).fill().map((_, i) => i);
+    this.rank = Array(n).fill(0);
+    this.count = n;
+    this.sizes = Array(n).fill(1);
+  }
+
+  find(x) {
+    if (this.parent[x] !== x) {
+      this.parent[x] = this.find(this.parent[x]);
+    }
+    return this.parent[x];
+  }
+
+  union(x, y) {
+    const rootX = this.find(x);
+    const rootY = this.find(y);
+
+    if (rootX === rootY) return false;
+
+    if (this.rank[rootX] < this.rank[rootY]) {
+      this.parent[rootX] = rootY;
+      this.sizes[rootY] += this.sizes[rootX];
+    } else if (this.rank[rootX] > this.rank[rootY]) {
+      this.parent[rootY] = rootX;
+      this.sizes[rootX] += this.sizes[rootY];
+    } else {
+      this.parent[rootY] = rootX;
+      this.rank[rootX]++;
+      this.sizes[rootX] += this.sizes[rootY];
+    }
+
+    this.count--;
+    return true;
+  }
+
+  getMaxSize() {
+    let maxSize = 0;
+    for (let i = 0; i < this.parent.length; i++) {
+      if (this.parent[i] === i) {
+        maxSize = Math.max(maxSize, this.sizes[i]);
+      }
+    }
+    return maxSize;
+  }
+}
diff --git a/solutions/2160-minimum-sum-of-four-digit-number-after-splitting-digits.js b/solutions/2160-minimum-sum-of-four-digit-number-after-splitting-digits.js
new file mode 100644
index 00000000..fa3b9b86
--- /dev/null
+++ b/solutions/2160-minimum-sum-of-four-digit-number-after-splitting-digits.js
@@ -0,0 +1,22 @@
+/**
+ * 2160. Minimum Sum of Four Digit Number After Splitting Digits
+ * https://leetcode.com/problems/minimum-sum-of-four-digit-number-after-splitting-digits/
+ * Difficulty: Easy
+ *
+ * You are given a positive integer num consisting of exactly four digits. Split num into two
+ * new integers new1 and new2 by using the digits found in num. Leading zeros are allowed in
+ * new1 and new2, and all the digits found in num must be used.
+ * - For example, given num = 2932, you have the following digits: two 2's, one 9 and one 3.
+ *   Some of the possible pairs [new1, new2] are [22, 93], [23, 92], [223, 9] and [2, 329].
+ *
+ * Return the minimum possible sum of new1 and new2.
+ */
+
+/**
+ * @param {number} num
+ * @return {number}
+ */
+var minimumSum = function(num) {
+  const digits = String(num).split('').map(Number).sort((a, b) => a - b);
+  return (digits[0] * 10 + digits[2]) + (digits[1] * 10 + digits[3]);
+};
diff --git a/solutions/2162-minimum-cost-to-set-cooking-time.js b/solutions/2162-minimum-cost-to-set-cooking-time.js
new file mode 100644
index 00000000..e1c51c1b
--- /dev/null
+++ b/solutions/2162-minimum-cost-to-set-cooking-time.js
@@ -0,0 +1,90 @@
+/**
+ * 2162. Minimum Cost to Set Cooking Time
+ * https://leetcode.com/problems/minimum-cost-to-set-cooking-time/
+ * Difficulty: Medium
+ *
+ * A generic microwave supports cooking times for:
+ * - at least 1 second.
+ * - at most 99 minutes and 99 seconds.
+ *
+ * To set the cooking time, you push at most four digits. The microwave normalizes what you push
+ * as four digits by prepending zeroes. It interprets the first two digits as the minutes and
+ * the last two digits as the seconds. It then adds them up as the cooking time. For example,
+ * - You push 9 5 4 (three digits). It is normalized as 0954 and interpreted as 9 minutes and
+ *   54 seconds.
+ * - You push 0 0 0 8 (four digits). It is interpreted as 0 minutes and 8 seconds.
+ * - You push 8 0 9 0. It is interpreted as 80 minutes and 90 seconds.
+ * - You push 8 1 3 0. It is interpreted as 81 minutes and 30 seconds.
+ *
+ * You are given integers startAt, moveCost, pushCost, and targetSeconds. Initially, your finger
+ * is on the digit startAt. Moving the finger above any specific digit costs moveCost units of
+ * fatigue. Pushing the digit below the finger once costs pushCost units of fatigue.
+ *
+ * There can be multiple ways to set the microwave to cook for targetSeconds seconds but you are
+ * interested in the way with the minimum cost.
+ *
+ * Return the minimum cost to set targetSeconds seconds of cooking time.
+ *
+ * Remember that one minute consists of 60 seconds.
+ */
+
+/**
+ * @param {number} startAt
+ * @param {number} moveCost
+ * @param {number} pushCost
+ * @param {number} targetSeconds
+ * @return {number}
+ */
+var minCostSetTime = function(startAt, moveCost, pushCost, targetSeconds) {
+  const minutes1 = Math.floor(targetSeconds / 60);
+  const seconds1 = targetSeconds % 60;
+  let result = Infinity;
+
+  if (minutes1 <= 99) {
+    const digits1 = getDigits(minutes1, seconds1);
+    result = Math.min(result, calculateCost(digits1));
+  }
+
+  const minutes2 = Math.floor(targetSeconds / 60) - 1;
+  const seconds2 = targetSeconds % 60 + 60;
+  if (minutes2 >= 0 && minutes2 <= 99 && seconds2 <= 99) {
+    const digits2 = getDigits(minutes2, seconds2);
+    result = Math.min(result, calculateCost(digits2));
+  }
+
+  return result;
+
+  function calculateCost(digits) {
+    let totalCost = 0;
+    let currentDigit = startAt;
+
+    for (const digit of digits) {
+      if (digit !== currentDigit) {
+        totalCost += moveCost;
+        currentDigit = digit;
+      }
+      totalCost += pushCost;
+    }
+
+    return totalCost;
+  }
+
+  function getDigits(minutes, seconds) {
+    const result = [];
+
+    if (minutes > 0) {
+      if (minutes >= 10) {
+        result.push(Math.floor(minutes / 10));
+      }
+      result.push(minutes % 10);
+    }
+
+    if (minutes > 0 || seconds >= 10) {
+      result.push(Math.floor(seconds / 10));
+    }
+
+    result.push(seconds % 10);
+
+    return result;
+  }
+};
diff --git a/solutions/2164-sort-even-and-odd-indices-independently.js b/solutions/2164-sort-even-and-odd-indices-independently.js
new file mode 100644
index 00000000..ed9322c6
--- /dev/null
+++ b/solutions/2164-sort-even-and-odd-indices-independently.js
@@ -0,0 +1,50 @@
+/**
+ * 2164. Sort Even and Odd Indices Independently
+ * https://leetcode.com/problems/sort-even-and-odd-indices-independently/
+ * Difficulty: Easy
+ *
+ * You are given a 0-indexed integer array nums. Rearrange the values of nums according to the
+ * following rules:
+ * 1. Sort the values at odd indices of nums in non-increasing order.
+ *    - For example, if nums = [4,1,2,3] before this step, it becomes [4,3,2,1] after. The values
+ *      at odd indices 1 and 3 are sorted in non-increasing order.
+ * 2. Sort the values at even indices of nums in non-decreasing order.
+ *    - For example, if nums = [4,1,2,3] before this step, it becomes [2,1,4,3] after. The values
+ *      at even indices 0 and 2 are sorted in non-decreasing order.
+ *
+ * Return the array formed after rearranging the values of nums.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var sortEvenOdd = function(nums) {
+  const evens = [];
+  const odds = [];
+
+  for (let i = 0; i < nums.length; i++) {
+    if (i % 2 === 0) {
+      evens.push(nums[i]);
+    } else {
+      odds.push(nums[i]);
+    }
+  }
+
+  evens.sort((a, b) => a - b);
+  odds.sort((a, b) => b - a);
+
+  const result = [];
+  let evenIndex = 0;
+  let oddIndex = 0;
+
+  for (let i = 0; i < nums.length; i++) {
+    if (i % 2 === 0) {
+      result.push(evens[evenIndex++]);
+    } else {
+      result.push(odds[oddIndex++]);
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2165-smallest-value-of-the-rearranged-number.js b/solutions/2165-smallest-value-of-the-rearranged-number.js
new file mode 100644
index 00000000..c3226062
--- /dev/null
+++ b/solutions/2165-smallest-value-of-the-rearranged-number.js
@@ -0,0 +1,34 @@
+/**
+ * 2165. Smallest Value of the Rearranged Number
+ * https://leetcode.com/problems/smallest-value-of-the-rearranged-number/
+ * Difficulty: Medium
+ *
+ * You are given an integer num. Rearrange the digits of num such that its value is minimized
+ * and it does not contain any leading zeros.
+ *
+ * Return the rearranged number with minimal value.
+ *
+ * Note that the sign of the number does not change after rearranging the digits.
+ */
+
+/**
+ * @param {number} num
+ * @return {number}
+ */
+var smallestNumber = function(num) {
+  const isNegative = num < 0;
+  const digits = Math.abs(num).toString().split('').map(Number);
+
+  if (isNegative) {
+    digits.sort((a, b) => b - a);
+    return -parseInt(digits.join(''), 10);
+  }
+
+  digits.sort((a, b) => a - b);
+  const firstNonZero = digits.findIndex(d => d !== 0);
+
+  if (firstNonZero === -1) return 0;
+
+  [digits[0], digits[firstNonZero]] = [digits[firstNonZero], digits[0]];
+  return parseInt(digits.join(''), 10);
+};
diff --git a/solutions/2166-design-bitset.js b/solutions/2166-design-bitset.js
new file mode 100644
index 00000000..155fe39b
--- /dev/null
+++ b/solutions/2166-design-bitset.js
@@ -0,0 +1,109 @@
+/**
+ * 2166. Design Bitset
+ * https://leetcode.com/problems/design-bitset/
+ * Difficulty: Medium
+ *
+ * A Bitset is a data structure that compactly stores bits.
+ *
+ * Implement the Bitset class:
+ * - Bitset(int size) Initializes the Bitset with size bits, all of which are 0.
+ * - void fix(int idx) Updates the value of the bit at the index idx to 1. If the value was
+ *   already 1, no change occurs.
+ * - void unfix(int idx) Updates the value of the bit at the index idx to 0. If the value
+ *   was already 0, no change occurs.
+ * - void flip() Flips the values of each bit in the Bitset. In other words, all bits with
+ *   value 0 will now have value 1 and vice versa.
+ * - boolean all() Checks if the value of each bit in the Bitset is 1. Returns true if it
+ *   satisfies the condition, false otherwise.
+ * - boolean one() Checks if there is at least one bit in the Bitset with value 1. Returns
+ *   true if it satisfies the condition, false otherwise.
+ * - int count() Returns the total number of bits in the Bitset which have value 1.
+ * - String toString() Returns the current composition of the Bitset. Note that in the
+ *   resultant string, the character at the ith index should coincide with the value at
+ *   the ith bit of the Bitset.
+ */
+
+/**
+ * @param {number} size
+ */
+var Bitset = function(size) {
+  this.bits = new Uint8Array(size);
+  this.ones = 0;
+  this.flipped = false;
+};
+
+/**
+ * @param {number} idx
+ * @return {void}
+ */
+Bitset.prototype.fix = function(idx) {
+  if (this.flipped) {
+    if (this.bits[idx] === 1) {
+      this.bits[idx] = 0;
+      this.ones++;
+    }
+  } else {
+    if (this.bits[idx] === 0) {
+      this.bits[idx] = 1;
+      this.ones++;
+    }
+  }
+};
+
+/**
+ * @param {number} idx
+ * @return {void}
+ */
+Bitset.prototype.unfix = function(idx) {
+  if (this.flipped) {
+    if (this.bits[idx] === 0) {
+      this.bits[idx] = 1;
+      this.ones--;
+    }
+  } else {
+    if (this.bits[idx] === 1) {
+      this.bits[idx] = 0;
+      this.ones--;
+    }
+  }
+};
+
+/**
+ * @return {void}
+ */
+Bitset.prototype.flip = function() {
+  this.flipped = !this.flipped;
+  this.ones = this.bits.length - this.ones;
+};
+
+/**
+ * @return {boolean}
+ */
+Bitset.prototype.all = function() {
+  return this.ones === this.bits.length;
+};
+
+/**
+ * @return {boolean}
+ */
+Bitset.prototype.one = function() {
+  return this.ones > 0;
+};
+
+/**
+ * @return {number}
+ */
+Bitset.prototype.count = function() {
+  return this.ones;
+};
+
+/**
+ * @return {string}
+ */
+Bitset.prototype.toString = function() {
+  let result = '';
+  for (let i = 0; i < this.bits.length; i++) {
+    result += this.flipped ? 1 - this.bits[i] : this.bits[i];
+  }
+  return result;
+};
diff --git a/solutions/2167-minimum-time-to-remove-all-cars-containing-illegal-goods.js b/solutions/2167-minimum-time-to-remove-all-cars-containing-illegal-goods.js
new file mode 100644
index 00000000..4877a0de
--- /dev/null
+++ b/solutions/2167-minimum-time-to-remove-all-cars-containing-illegal-goods.js
@@ -0,0 +1,37 @@
+/**
+ * 2167. Minimum Time to Remove All Cars Containing Illegal Goods
+ * https://leetcode.com/problems/minimum-time-to-remove-all-cars-containing-illegal-goods/
+ * Difficulty: Hard
+ *
+ * You are given a 0-indexed binary string s which represents a sequence of train cars. s[i] = '0'
+ * denotes that the ith car does not contain illegal goods and s[i] = '1' denotes that the ith car
+ * does contain illegal goods.
+ *
+ * As the train conductor, you would like to get rid of all the cars containing illegal goods.
+ * You can do any of the following three operations any number of times:
+ * 1. Remove a train car from the left end (i.e., remove s[0]) which takes 1 unit of time.
+ * 2. Remove a train car from the right end (i.e., remove s[s.length - 1]) which takes 1 unit
+ *    of time.
+ * 3. Remove a train car from anywhere in the sequence which takes 2 units of time.
+ *
+ * Return the minimum time to remove all the cars containing illegal goods.
+ *
+ * Note that an empty sequence of cars is considered to have no cars containing illegal goods.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var minimumTime = function(s) {
+  let leftCost = 0;
+  let result = s.length;
+
+  for (let i = 0; i < s.length; i++) {
+    leftCost = Math.min(leftCost + (s[i] === '1' ? 2 : 0), i + 1);
+    const rightCost = s.length - i - 1;
+    result = Math.min(result, leftCost + rightCost);
+  }
+
+  return result;
+};
diff --git a/solutions/2169-count-operations-to-obtain-zero.js b/solutions/2169-count-operations-to-obtain-zero.js
new file mode 100644
index 00000000..428bd42a
--- /dev/null
+++ b/solutions/2169-count-operations-to-obtain-zero.js
@@ -0,0 +1,34 @@
+/**
+ * 2169. Count Operations to Obtain Zero
+ * https://leetcode.com/problems/count-operations-to-obtain-zero/
+ * Difficulty: Easy
+ *
+ * You are given two non-negative integers num1 and num2.
+ *
+ * In one operation, if num1 >= num2, you must subtract num2 from num1, otherwise subtract num1
+ * from num2.
+ * - For example, if num1 = 5 and num2 = 4, subtract num2 from num1, thus obtaining num1 = 1 and
+ *   num2 = 4. However, if num1 = 4 and num2 = 5, after one operation, num1 = 4 and num2 = 1.
+ *
+ * Return the number of operations required to make either num1 = 0 or num2 = 0.
+ */
+
+/**
+ * @param {number} num1
+ * @param {number} num2
+ * @return {number}
+ */
+var countOperations = function(num1, num2) {
+  let operations = 0;
+
+  while (num1 !== 0 && num2 !== 0) {
+    if (num1 >= num2) {
+      num1 -= num2;
+    } else {
+      num2 -= num1;
+    }
+    operations++;
+  }
+
+  return operations;
+};
diff --git a/solutions/2170-minimum-operations-to-make-the-array-alternating.js b/solutions/2170-minimum-operations-to-make-the-array-alternating.js
new file mode 100644
index 00000000..770b6d62
--- /dev/null
+++ b/solutions/2170-minimum-operations-to-make-the-array-alternating.js
@@ -0,0 +1,49 @@
+/**
+ * 2170. Minimum Operations to Make the Array Alternating
+ * https://leetcode.com/problems/minimum-operations-to-make-the-array-alternating/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed array nums consisting of n positive integers.
+ *
+ * The array nums is called alternating if:
+ * - nums[i - 2] == nums[i], where 2 <= i <= n - 1.
+ * - nums[i - 1] != nums[i], where 1 <= i <= n - 1.
+ *
+ * In one operation, you can choose an index i and change nums[i] into any positive integer.
+ *
+ * Return the minimum number of operations required to make the array alternating.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var minimumOperations = function(nums) {
+  const evenFreq = new Map();
+  const oddFreq = new Map();
+
+  if (nums.length <= 1) return 0;
+  for (let i = 0; i < nums.length; i++) {
+    if (i % 2 === 0) {
+      evenFreq.set(nums[i], (evenFreq.get(nums[i]) || 0) + 1);
+    } else {
+      oddFreq.set(nums[i], (oddFreq.get(nums[i]) || 0) + 1);
+    }
+  }
+
+  const evenTop = [...evenFreq.entries()].sort((a, b) => b[1] - a[1]).slice(0, 2);
+  const oddTop = [...oddFreq.entries()].sort((a, b) => b[1] - a[1]).slice(0, 2);
+  const evenTotal = Math.ceil(nums.length / 2);
+  const oddTotal = Math.floor(nums.length / 2);
+  let minOps = nums.length;
+
+  for (const [evenNum, evenCount] of evenTop.length ? evenTop : [[0, 0]]) {
+    for (const [oddNum, oddCount] of oddTop.length ? oddTop : [[0, 0]]) {
+      if (evenNum !== oddNum || evenNum === 0) {
+        minOps = Math.min(minOps, (evenTotal - evenCount) + (oddTotal - oddCount));
+      }
+    }
+  }
+
+  return minOps === nums.length ? Math.min(evenTotal, oddTotal) : minOps;
+};
diff --git a/solutions/2171-removing-minimum-number-of-magic-beans.js b/solutions/2171-removing-minimum-number-of-magic-beans.js
new file mode 100644
index 00000000..23fd2236
--- /dev/null
+++ b/solutions/2171-removing-minimum-number-of-magic-beans.js
@@ -0,0 +1,33 @@
+/**
+ * 2171. Removing Minimum Number of Magic Beans
+ * https://leetcode.com/problems/removing-minimum-number-of-magic-beans/
+ * Difficulty: Medium
+ *
+ * You are given an array of positive integers beans, where each integer represents the number
+ * of magic beans found in a particular magic bag.
+ *
+ * Remove any number of beans (possibly none) from each bag such that the number of beans in
+ * each remaining non-empty bag (still containing at least one bean) is equal. Once a bean
+ * has been removed from a bag, you are not allowed to return it to any of the bags.
+ *
+ * Return the minimum number of magic beans that you have to remove.
+ */
+
+/**
+ * @param {number[]} beans
+ * @return {number}
+ */
+var minimumRemoval = function(beans) {
+  const sortedBeans = beans.sort((a, b) => a - b);
+  const totalBeans = sortedBeans.reduce((sum, bean) => sum + bean, 0);
+  let result = totalBeans;
+  let remaining = totalBeans;
+
+  for (let i = 0; i < sortedBeans.length; i++) {
+    remaining -= sortedBeans[i];
+    const equalBags = sortedBeans.length - i;
+    result = Math.min(result, totalBeans - sortedBeans[i] * equalBags);
+  }
+
+  return result;
+};
diff --git a/solutions/2172-maximum-and-sum-of-array.js b/solutions/2172-maximum-and-sum-of-array.js
new file mode 100644
index 00000000..6c9020d6
--- /dev/null
+++ b/solutions/2172-maximum-and-sum-of-array.js
@@ -0,0 +1,46 @@
+/**
+ * 2172. Maximum AND Sum of Array
+ * https://leetcode.com/problems/maximum-and-sum-of-array/
+ * Difficulty: Hard
+ *
+ * You are given an integer array nums of length n and an integer numSlots such that
+ * 2 * numSlots >= n. There are numSlots slots numbered from 1 to numSlots.
+ *
+ * You have to place all n integers into the slots such that each slot contains at most
+ * two numbers. The AND sum of a given placement is the sum of the bitwise AND of every
+ * number with its respective slot number.
+ * - For example, the AND sum of placing the numbers [1, 3] into slot 1 and [4, 6] into
+ *   slot 2 is equal to (1 AND 1) + (3 AND 1) + (4 AND 2) + (6 AND 2) = 1 + 1 + 0 + 2 = 4.
+ *
+ * Return the maximum possible AND sum of nums given numSlots slots.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} numSlots
+ * @return {number}
+ */
+var maximumANDSum = function(nums, numSlots) {
+  const map = new Map();
+
+  return dp(0, new Array(numSlots).fill(0));
+
+  function dp(index, slots) {
+    if (index >= nums.length) return 0;
+
+    const key = `${index},${slots.join(',')}`;
+    if (map.has(key)) return map.get(key);
+
+    let result = 0;
+    for (let i = 0; i < numSlots; i++) {
+      if (slots[i] < 2) {
+        slots[i]++;
+        result = Math.max(result, (nums[index] & (i + 1)) + dp(index + 1, slots));
+        slots[i]--;
+      }
+    }
+
+    map.set(key, result);
+    return result;
+  }
+};
diff --git a/solutions/2177-find-three-consecutive-integers-that-sum-to-a-given-number.js b/solutions/2177-find-three-consecutive-integers-that-sum-to-a-given-number.js
new file mode 100644
index 00000000..f77b1221
--- /dev/null
+++ b/solutions/2177-find-three-consecutive-integers-that-sum-to-a-given-number.js
@@ -0,0 +1,19 @@
+/**
+ * 2177. Find Three Consecutive Integers That Sum to a Given Number
+ * https://leetcode.com/problems/find-three-consecutive-integers-that-sum-to-a-given-number/
+ * Difficulty: Medium
+ *
+ * Given an integer num, return three consecutive integers (as a sorted array) that sum to num.
+ * If num cannot be expressed as the sum of three consecutive integers, return an empty array.
+ */
+
+/**
+ * @param {number} num
+ * @return {number[]}
+ */
+var sumOfThree = function(num) {
+  if (num % 3 !== 0) return [];
+
+  const middle = num / 3;
+  return [middle - 1, middle, middle + 1];
+};
diff --git a/solutions/2178-maximum-split-of-positive-even-integers.js b/solutions/2178-maximum-split-of-positive-even-integers.js
new file mode 100644
index 00000000..602b2ab7
--- /dev/null
+++ b/solutions/2178-maximum-split-of-positive-even-integers.js
@@ -0,0 +1,40 @@
+/**
+ * 2178. Maximum Split of Positive Even Integers
+ * https://leetcode.com/problems/maximum-split-of-positive-even-integers/
+ * Difficulty: Medium
+ *
+ * You are given an integer finalSum. Split it into a sum of a maximum number of unique
+ * positive even integers.
+ * - For example, given finalSum = 12, the following splits are valid (unique positive even
+ *   integers summing up to finalSum): (12), (2 + 10), (2 + 4 + 6), and (4 + 8). Among them,
+ *   (2 + 4 + 6) contains the maximum number of integers. Note that finalSum cannot be split
+ *   into (2 + 2 + 4 + 4) as all the numbers should be unique.
+ *
+ * Return a list of integers that represent a valid split containing a maximum number of
+ * integers. If no valid split exists for finalSum, return an empty list. You may return
+ * the integers in any order.
+ */
+
+/**
+ * @param {number} finalSum
+ * @return {number[]}
+ */
+var maximumEvenSplit = function(finalSum) {
+  if (finalSum % 2 !== 0) return [];
+
+  const result = [];
+  let current = 2;
+  let remaining = finalSum;
+
+  while (remaining >= current) {
+    if (remaining - current <= current) {
+      result.push(remaining);
+      return result;
+    }
+    result.push(current);
+    remaining -= current;
+    current += 2;
+  }
+
+  return result;
+};
diff --git a/solutions/2180-count-integers-with-even-digit-sum.js b/solutions/2180-count-integers-with-even-digit-sum.js
new file mode 100644
index 00000000..c1a077e6
--- /dev/null
+++ b/solutions/2180-count-integers-with-even-digit-sum.js
@@ -0,0 +1,32 @@
+/**
+ * 2180. Count Integers With Even Digit Sum
+ * https://leetcode.com/problems/count-integers-with-even-digit-sum/
+ * Difficulty: Easy
+ *
+ * Given a positive integer num, return the number of positive integers less than or equal to num
+ * whose digit sums are even.
+ *
+ * The digit sum of a positive integer is the sum of all its digits.
+ */
+
+/**
+ * @param {number} num
+ * @return {number}
+ */
+var countEven = function(num) {
+  let result = 0;
+
+  for (let i = 1; i <= num; i++) {
+    let sum = 0;
+    let current = i;
+    while (current > 0) {
+      sum += current % 10;
+      current = Math.floor(current / 10);
+    }
+    if (sum % 2 === 0) {
+      result++;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2181-merge-nodes-in-between-zeros.js b/solutions/2181-merge-nodes-in-between-zeros.js
new file mode 100644
index 00000000..f5993c27
--- /dev/null
+++ b/solutions/2181-merge-nodes-in-between-zeros.js
@@ -0,0 +1,46 @@
+/**
+ * 2181. Merge Nodes in Between Zeros
+ * https://leetcode.com/problems/merge-nodes-in-between-zeros/
+ * Difficulty: Medium
+ *
+ * You are given the head of a linked list, which contains a series of integers separated
+ * by 0's. The beginning and end of the linked list will have Node.val == 0.
+ *
+ * For every two consecutive 0's, merge all the nodes lying in between them into a single
+ * node whose value is the sum of all the merged nodes. The modified list should not contain
+ * any 0's.
+ *
+ * Return the head of the modified linked list.
+ */
+
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @return {ListNode}
+ */
+var mergeNodes = function(head) {
+  const result = new ListNode(0);
+  let current = result;
+  let sum = 0;
+
+  head = head.next;
+
+  while (head) {
+    if (head.val === 0) {
+      current.next = new ListNode(sum);
+      current = current.next;
+      sum = 0;
+    } else {
+      sum += head.val;
+    }
+    head = head.next;
+  }
+
+  return result.next;
+};
diff --git a/solutions/2183-count-array-pairs-divisible-by-k.js b/solutions/2183-count-array-pairs-divisible-by-k.js
new file mode 100644
index 00000000..96f923c4
--- /dev/null
+++ b/solutions/2183-count-array-pairs-divisible-by-k.js
@@ -0,0 +1,39 @@
+/**
+ * 2183. Count Array Pairs Divisible by K
+ * https://leetcode.com/problems/count-array-pairs-divisible-by-k/
+ * Difficulty: Hard
+ *
+ * Given a 0-indexed integer array nums of length n and an integer k, return the number
+ * of pairs (i, j) such that:
+ * - 0 <= i < j <= n - 1 and
+ * - nums[i] * nums[j] is divisible by k.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+var countPairs = function(nums, k) {
+  const map = new Map();
+  let pairs = 0;
+
+  for (const num of nums) {
+    const gcd1 = gcd(num, k);
+    for (const [gcd2, count] of map) {
+      if ((gcd1 * gcd2) % k === 0) {
+        pairs += count;
+      }
+    }
+    map.set(gcd1, (map.get(gcd1) || 0) + 1);
+  }
+
+  return pairs;
+};
+
+function gcd(a, b) {
+  while (b) {
+    [a, b] = [b, a % b];
+  }
+  return a;
+}
diff --git a/solutions/2186-minimum-number-of-steps-to-make-two-strings-anagram-ii.js b/solutions/2186-minimum-number-of-steps-to-make-two-strings-anagram-ii.js
new file mode 100644
index 00000000..2501cbd5
--- /dev/null
+++ b/solutions/2186-minimum-number-of-steps-to-make-two-strings-anagram-ii.js
@@ -0,0 +1,36 @@
+/**
+ * 2186. Minimum Number of Steps to Make Two Strings Anagram II
+ * https://leetcode.com/problems/minimum-number-of-steps-to-make-two-strings-anagram-ii/
+ * Difficulty: Medium
+ *
+ * You are given two strings s and t. In one step, you can append any character to either s or t.
+ *
+ * Return the minimum number of steps to make s and t anagrams of each other.
+ *
+ * An anagram of a string is a string that contains the same characters with a different (or the
+ * same) ordering.
+ */
+
+/**
+ * @param {string} s
+ * @param {string} t
+ * @return {number}
+ */
+var minSteps = function(s, t) {
+  const charCount = new Array(26).fill(0);
+
+  for (const char of s) {
+    charCount[char.charCodeAt(0) - 97]++;
+  }
+
+  for (const char of t) {
+    charCount[char.charCodeAt(0) - 97]--;
+  }
+
+  let result = 0;
+  for (const count of charCount) {
+    result += Math.abs(count);
+  }
+
+  return result;
+};
diff --git a/solutions/2187-minimum-time-to-complete-trips.js b/solutions/2187-minimum-time-to-complete-trips.js
new file mode 100644
index 00000000..305712f8
--- /dev/null
+++ b/solutions/2187-minimum-time-to-complete-trips.js
@@ -0,0 +1,39 @@
+/**
+ * 2187. Minimum Time to Complete Trips
+ * https://leetcode.com/problems/minimum-time-to-complete-trips/
+ * Difficulty: Medium
+ *
+ * You are given an array time where time[i] denotes the time taken by the ith bus to complete
+ * one trip.
+ *
+ * Each bus can make multiple trips successively; that is, the next trip can start immediately
+ * after completing the current trip. Also, each bus operates independently; that is, the trips
+ * of one bus do not influence the trips of any other bus.
+ *
+ * You are also given an integer totalTrips, which denotes the number of trips all buses should
+ * make in total. Return the minimum time required for all buses to complete at least totalTrips
+ * trips.
+ */
+
+/**
+ * @param {number[]} time
+ * @param {number} totalTrips
+ * @return {number}
+ */
+var minimumTime = function(time, totalTrips) {
+  let left = 1;
+  let right = Math.min(...time) * totalTrips;
+
+  while (left < right) {
+    const mid = Math.floor((left + right) / 2);
+    const trips = time.reduce((sum, t) => sum + Math.floor(mid / t), 0);
+
+    if (trips >= totalTrips) {
+      right = mid;
+    } else {
+      left = mid + 1;
+    }
+  }
+
+  return left;
+};
diff --git a/solutions/2188-minimum-time-to-finish-the-race.js b/solutions/2188-minimum-time-to-finish-the-race.js
new file mode 100644
index 00000000..fbce3d25
--- /dev/null
+++ b/solutions/2188-minimum-time-to-finish-the-race.js
@@ -0,0 +1,56 @@
+/**
+ * 2188. Minimum Time to Finish the Race
+ * https://leetcode.com/problems/minimum-time-to-finish-the-race/
+ * Difficulty: Hard
+ *
+ * You are given a 0-indexed 2D integer array tires where tires[i] = [fi, ri] indicates that
+ * the ith tire can finish its xth successive lap in fi * ri(x-1) seconds.
+ * - For example, if fi = 3 and ri = 2, then the tire would finish its 1st lap in 3 seconds,
+ *   its 2nd lap in 3 * 2 = 6 seconds, its 3rd lap in 3 * 22 = 12 seconds, etc.
+ *
+ * You are also given an integer changeTime and an integer numLaps.
+ *
+ * The race consists of numLaps laps and you may start the race with any tire. You have an
+ * unlimited supply of each tire and after every lap, you may change to any given tire (including
+ * the current tire type) if you wait changeTime seconds.
+ *
+ * Return the minimum time to finish the race.
+ */
+
+/**
+ * @param {number[][]} tires
+ * @param {number} changeTime
+ * @param {number} numLaps
+ * @return {number}
+ */
+var minimumFinishTime = function(tires, changeTime, numLaps) {
+  const minTimes = new Array(18).fill(Infinity);
+  const bestTime = new Array(numLaps + 1).fill(Infinity);
+
+  for (const [baseTime, multiplier] of tires) {
+    let currentTime = baseTime;
+    let totalTime = baseTime;
+
+    for (let lap = 1; lap <= Math.min(numLaps, 17); lap++) {
+      if (currentTime > changeTime + baseTime) break;
+      minTimes[lap] = Math.min(minTimes[lap], totalTime);
+      currentTime *= multiplier;
+      totalTime += currentTime;
+    }
+  }
+
+  bestTime[0] = 0;
+
+  for (let lap = 1; lap <= numLaps; lap++) {
+    for (let prev = 1; prev <= Math.min(lap, 17); prev++) {
+      if (minTimes[prev] !== Infinity) {
+        bestTime[lap] = Math.min(
+          bestTime[lap],
+          bestTime[lap - prev] + minTimes[prev] + (lap === prev ? 0 : changeTime)
+        );
+      }
+    }
+  }
+
+  return bestTime[numLaps];
+};
diff --git a/solutions/2190-most-frequent-number-following-key-in-an-array.js b/solutions/2190-most-frequent-number-following-key-in-an-array.js
new file mode 100644
index 00000000..b5485503
--- /dev/null
+++ b/solutions/2190-most-frequent-number-following-key-in-an-array.js
@@ -0,0 +1,43 @@
+/**
+ * 2190. Most Frequent Number Following Key In an Array
+ * https://leetcode.com/problems/most-frequent-number-following-key-in-an-array/
+ * Difficulty: Easy
+ *
+ * You are given a 0-indexed integer array nums. You are also given an integer key, which is
+ * present in nums.
+ *
+ * For every unique integer target in nums, count the number of times target immediately follows
+ * an occurrence of key in nums. In other words, count the number of indices i such that:
+ * - 0 <= i <= nums.length - 2,
+ * - nums[i] == key and,
+ * - nums[i + 1] == target.
+ *
+ * Return the target with the maximum count. The test cases will be generated such that the target
+ * with maximum count is unique.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} key
+ * @return {number}
+ */
+var mostFrequent = function(nums, key) {
+  const frequency = new Map();
+  let maxCount = 0;
+  let result = 0;
+
+  for (let i = 0; i < nums.length - 1; i++) {
+    if (nums[i] === key) {
+      const target = nums[i + 1];
+      const count = (frequency.get(target) || 0) + 1;
+      frequency.set(target, count);
+
+      if (count > maxCount) {
+        maxCount = count;
+        result = target;
+      }
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2191-sort-the-jumbled-numbers.js b/solutions/2191-sort-the-jumbled-numbers.js
new file mode 100644
index 00000000..490e1d17
--- /dev/null
+++ b/solutions/2191-sort-the-jumbled-numbers.js
@@ -0,0 +1,53 @@
+/**
+ * 2191. Sort the Jumbled Numbers
+ * https://leetcode.com/problems/sort-the-jumbled-numbers/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed integer array mapping which represents the mapping rule of a shuffled
+ * decimal system. mapping[i] = j means digit i should be mapped to digit j in this system.
+ *
+ * The mapped value of an integer is the new integer obtained by replacing each occurrence of digit
+ * i in the integer with mapping[i] for all 0 <= i <= 9.
+ *
+ * You are also given another integer array nums. Return the array nums sorted in non-decreasing
+ * order based on the mapped values of its elements.
+ *
+ * Notes:
+ * - Elements with the same mapped values should appear in the same relative order as in the input.
+ * - The elements of nums should only be sorted based on their mapped values and not be replaced by
+ *   them.
+ */
+
+/**
+ * @param {number[]} mapping
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var sortJumbled = function(mapping, nums) {
+  const mapped = nums.map((num, index) => {
+    let mappedNum = 0;
+    let temp = num;
+
+    if (temp === 0) {
+      mappedNum = mapping[0];
+    } else {
+      const digits = [];
+      while (temp > 0) {
+        digits.push(mapping[temp % 10]);
+        temp = Math.floor(temp / 10);
+      }
+      while (digits.length > 0) {
+        mappedNum = mappedNum * 10 + digits.pop();
+      }
+    }
+
+    return { original: num, mapped: mappedNum, index };
+  });
+
+  mapped.sort((a, b) => {
+    if (a.mapped === b.mapped) return a.index - b.index;
+    return a.mapped - b.mapped;
+  });
+
+  return mapped.map(item => item.original);
+};
diff --git a/solutions/2192-all-ancestors-of-a-node-in-a-directed-acyclic-graph.js b/solutions/2192-all-ancestors-of-a-node-in-a-directed-acyclic-graph.js
new file mode 100644
index 00000000..9eeb8509
--- /dev/null
+++ b/solutions/2192-all-ancestors-of-a-node-in-a-directed-acyclic-graph.js
@@ -0,0 +1,48 @@
+/**
+ * 2192. All Ancestors of a Node in a Directed Acyclic Graph
+ * https://leetcode.com/problems/all-ancestors-of-a-node-in-a-directed-acyclic-graph/
+ * Difficulty: Medium
+ *
+ * You are given a positive integer n representing the number of nodes of a Directed Acyclic
+ * Graph (DAG). The nodes are numbered from 0 to n - 1 (inclusive).
+ *
+ * You are also given a 2D integer array edges, where edges[i] = [fromi, toi] denotes that there
+ * is a unidirectional edge from fromi to toi in the graph.
+ *
+ * Return a list answer, where answer[i] is the list of ancestors of the ith node, sorted in
+ * ascending order.
+ *
+ * A node u is an ancestor of another node v if u can reach v via a set of edges.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} edges
+ * @return {number[][]}
+ */
+var getAncestors = function(n, edges) {
+  const graph = Array.from({ length: n }, () => []);
+  const ancestors = Array.from({ length: n }, () => new Set());
+
+  for (const [from, to] of edges) {
+    graph[to].push(from);
+  }
+
+  for (let i = 0; i < n; i++) {
+    findAncestors(i);
+  }
+
+  return ancestors.map(set => [...set].sort((a, b) => a - b));
+
+  function findAncestors(node) {
+    if (ancestors[node].size > 0) return;
+
+    for (const parent of graph[node]) {
+      ancestors[node].add(parent);
+      findAncestors(parent);
+      for (const ancestor of ancestors[parent]) {
+        ancestors[node].add(ancestor);
+      }
+    }
+  }
+};
diff --git a/solutions/2193-minimum-number-of-moves-to-make-palindrome.js b/solutions/2193-minimum-number-of-moves-to-make-palindrome.js
new file mode 100644
index 00000000..c25800e2
--- /dev/null
+++ b/solutions/2193-minimum-number-of-moves-to-make-palindrome.js
@@ -0,0 +1,42 @@
+/**
+ * 2193. Minimum Number of Moves to Make Palindrome
+ * https://leetcode.com/problems/minimum-number-of-moves-to-make-palindrome/
+ * Difficulty: Hard
+ *
+ * You are given a string s consisting only of lowercase English letters.
+ *
+ * In one move, you can select any two adjacent characters of s and swap them.
+ *
+ * Return the minimum number of moves needed to make s a palindrome.
+ *
+ * Note that the input will be generated such that s can always be converted to a palindrome.
+ */
+
+/**
+* @param {string} s
+* @return {number}
+*/
+var minMovesToMakePalindrome = function(s) {
+  const chars = s.split('');
+  let moves = 0;
+
+  while (chars.length > 1) {
+    const matchIndex = chars.lastIndexOf(chars[0]);
+
+    if (matchIndex === 0) {
+      const middlePos = Math.floor(chars.length / 2);
+      moves += middlePos;
+      chars.splice(0, 1);
+    } else {
+      for (let i = matchIndex; i < chars.length - 1; i++) {
+        [chars[i], chars[i + 1]] = [chars[i + 1], chars[i]];
+        moves++;
+      }
+
+      chars.pop();
+      chars.shift();
+    }
+  }
+
+  return moves;
+};
diff --git a/solutions/2194-cells-in-a-range-on-an-excel-sheet.js b/solutions/2194-cells-in-a-range-on-an-excel-sheet.js
new file mode 100644
index 00000000..194c60cc
--- /dev/null
+++ b/solutions/2194-cells-in-a-range-on-an-excel-sheet.js
@@ -0,0 +1,37 @@
+/**
+ * 2194. Cells in a Range on an Excel Sheet
+ * https://leetcode.com/problems/cells-in-a-range-on-an-excel-sheet/
+ * Difficulty: Easy
+ *
+ * A cell (r, c) of an excel sheet is represented as a string "" where:
+ * -  denotes the column number c of the cell. It is represented by alphabetical letters.
+ *   - For example, the 1st column is denoted by 'A', the 2nd by 'B', the 3rd by 'C', and so on.
+ * -  is the row number r of the cell. The rth row is represented by the integer r.
+ *   - You are given a string s in the format ":", where  represents
+ *     the column c1,  represents the row r1,  represents the column c2, and 
+ *     represents the row r2, such that r1 <= r2 and c1 <= c2.
+ *
+ * Return the list of cells (x, y) such that r1 <= x <= r2 and c1 <= y <= c2. The cells should
+ * be represented as strings in the format mentioned above and be sorted in non-decreasing order
+ * first by columns and then by rows.
+ */
+
+/**
+ * @param {string} s
+ * @return {string[]}
+ */
+var cellsInRange = function(s) {
+  const result = [];
+  const startCol = s.charCodeAt(0);
+  const endCol = s.charCodeAt(3);
+  const startRow = parseInt(s[1], 10);
+  const endRow = parseInt(s[4], 10);
+
+  for (let col = startCol; col <= endCol; col++) {
+    for (let row = startRow; row <= endRow; row++) {
+      result.push(String.fromCharCode(col) + row);
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2195-append-k-integers-with-minimal-sum.js b/solutions/2195-append-k-integers-with-minimal-sum.js
new file mode 100644
index 00000000..bc873455
--- /dev/null
+++ b/solutions/2195-append-k-integers-with-minimal-sum.js
@@ -0,0 +1,40 @@
+/**
+ * 2195. Append K Integers With Minimal Sum
+ * https://leetcode.com/problems/append-k-integers-with-minimal-sum/
+ * Difficulty: Medium
+ *
+ * You are given an integer array nums and an integer k. Append k unique positive integers that do
+ * not appear in nums to nums such that the resulting total sum is minimum.
+ *
+ * Return the sum of the k integers appended to nums.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+var minimalKSum = function(nums, k) {
+  const sortedUnique = [...new Set(nums)].sort((a, b) => a - b);
+  let sum = BigInt(0);
+  let current = 1;
+  let i = 0;
+
+  while (k > 0 && i < sortedUnique.length) {
+    if (current < sortedUnique[i]) {
+      const count = Math.min(k, sortedUnique[i] - current);
+      sum += (BigInt(current) + BigInt(current + count - 1)) * BigInt(count) / BigInt(2);
+      k -= count;
+      current += count;
+    } else {
+      current = sortedUnique[i] + 1;
+      i++;
+    }
+  }
+
+  if (k > 0) {
+    sum += (BigInt(current) + BigInt(current + k - 1)) * BigInt(k) / BigInt(2);
+  }
+
+  return Number(sum);
+};
diff --git a/solutions/2196-create-binary-tree-from-descriptions.js b/solutions/2196-create-binary-tree-from-descriptions.js
new file mode 100644
index 00000000..67af6416
--- /dev/null
+++ b/solutions/2196-create-binary-tree-from-descriptions.js
@@ -0,0 +1,54 @@
+/**
+ * 2196. Create Binary Tree From Descriptions
+ * https://leetcode.com/problems/create-binary-tree-from-descriptions/
+ * Difficulty: Medium
+ *
+ * You are given a 2D integer array descriptions where descriptions[i] = [parenti, childi, isLefti]
+ * indicates that parenti is the parent of childi in a binary tree of unique values. Furthermore,
+ * - If isLefti == 1, then childi is the left child of parenti.
+ * - If isLefti == 0, then childi is the right child of parenti.
+ *
+ * Construct the binary tree described by descriptions and return its root.
+ *
+ * The test cases will be generated such that the binary tree is valid.
+ */
+
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {number[][]} descriptions
+ * @return {TreeNode}
+ */
+var createBinaryTree = function(descriptions) {
+  const nodes = new Map();
+  const children = new Set();
+
+  for (const [parent, child, isLeft] of descriptions) {
+    if (!nodes.has(parent)) {
+      nodes.set(parent, new TreeNode(parent));
+    }
+    if (!nodes.has(child)) {
+      nodes.set(child, new TreeNode(child));
+    }
+    children.add(child);
+    if (isLeft) {
+      nodes.get(parent).left = nodes.get(child);
+    } else {
+      nodes.get(parent).right = nodes.get(child);
+    }
+  }
+
+  for (const [val, node] of nodes) {
+    if (!children.has(val)) {
+      return node;
+    }
+  }
+
+  return null;
+};
diff --git a/solutions/2197-replace-non-coprime-numbers-in-array.js b/solutions/2197-replace-non-coprime-numbers-in-array.js
new file mode 100644
index 00000000..2bc8e2d5
--- /dev/null
+++ b/solutions/2197-replace-non-coprime-numbers-in-array.js
@@ -0,0 +1,48 @@
+/**
+ * 2197. Replace Non-Coprime Numbers in Array
+ * https://leetcode.com/problems/replace-non-coprime-numbers-in-array/
+ * Difficulty: Hard
+ *
+ * You are given an array of integers nums. Perform the following steps:
+ * 1. Find any two adjacent numbers in nums that are non-coprime.
+ * 2. If no such numbers are found, stop the process.
+ * 3. Otherwise, delete the two numbers and replace them with their LCM (Least Common Multiple).
+ * 4. Repeat this process as long as you keep finding two adjacent non-coprime numbers.
+ *
+ * Return the final modified array. It can be shown that replacing adjacent non-coprime numbers in
+ * any arbitrary order will lead to the same result.
+ *
+ * The test cases are generated such that the values in the final array are less than or equal to
+ * 108.
+ *
+ * Two values x and y are non-coprime if GCD(x, y) > 1 where GCD(x, y) is the Greatest Common
+ * Divisor of x and y.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var replaceNonCoprimes = function(nums) {
+  const result = [];
+
+  for (let num of nums) {
+    while (result.length > 0) {
+      const last = result[result.length - 1];
+      const gcdValue = gcd(last, num);
+      if (gcdValue === 1) break;
+      result.pop();
+      num = (last / gcdValue) * num;
+    }
+    result.push(num);
+  }
+
+  return result;
+};
+
+function gcd(a, b) {
+  while (b) {
+    [a, b] = [b, a % b];
+  }
+  return a;
+}
diff --git a/solutions/2200-find-all-k-distant-indices-in-an-array.js b/solutions/2200-find-all-k-distant-indices-in-an-array.js
new file mode 100644
index 00000000..5a2cd1bd
--- /dev/null
+++ b/solutions/2200-find-all-k-distant-indices-in-an-array.js
@@ -0,0 +1,31 @@
+/**
+ * 2200. Find All K-Distant Indices in an Array
+ * https://leetcode.com/problems/find-all-k-distant-indices-in-an-array/
+ * Difficulty: Easy
+ *
+ * You are given a 0-indexed integer array nums and two integers key and k. A k-distant index
+ * is an index i of nums for which there exists at least one index j such that |i - j| <= k
+ * and nums[j] == key.
+ *
+ * Return a list of all k-distant indices sorted in increasing order.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} key
+ * @param {number} k
+ * @return {number[]}
+ */
+var findKDistantIndices = function(nums, key, k) {
+  const result = new Set();
+
+  for (let j = 0; j < nums.length; j++) {
+    if (nums[j] === key) {
+      for (let i = Math.max(0, j - k); i <= Math.min(nums.length - 1, j + k); i++) {
+        result.add(i);
+      }
+    }
+  }
+
+  return [...result].sort((a, b) => a - b);
+};
diff --git a/solutions/2201-count-artifacts-that-can-be-extracted.js b/solutions/2201-count-artifacts-that-can-be-extracted.js
new file mode 100644
index 00000000..cd5f5627
--- /dev/null
+++ b/solutions/2201-count-artifacts-that-can-be-extracted.js
@@ -0,0 +1,57 @@
+/**
+ * 2201. Count Artifacts That Can Be Extracted
+ * https://leetcode.com/problems/count-artifacts-that-can-be-extracted/
+ * Difficulty: Medium
+ *
+ * There is an n x n 0-indexed grid with some artifacts buried in it. You are given the integer
+ * n and a 0-indexed 2D integer array artifacts describing the positions of the rectangular
+ * artifacts where artifacts[i] = [r1i, c1i, r2i, c2i] denotes that the ith artifact is buried
+ * in the subgrid where:
+ * - (r1i, c1i) is the coordinate of the top-left cell of the ith artifact and
+ * - (r2i, c2i) is the coordinate of the bottom-right cell of the ith artifact.
+ *
+ * You will excavate some cells of the grid and remove all the mud from them. If the cell has a
+ * part of an artifact buried underneath, it will be uncovered. If all the parts of an artifact
+ * are uncovered, you can extract it.
+ *
+ * Given a 0-indexed 2D integer array dig where dig[i] = [ri, ci] indicates that you will excavate
+ * the cell (ri, ci), return the number of artifacts that you can extract.
+ *
+ * The test cases are generated such that:
+ * - No two artifacts overlap.
+ * - Each artifact only covers at most 4 cells.
+ * - The entries of dig are unique.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} artifacts
+ * @param {number[][]} dig
+ * @return {number}
+ */
+var digArtifacts = function(n, artifacts, dig) {
+  const excavated = new Set();
+  let result = 0;
+
+  for (const [row, col] of dig) {
+    excavated.add(`${row},${col}`);
+  }
+
+  for (const [r1, c1, r2, c2] of artifacts) {
+    let allUncovered = true;
+
+    for (let r = r1; r <= r2; r++) {
+      for (let c = c1; c <= c2; c++) {
+        if (!excavated.has(`${r},${c}`)) {
+          allUncovered = false;
+          break;
+        }
+      }
+      if (!allUncovered) break;
+    }
+
+    if (allUncovered) result++;
+  }
+
+  return result;
+};
diff --git a/solutions/2202-maximize-the-topmost-element-after-k-moves.js b/solutions/2202-maximize-the-topmost-element-after-k-moves.js
new file mode 100644
index 00000000..8b1554b2
--- /dev/null
+++ b/solutions/2202-maximize-the-topmost-element-after-k-moves.js
@@ -0,0 +1,42 @@
+/**
+ * 2202. Maximize the Topmost Element After K Moves
+ * https://leetcode.com/problems/maximize-the-topmost-element-after-k-moves/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed integer array nums representing the contents of a pile, where nums[0]
+ * is the topmost element of the pile.
+ *
+ * In one move, you can perform either of the following:
+ * - If the pile is not empty, remove the topmost element of the pile.
+ * - If there are one or more removed elements, add any one of them back onto the pile. This element
+ *   becomes the new topmost element.
+ *
+ * You are also given an integer k, which denotes the total number of moves to be made.
+ *
+ * Return the maximum value of the topmost element of the pile possible after exactly k moves. In
+ * case it is not possible to obtain a non-empty pile after k moves, return -1.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+var maximumTop = function(nums, k) {
+  const n = nums.length;
+
+  if (n === 1 && k % 2 === 1) return -1;
+  if (k === 0) return nums[0];
+  if (k === 1) return n > 1 ? nums[1] : -1;
+
+  let result = 0;
+  for (let i = 0; i < Math.min(k - 1, n); i++) {
+    result = Math.max(result, nums[i]);
+  }
+
+  if (k < n) {
+    result = Math.max(result, nums[k]);
+  }
+
+  return result;
+};
diff --git a/solutions/2203-minimum-weighted-subgraph-with-the-required-paths.js b/solutions/2203-minimum-weighted-subgraph-with-the-required-paths.js
new file mode 100644
index 00000000..4ace48ff
--- /dev/null
+++ b/solutions/2203-minimum-weighted-subgraph-with-the-required-paths.js
@@ -0,0 +1,73 @@
+/**
+ * 2203. Minimum Weighted Subgraph With the Required Paths
+ * https://leetcode.com/problems/minimum-weighted-subgraph-with-the-required-paths/
+ * Difficulty: Hard
+ *
+ * You are given an integer n denoting the number of nodes of a weighted directed graph. The nodes
+ * are numbered from 0 to n - 1.
+ *
+ * You are also given a 2D integer array edges where edges[i] = [fromi, toi, weighti] denotes that
+ * there exists a directed edge from fromi to toi with weight weighti.
+ *
+ * Lastly, you are given three distinct integers src1, src2, and dest denoting three distinct nodes
+ * of the graph.
+ *
+ * Return the minimum weight of a subgraph of the graph such that it is possible to reach dest from
+ * both src1 and src2 via a set of edges of this subgraph. In case such a subgraph does not exist,
+ * return -1.
+ *
+ * A subgraph is a graph whose vertices and edges are subsets of the original graph. The weight of
+ * a subgraph is the sum of weights of its constituent edges.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} edges
+ * @param {number} src1
+ * @param {number} src2
+ * @param {number} dest
+ * @return {number}
+ */
+var minimumWeight = function(n, edges, src1, src2, dest) {
+  const forwardGraph = Array.from({ length: n }, () => []);
+  const reverseGraph = Array.from({ length: n }, () => []);
+
+  for (const [from, to, weight] of edges) {
+    forwardGraph[from].push([to, weight]);
+    reverseGraph[to].push([from, weight]);
+  }
+
+  const distFromSrc1 = dijkstra(forwardGraph, src1);
+  const distFromSrc2 = dijkstra(forwardGraph, src2);
+  const distToDest = dijkstra(reverseGraph, dest);
+  let minWeight = Infinity;
+  for (let i = 0; i < n; i++) {
+    if (distFromSrc1[i] !== Infinity && distFromSrc2[i] !== Infinity
+        && distToDest[i] !== Infinity) {
+      minWeight = Math.min(minWeight, distFromSrc1[i] + distFromSrc2[i] + distToDest[i]);
+    }
+  }
+
+  return minWeight === Infinity ? -1 : minWeight;
+
+  function dijkstra(graph, start) {
+    const distances = new Array(n).fill(Infinity);
+    distances[start] = 0;
+    const pq = new PriorityQueue((a, b) => a[0] - b[0]);
+    pq.enqueue([0, start]);
+
+    while (!pq.isEmpty()) {
+      const [dist, node] = pq.dequeue();
+      if (dist > distances[node]) continue;
+
+      for (const [next, weight] of graph[node]) {
+        if (distances[next] > dist + weight) {
+          distances[next] = dist + weight;
+          pq.enqueue([distances[next], next]);
+        }
+      }
+    }
+
+    return distances;
+  }
+};
diff --git a/solutions/2207-maximize-number-of-subsequences-in-a-string.js b/solutions/2207-maximize-number-of-subsequences-in-a-string.js
new file mode 100644
index 00000000..d7ad35a6
--- /dev/null
+++ b/solutions/2207-maximize-number-of-subsequences-in-a-string.js
@@ -0,0 +1,41 @@
+/**
+ * 2207. Maximize Number of Subsequences in a String
+ * https://leetcode.com/problems/maximize-number-of-subsequences-in-a-string/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed string text and another 0-indexed string pattern of length 2, both
+ * of which consist of only lowercase English letters.
+ *
+ * You can add either pattern[0] or pattern[1] anywhere in text exactly once. Note that the
+ * character can be added even at the beginning or at the end of text.
+ *
+ * Return the maximum number of times pattern can occur as a subsequence of the modified text.
+ *
+ * A subsequence is a string that can be derived from another string by deleting some or no
+ * characters without changing the order of the remaining characters.
+ */
+
+/**
+ * @param {string} text
+ * @param {string} pattern
+ * @return {number}
+ */
+var maximumSubsequenceCount = function(text, pattern) {
+  const firstChar = pattern[0];
+  const secondChar = pattern[1];
+  let firstCount = 0;
+  let secondCount = 0;
+  let subsequences = 0;
+
+  for (const char of text) {
+    if (char === secondChar) {
+      subsequences += firstCount;
+      secondCount++;
+    }
+    if (char === firstChar) {
+      firstCount++;
+    }
+  }
+
+  return subsequences + Math.max(firstCount, secondCount);
+};
diff --git a/solutions/2209-minimum-white-tiles-after-covering-with-carpets.js b/solutions/2209-minimum-white-tiles-after-covering-with-carpets.js
new file mode 100644
index 00000000..08fb5fd3
--- /dev/null
+++ b/solutions/2209-minimum-white-tiles-after-covering-with-carpets.js
@@ -0,0 +1,36 @@
+/**
+ * 2209. Minimum White Tiles After Covering With Carpets
+ * https://leetcode.com/problems/minimum-white-tiles-after-covering-with-carpets/
+ * Difficulty: Hard
+ *
+ * You are given a 0-indexed binary string floor, which represents the colors of tiles on a floor:
+ * - floor[i] = '0' denotes that the ith tile of the floor is colored black.
+ * - On the other hand, floor[i] = '1' denotes that the ith tile of the floor is colored white.
+ *
+ * You are also given numCarpets and carpetLen. You have numCarpets black carpets, each of length
+ * carpetLen tiles. Cover the tiles with the given carpets such that the number of white tiles
+ * still visible is minimum. Carpets may overlap one another.
+ *
+ * Return the minimum number of white tiles still visible.
+ */
+
+/**
+ * @param {string} floor
+ * @param {number} numCarpets
+ * @param {number} carpetLen
+ * @return {number}
+ */
+var minimumWhiteTiles = function(floor, numCarpets, carpetLen) {
+  const n = floor.length;
+  const dp = Array.from({ length: n + 1 }, () => Array(numCarpets + 1).fill(0));
+
+  for (let i = 1; i <= n; i++) {
+    for (let j = 0; j <= numCarpets; j++) {
+      const skip = dp[i - 1][j] + (floor[i - 1] === '1' ? 1 : 0);
+      const cover = j > 0 ? dp[Math.max(0, i - carpetLen)][j - 1] : Infinity;
+      dp[i][j] = Math.min(skip, cover);
+    }
+  }
+
+  return dp[n][numCarpets];
+};
diff --git a/solutions/2210-count-hills-and-valleys-in-an-array.js b/solutions/2210-count-hills-and-valleys-in-an-array.js
new file mode 100644
index 00000000..388d78f5
--- /dev/null
+++ b/solutions/2210-count-hills-and-valleys-in-an-array.js
@@ -0,0 +1,38 @@
+/**
+ * 2210. Count Hills and Valleys in an Array
+ * https://leetcode.com/problems/count-hills-and-valleys-in-an-array/
+ * Difficulty: Easy
+ *
+ * You are given a 0-indexed integer array nums. An index i is part of a hill in nums if the
+ * closest non-equal neighbors of i are smaller than nums[i]. Similarly, an index i is part
+ * of a valley in nums if the closest non-equal neighbors of i are larger than nums[i].
+ * Adjacent indices i and j are part of the same hill or valley if nums[i] == nums[j].
+ *
+ * Note that for an index to be part of a hill or valley, it must have a non-equal neighbor
+ * on both the left and right of the index.
+ *
+ * Return the number of hills and valleys in nums.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var countHillValley = function(nums) {
+  let count = 0;
+  let prev = nums[0];
+
+  for (let i = 1; i < nums.length - 1; i++) {
+    if (nums[i] === nums[i + 1]) continue;
+    const left = prev;
+    const right = nums[i + 1];
+
+    if ((nums[i] > left && nums[i] > right) || (nums[i] < left && nums[i] < right)) {
+      count++;
+    }
+
+    prev = nums[i];
+  }
+
+  return count;
+};
diff --git a/solutions/2211-count-collisions-on-a-road.js b/solutions/2211-count-collisions-on-a-road.js
new file mode 100644
index 00000000..acbd8920
--- /dev/null
+++ b/solutions/2211-count-collisions-on-a-road.js
@@ -0,0 +1,41 @@
+/**
+ * 2211. Count Collisions on a Road
+ * https://leetcode.com/problems/count-collisions-on-a-road/
+ * Difficulty: Medium
+ *
+ * There are n cars on an infinitely long road. The cars are numbered from 0 to n - 1 from left
+ * to right and each car is present at a unique point.
+ *
+ * You are given a 0-indexed string directions of length n. directions[i] can be either 'L', 'R',
+ * or 'S' denoting whether the ith car is moving towards the left, towards the right, or staying
+ * at its current point respectively. Each moving car has the same speed.
+ *
+ * The number of collisions can be calculated as follows:
+ * - When two cars moving in opposite directions collide with each other, the number of collisions
+ *   increases by 2.
+ * - When a moving car collides with a stationary car, the number of collisions increases by 1.
+ *
+ * After a collision, the cars involved can no longer move and will stay at the point where they
+ * collided. Other than that, cars cannot change their state or direction of motion.
+ *
+ * Return the total number of collisions that will happen on the road.
+ */
+
+/**
+ * @param {string} directions
+ * @return {number}
+ */
+var countCollisions = function(directions) {
+  let result = 0;
+  let left = 0;
+  let right = directions.length - 1;
+
+  while (left < directions.length && directions[left] === 'L') left++;
+  while (right >= 0 && directions[right] === 'R') right--;
+
+  for (let i = left; i <= right; i++) {
+    if (directions[i] !== 'S') result++;
+  }
+
+  return result;
+};
diff --git a/solutions/2212-maximum-points-in-an-archery-competition.js b/solutions/2212-maximum-points-in-an-archery-competition.js
new file mode 100644
index 00000000..2c09fc29
--- /dev/null
+++ b/solutions/2212-maximum-points-in-an-archery-competition.js
@@ -0,0 +1,60 @@
+/**
+ * 2212. Maximum Points in an Archery Competition
+ * https://leetcode.com/problems/maximum-points-in-an-archery-competition/
+ * Difficulty: Medium
+ *
+ * Alice and Bob are opponents in an archery competition. The competition has set the
+ * following rules:
+ * 1. Alice first shoots numArrows arrows and then Bob shoots numArrows arrows.
+ * 2. The points are then calculated as follows:
+ *    1. The target has integer scoring sections ranging from 0 to 11 inclusive.
+ *    2. For each section of the target with score k (in between 0 to 11), say Alice and Bob have
+ *       shot ak and bk arrows on that section respectively. If ak >= bk, then Alice takes k points.
+ *       If ak < bk, then Bob takes k points.
+ *    3. However, if ak == bk == 0, then nobody takes k points.
+ * - For example, if Alice and Bob both shot 2 arrows on the section with score 11, then Alice takes
+ *   11 points. On the other hand, if Alice shot 0 arrows on the section with score 11 and Bob shot
+ *   2 arrows on that same section, then Bob takes 11 points.
+ *
+ * You are given the integer numArrows and an integer array aliceArrows of size 12, which represents
+ * the number of arrows Alice shot on each scoring section from 0 to 11. Now, Bob wants to maximize
+ * the total number of points he can obtain.
+ *
+ * Return the array bobArrows which represents the number of arrows Bob shot on each scoring section
+ * from 0 to 11. The sum of the values in bobArrows should equal numArrows.
+ *
+ * If there are multiple ways for Bob to earn the maximum total points, return any one of them.
+ */
+
+/**
+ * @param {number} numArrows
+ * @param {number[]} aliceArrows
+ * @return {number[]}
+ */
+var maximumBobPoints = function(numArrows, aliceArrows) {
+  let maxScore = 0;
+  let bestConfig = new Array(12).fill(0);
+
+  backtrack(1, numArrows, 0, new Array(12).fill(0));
+  return bestConfig;
+
+  function backtrack(index, arrowsLeft, score, config) {
+    if (index === 12 || arrowsLeft === 0) {
+      if (score > maxScore) {
+        maxScore = score;
+        bestConfig = [...config];
+        bestConfig[0] += arrowsLeft;
+      }
+      return;
+    }
+
+    const needed = aliceArrows[index] + 1;
+    if (arrowsLeft >= needed) {
+      config[index] = needed;
+      backtrack(index + 1, arrowsLeft - needed, score + index, config);
+      config[index] = 0;
+    }
+
+    backtrack(index + 1, arrowsLeft, score, config);
+  }
+};
diff --git a/solutions/2213-longest-substring-of-one-repeating-character.js b/solutions/2213-longest-substring-of-one-repeating-character.js
new file mode 100644
index 00000000..ca5e006d
--- /dev/null
+++ b/solutions/2213-longest-substring-of-one-repeating-character.js
@@ -0,0 +1,117 @@
+/**
+ * 2213. Longest Substring of One Repeating Character
+ * https://leetcode.com/problems/longest-substring-of-one-repeating-character/
+ * Difficulty: Hard
+ *
+ * You are given a 0-indexed string s. You are also given a 0-indexed string queryCharacters of
+ * length k and a 0-indexed array of integer indices queryIndices of length k, both of which are
+ * used to describe k queries.
+ *
+ * The ith query updates the character in s at index queryIndices[i] to the character
+ * queryCharacters[i].
+ *
+ * Return an array lengths of length k where lengths[i] is the length of the longest substring of
+ * s consisting of only one repeating character after the ith query is performed.
+ */
+
+/**
+* @param {string} s
+* @param {string} queryCharacters
+* @param {number[]} queryIndices
+* @return {number[]}
+*/
+var longestRepeating = function(s, queryCharacters, queryIndices) {
+  const chars = s.split('');
+  const n = chars.length;
+  const k = queryIndices.length;
+  const result = [];
+
+  class Node {
+    constructor() {
+      this.left = 0;
+      this.right = 0;
+      this.max = 0;
+      this.total = 0;
+    }
+  }
+
+  const tree = new Array(4 * n);
+  for (let i = 0; i < 4 * n; i++) {
+    tree[i] = new Node();
+  }
+
+  function buildTree(node, start, end) {
+    if (start === end) {
+      tree[node].left = 1;
+      tree[node].right = 1;
+      tree[node].max = 1;
+      tree[node].total = 1;
+      return;
+    }
+
+    const mid = Math.floor((start + end) / 2);
+    buildTree(2 * node, start, mid);
+    buildTree(2 * node + 1, mid + 1, end);
+
+    updateNode(node, start, end);
+  }
+
+  function updateNode(node, start, end) {
+    const leftChild = 2 * node;
+    const rightChild = 2 * node + 1;
+    const mid = Math.floor((start + end) / 2);
+
+    tree[node].total = tree[leftChild].total + tree[rightChild].total;
+
+    if (tree[leftChild].total === tree[leftChild].left
+        && mid + 1 <= end && chars[mid] === chars[mid + 1]) {
+      tree[node].left = tree[leftChild].total + tree[rightChild].left;
+    } else {
+      tree[node].left = tree[leftChild].left;
+    }
+
+    if (tree[rightChild].total === tree[rightChild].right
+        && mid >= start && chars[mid] === chars[mid + 1]) {
+      tree[node].right = tree[rightChild].total + tree[leftChild].right;
+    } else {
+      tree[node].right = tree[rightChild].right;
+    }
+
+    tree[node].max = Math.max(tree[leftChild].max, tree[rightChild].max);
+
+    if (mid >= start && mid + 1 <= end && chars[mid] === chars[mid + 1]) {
+      tree[node].max = Math.max(tree[node].max,
+        tree[leftChild].right + tree[rightChild].left);
+    }
+  }
+
+  function update(node, start, end, index) {
+    if (index < start || index > end) {
+      return;
+    }
+
+    if (start === end) {
+      return;
+    }
+
+    const mid = Math.floor((start + end) / 2);
+    update(2 * node, start, mid, index);
+    update(2 * node + 1, mid + 1, end, index);
+
+    updateNode(node, start, end);
+  }
+
+  buildTree(1, 0, n - 1);
+
+  for (let q = 0; q < k; q++) {
+    const index = queryIndices[q];
+    const newChar = queryCharacters[q];
+
+    chars[index] = newChar;
+    update(1, 0, n - 1, index);
+
+    result.push(tree[1].max);
+  }
+
+  return result;
+};
diff --git a/solutions/2216-minimum-deletions-to-make-array-beautiful.js b/solutions/2216-minimum-deletions-to-make-array-beautiful.js
new file mode 100644
index 00000000..7cb1012e
--- /dev/null
+++ b/solutions/2216-minimum-deletions-to-make-array-beautiful.js
@@ -0,0 +1,41 @@
+/**
+ * 2216. Minimum Deletions to Make Array Beautiful
+ * https://leetcode.com/problems/minimum-deletions-to-make-array-beautiful/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed integer array nums. The array nums is beautiful if:
+ * - nums.length is even.
+ * - nums[i] != nums[i + 1] for all i % 2 == 0.
+ *
+ * Note that an empty array is considered beautiful.
+ *
+ * You can delete any number of elements from nums. When you delete an element, all the elements
+ * to the right of the deleted element will be shifted one unit to the left to fill the gap
+ * created and all the elements to the left of the deleted element will remain unchanged.
+ *
+ * Return the minimum number of elements to delete from nums to make it beautiful.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var minDeletion = function(nums) {
+  let result = 0;
+  let i = 0;
+
+  while (i < nums.length - 1) {
+    if (nums[i] === nums[i + 1]) {
+      result++;
+      i++;
+    } else {
+      i += 2;
+    }
+  }
+
+  if ((nums.length - result) % 2 !== 0) {
+    result++;
+  }
+
+  return result;
+};
diff --git a/solutions/2217-find-palindrome-with-fixed-length.js b/solutions/2217-find-palindrome-with-fixed-length.js
new file mode 100644
index 00000000..4e6a83aa
--- /dev/null
+++ b/solutions/2217-find-palindrome-with-fixed-length.js
@@ -0,0 +1,40 @@
+/**
+ * 2217. Find Palindrome With Fixed Length
+ * https://leetcode.com/problems/find-palindrome-with-fixed-length/
+ * Difficulty: Medium
+ *
+ * Given an integer array queries and a positive integer intLength, return an array answer where
+ * answer[i] is either the queries[i]th smallest positive palindrome of length intLength or -1
+ * if no such palindrome exists.
+ *
+ * A palindrome is a number that reads the same backwards and forwards. Palindromes cannot have
+ * leading zeros.
+ */
+
+/**
+ * @param {number[]} queries
+ * @param {number} intLength
+ * @return {number[]}
+ */
+var kthPalindrome = function(queries, intLength) {
+  const halfLength = Math.ceil(intLength / 2);
+  const maxPalindromes = Math.pow(10, halfLength - 1) * 9;
+
+  return queries.map(generatePalindrome);
+
+  function generatePalindrome(query) {
+    if (query > maxPalindromes) return -1;
+
+    let firstHalf = Math.pow(10, halfLength - 1) + query - 1;
+    let result = firstHalf;
+
+    if (intLength % 2 === 1) firstHalf = Math.floor(firstHalf / 10);
+
+    while (firstHalf > 0) {
+      result = result * 10 + (firstHalf % 10);
+      firstHalf = Math.floor(firstHalf / 10);
+    }
+
+    return result;
+  }
+};
diff --git a/solutions/2218-maximum-value-of-k-coins-from-piles.js b/solutions/2218-maximum-value-of-k-coins-from-piles.js
new file mode 100644
index 00000000..743b4b17
--- /dev/null
+++ b/solutions/2218-maximum-value-of-k-coins-from-piles.js
@@ -0,0 +1,44 @@
+/**
+ * 2218. Maximum Value of K Coins From Piles
+ * https://leetcode.com/problems/maximum-value-of-k-coins-from-piles/
+ * Difficulty: Hard
+ *
+ * There are n piles of coins on a table. Each pile consists of a positive number of coins of
+ * assorted denominations.
+ *
+ * In one move, you can choose any coin on top of any pile, remove it, and add it to your wallet.
+ *
+ * Given a list piles, where piles[i] is a list of integers denoting the composition of the ith
+ * pile from top to bottom, and a positive integer k, return the maximum total value of coins
+ * you can have in your wallet if you choose exactly k coins optimally.
+ */
+
+/**
+ * @param {number[][]} piles
+ * @param {number} k
+ * @return {number}
+ */
+var maxValueOfCoins = function(piles, k) {
+  const n = piles.length;
+  const dp = Array.from({ length: n + 1 }, () => new Array(k + 1).fill(-1));
+
+  return maximize(0, k);
+
+  function maximize(pileIndex, remainingCoins) {
+    if (pileIndex === n || remainingCoins === 0) return 0;
+    if (dp[pileIndex][remainingCoins] !== -1) return dp[pileIndex][remainingCoins];
+
+    let maxValue = maximize(pileIndex + 1, remainingCoins);
+    let currentSum = 0;
+
+    for (let i = 0; i < Math.min(piles[pileIndex].length, remainingCoins); i++) {
+      currentSum += piles[pileIndex][i];
+      maxValue = Math.max(
+        maxValue,
+        currentSum + maximize(pileIndex + 1, remainingCoins - (i + 1))
+      );
+    }
+
+    return dp[pileIndex][remainingCoins] = maxValue;
+  }
+};
diff --git a/solutions/2220-minimum-bit-flips-to-convert-number.js b/solutions/2220-minimum-bit-flips-to-convert-number.js
new file mode 100644
index 00000000..28e0a9da
--- /dev/null
+++ b/solutions/2220-minimum-bit-flips-to-convert-number.js
@@ -0,0 +1,24 @@
+/**
+ * 2220. Minimum Bit Flips to Convert Number
+ * https://leetcode.com/problems/minimum-bit-flips-to-convert-number/
+ * Difficulty: Easy
+ *
+ * A bit flip of a number x is choosing a bit in the binary representation of x and flipping it
+ * from either 0 to 1 or 1 to 0.
+ * - For example, for x = 7, the binary representation is 111 and we may choose any bit (including
+ *   any leading zeros not shown) and flip it. We can flip the first bit from the right to get 110,
+ *   flip the second bit from the right to get 101, flip the fifth bit from the right (a leading
+ *   zero) to get 10111, etc.
+ *
+ * Given two integers start and goal, return the minimum number of bit flips to convert start to
+ * goal.
+ */
+
+/**
+ * @param {number} start
+ * @param {number} goal
+ * @return {number}
+ */
+var minBitFlips = function(start, goal) {
+  return (start ^ goal).toString(2).replace(/0+/g, '').length;
+};
diff --git a/solutions/2221-find-triangular-sum-of-an-array.js b/solutions/2221-find-triangular-sum-of-an-array.js
new file mode 100644
index 00000000..3ef02fcd
--- /dev/null
+++ b/solutions/2221-find-triangular-sum-of-an-array.js
@@ -0,0 +1,34 @@
+/**
+ * 2221. Find Triangular Sum of an Array
+ * https://leetcode.com/problems/find-triangular-sum-of-an-array/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed integer array nums, where nums[i] is a digit between 0 and
+ * 9 (inclusive).
+ *
+ * The triangular sum of nums is the value of the only element present in nums after the
+ * following process terminates:
+ * 1. Let nums comprise of n elements. If n == 1, end the process. Otherwise, create a new
+ *    0-indexed integer array newNums of length n - 1.
+ * 2. For each index i, where 0 <= i < n - 1, assign the value of newNums[i] as
+ *    (nums[i] + nums[i+1]) % 10, where % denotes modulo operator.
+ * 3. Replace the array nums with newNums.
+ * 4. Repeat the entire process starting from step 1.
+ *
+ * Return the triangular sum of nums.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var triangularSum = function(nums) {
+  while (nums.length > 1) {
+    const updated = [];
+    for (let i = 0; i < nums.length - 1; i++) {
+      updated.push((nums[i] + nums[i + 1]) % 10);
+    }
+    nums = updated;
+  }
+  return nums[0];
+};
diff --git a/solutions/2222-number-of-ways-to-select-buildings.js b/solutions/2222-number-of-ways-to-select-buildings.js
new file mode 100644
index 00000000..012573fc
--- /dev/null
+++ b/solutions/2222-number-of-ways-to-select-buildings.js
@@ -0,0 +1,49 @@
+/**
+ * 2222. Number of Ways to Select Buildings
+ * https://leetcode.com/problems/number-of-ways-to-select-buildings/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed binary string s which represents the types of buildings along
+ * a street where:
+ * - s[i] = '0' denotes that the ith building is an office and
+ * - s[i] = '1' denotes that the ith building is a restaurant.
+ *
+ * As a city official, you would like to select 3 buildings for random inspection. However, to
+ * ensure variety, no two consecutive buildings out of the selected buildings can be of the same
+ * type.
+ * - For example, given s = "001101", we cannot select the 1st, 3rd, and 5th buildings as that
+ *   would form "011" which is not allowed due to having two consecutive buildings of the same type.
+ *
+ * Return the number of valid ways to select 3 buildings.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var numberOfWays = function(s) {
+  const prefixCounts = [[0, 0]];
+  let zeros = 0;
+  let ones = 0;
+
+  for (const char of s) {
+    if (char === '0') zeros++;
+    else ones++;
+    prefixCounts.push([zeros, ones]);
+  }
+
+  let result = 0;
+  for (let i = 1; i < s.length - 1; i++) {
+    if (s[i] === '0') {
+      const leftOnes = prefixCounts[i][1];
+      const rightOnes = prefixCounts[s.length][1] - prefixCounts[i + 1][1];
+      result += leftOnes * rightOnes;
+    } else {
+      const leftZeros = prefixCounts[i][0];
+      const rightZeros = prefixCounts[s.length][0] - prefixCounts[i + 1][0];
+      result += leftZeros * rightZeros;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2223-sum-of-scores-of-built-strings.js b/solutions/2223-sum-of-scores-of-built-strings.js
new file mode 100644
index 00000000..4b5110cc
--- /dev/null
+++ b/solutions/2223-sum-of-scores-of-built-strings.js
@@ -0,0 +1,40 @@
+/**
+ * 2223. Sum of Scores of Built Strings
+ * https://leetcode.com/problems/sum-of-scores-of-built-strings/
+ * Difficulty: Hard
+ *
+ * You are building a string s of length n one character at a time, prepending each new character
+ * to the front of the string. The strings are labeled from 1 to n, where the string with length
+ * i is labeled si.
+ * - For example, for s = "abaca", s1 == "a", s2 == "ca", s3 == "aca", etc.
+ *
+ * The score of si is the length of the longest common prefix between si and sn (Note that s == sn).
+ *
+ * Given the final string s, return the sum of the score of every si.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var sumScores = function(s) {
+  const n = s.length;
+  const z = new Array(n).fill(0);
+  let left = 0;
+  let right = 0;
+
+  for (let i = 1; i < n; i++) {
+    if (i <= right) {
+      z[i] = Math.min(right - i + 1, z[i - left]);
+    }
+    while (i + z[i] < n && s[z[i]] === s[i + z[i]]) {
+      z[i]++;
+    }
+    if (i + z[i] - 1 > right) {
+      left = i;
+      right = i + z[i] - 1;
+    }
+  }
+
+  return z.reduce((sum, val) => sum + val, n);
+};
diff --git a/solutions/2224-minimum-number-of-operations-to-convert-time.js b/solutions/2224-minimum-number-of-operations-to-convert-time.js
new file mode 100644
index 00000000..95b7a154
--- /dev/null
+++ b/solutions/2224-minimum-number-of-operations-to-convert-time.js
@@ -0,0 +1,38 @@
+/**
+ * 2224. Minimum Number of Operations to Convert Time
+ * https://leetcode.com/problems/minimum-number-of-operations-to-convert-time/
+ * Difficulty: Easy
+ *
+ * You are given two strings current and correct representing two 24-hour times.
+ *
+ * 24-hour times are formatted as "HH:MM", where HH is between 00 and 23, and MM is between 00
+ * and 59. The earliest 24-hour time is 00:00, and the latest is 23:59.
+ *
+ * In one operation you can increase the time current by 1, 5, 15, or 60 minutes. You can perform
+ * this operation any number of times.
+ *
+ * Return the minimum number of operations needed to convert current to correct.
+ */
+
+/**
+ * @param {string} current
+ * @param {string} correct
+ * @return {number}
+ */
+var convertTime = function(current, correct) {
+  const toMinutes = time => {
+    const [hours, minutes] = time.split(':').map(Number);
+    return hours * 60 + minutes;
+  };
+
+  let diff = toMinutes(correct) - toMinutes(current);
+  const increments = [60, 15, 5, 1];
+  let result = 0;
+
+  for (const increment of increments) {
+    result += Math.floor(diff / increment);
+    diff %= increment;
+  }
+
+  return result;
+};
diff --git a/solutions/2225-find-players-with-zero-or-one-losses.js b/solutions/2225-find-players-with-zero-or-one-losses.js
new file mode 100644
index 00000000..3182523e
--- /dev/null
+++ b/solutions/2225-find-players-with-zero-or-one-losses.js
@@ -0,0 +1,44 @@
+/**
+ * 2225. Find Players With Zero or One Losses
+ * https://leetcode.com/problems/find-players-with-zero-or-one-losses/
+ * Difficulty: Medium
+ *
+ * You are given an integer array matches where matches[i] = [winneri, loseri] indicates that
+ * the player winneri defeated player loseri in a match.
+ *
+ * Return a list answer of size 2 where:
+ * - answer[0] is a list of all players that have not lost any matches.
+ * - answer[1] is a list of all players that have lost exactly one match.
+ *
+ * The values in the two lists should be returned in increasing order.
+ *
+ * Note:
+ * - You should only consider the players that have played at least one match.
+ * - The testcases will be generated such that no two matches will have the same outcome.
+ */
+
+/**
+ * @param {number[][]} matches
+ * @return {number[][]}
+ */
+var findWinners = function(matches) {
+  const lossCount = new Map();
+
+  for (const [winner, loser] of matches) {
+    lossCount.set(winner, lossCount.get(winner) || 0);
+    lossCount.set(loser, (lossCount.get(loser) || 0) + 1);
+  }
+
+  const noLosses = [];
+  const oneLoss = [];
+
+  for (const [player, losses] of lossCount) {
+    if (losses === 0) noLosses.push(player);
+    else if (losses === 1) oneLoss.push(player);
+  }
+
+  return [
+    noLosses.sort((a, b) => a - b),
+    oneLoss.sort((a, b) => a - b)
+  ];
+};
diff --git a/solutions/2227-encrypt-and-decrypt-strings.js b/solutions/2227-encrypt-and-decrypt-strings.js
new file mode 100644
index 00000000..fd773ffb
--- /dev/null
+++ b/solutions/2227-encrypt-and-decrypt-strings.js
@@ -0,0 +1,77 @@
+/**
+ * 2227. Encrypt and Decrypt Strings
+ * https://leetcode.com/problems/encrypt-and-decrypt-strings/
+ * Difficulty: Hard
+ *
+ * You are given a character array keys containing unique characters and a string array values
+ * containing strings of length 2. You are also given another string array dictionary that
+ * contains all permitted original strings after decryption. You should implement a data structure
+ * that can encrypt or decrypt a 0-indexed string.
+ *
+ * A string is encrypted with the following process:
+ * 1. For each character c in the string, we find the index i satisfying keys[i] == c in keys.
+ * 2. Replace c with values[i] in the string.
+ *
+ * Note that in case a character of the string is not present in keys, the encryption process cannot
+ * be carried out, and an empty string "" is returned.
+ *
+ * A string is decrypted with the following process:
+ * 1. For each substring s of length 2 occurring at an even index in the string, we find an i such
+ *    that values[i] == s. If there are multiple valid i, we choose any one of them. This means a
+ *    string could have multiple possible strings it can decrypt to.
+ * 2. Replace s with keys[i] in the string.
+ *
+ * Implement the Encrypter class:
+ * - Encrypter(char[] keys, String[] values, String[] dictionary) Initializes the Encrypter class
+ *   with keys, values, and dictionary.
+ * - String encrypt(String word1) Encrypts word1 with the encryption process described above and
+ *   returns the encrypted string.
+ * - int decrypt(String word2) Returns the number of possible strings word2 could decrypt to that
+ *   also appear in dictionary.
+ */
+
+/**
+ * @param {character[]} keys
+ * @param {string[]} values
+ * @param {string[]} dictionary
+ */
+var Encrypter = function(keys, values, dictionary) {
+  this.encryptMap = new Map();
+  this.validEncryptions = new Map();
+
+  for (let i = 0; i < keys.length; i++) {
+    this.encryptMap.set(keys[i], values[i]);
+  }
+
+  for (const word of dictionary) {
+    const encrypted = this.encrypt(word);
+    if (encrypted !== '') {
+      this.validEncryptions.set(encrypted, (this.validEncryptions.get(encrypted) || 0) + 1);
+    }
+  }
+};
+
+/**
+ * @param {string} word1
+ * @return {string}
+ */
+Encrypter.prototype.encrypt = function(word1) {
+  let result = '';
+
+  for (const char of word1) {
+    if (!this.encryptMap.has(char)) {
+      return '';
+    }
+    result += this.encryptMap.get(char);
+  }
+
+  return result;
+};
+
+/**
+ * @param {string} word2
+ * @return {number}
+ */
+Encrypter.prototype.decrypt = function(word2) {
+  return this.validEncryptions.get(word2) || 0;
+};
diff --git a/solutions/2232-minimize-result-by-adding-parentheses-to-expression.js b/solutions/2232-minimize-result-by-adding-parentheses-to-expression.js
new file mode 100644
index 00000000..f32f3fd9
--- /dev/null
+++ b/solutions/2232-minimize-result-by-adding-parentheses-to-expression.js
@@ -0,0 +1,50 @@
+/**
+ * 2232. Minimize Result by Adding Parentheses to Expression
+ * https://leetcode.com/problems/minimize-result-by-adding-parentheses-to-expression/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed string expression of the form "+" where 
+ * and  represent positive integers.
+ *
+ * Add a pair of parentheses to expression such that after the addition of parentheses,
+ * expression is a valid mathematical expression and evaluates to the smallest possible
+ * value. The left parenthesis must be added to the left of '+' and the right parenthesis
+ * must be added to the right of '+'.
+ *
+ * Return expression after adding a pair of parentheses such that expression evaluates to the
+ * smallest possible value. If there are multiple answers that yield the same result, return
+ * any of them.
+ *
+ * The input has been generated such that the original value of expression, and the value of
+ * expression after adding any pair of parentheses that meets the requirements fits within
+ * a signed 32-bit integer.
+ */
+
+/**
+ * @param {string} expression
+ * @return {string}
+ */
+var minimizeResult = function(expression) {
+  const plusIndex = expression.indexOf('+');
+  const left = expression.slice(0, plusIndex);
+  const right = expression.slice(plusIndex + 1);
+  let minValue = Infinity;
+  let result = '';
+
+  for (let i = 0; i < left.length; i++) {
+    for (let j = 1; j <= right.length; j++) {
+      const leftPrefix = i === 0 ? 1 : parseInt(left.slice(0, i));
+      const leftNum = parseInt(left.slice(i), 10);
+      const rightNum = parseInt(right.slice(0, j), 10);
+      const rightSuffix = j === right.length ? 1 : parseInt(right.slice(j), 10);
+      const value = leftPrefix * (leftNum + rightNum) * rightSuffix;
+
+      if (value < minValue) {
+        minValue = value;
+        result = `${left.slice(0, i)}(${left.slice(i)}+${right.slice(0, j)})${right.slice(j)}`;
+      }
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2234-maximum-total-beauty-of-the-gardens.js b/solutions/2234-maximum-total-beauty-of-the-gardens.js
new file mode 100644
index 00000000..54985527
--- /dev/null
+++ b/solutions/2234-maximum-total-beauty-of-the-gardens.js
@@ -0,0 +1,99 @@
+/**
+ * 2234. Maximum Total Beauty of the Gardens
+ * https://leetcode.com/problems/maximum-total-beauty-of-the-gardens/
+ * Difficulty: Hard
+ *
+ * Alice is a caretaker of n gardens and she wants to plant flowers to maximize the total
+ * beauty of all her gardens.
+ *
+ * You are given a 0-indexed integer array flowers of size n, where flowers[i] is the number
+ * of flowers already planted in the ith garden. Flowers that are already planted cannot be
+ * removed. You are then given another integer newFlowers, which is the maximum number of
+ * flowers that Alice can additionally plant. You are also given the integers target, full,
+ * and partial.
+ *
+ * A garden is considered complete if it has at least target flowers. The total beauty of the
+ * gardens is then determined as the sum of the following:
+ * - The number of complete gardens multiplied by full.
+ * - The minimum number of flowers in any of the incomplete gardens multiplied by partial.
+ *   If there are no incomplete gardens, then this value will be 0.
+ *
+ * Return the maximum total beauty that Alice can obtain after planting at most newFlowers flowers.
+ */
+
+/**
+ * @param {number[]} flowers
+ * @param {number} newFlowers
+ * @param {number} target
+ * @param {number} full
+ * @param {number} partial
+ * @return {number}
+ */
+var maximumBeauty = function(flowers, newFlowers, target, full, partial) {
+  const n = flowers.length;
+  flowers.sort((a, b) => b - a);
+
+  const suffixCost = Array(n + 1).fill(0);
+  let lastUniqueIndex = n - 1;
+  let minIndex = n - 2;
+
+  for (; minIndex >= 0; --minIndex) {
+    if (flowers[minIndex] >= target) break;
+
+    const flowerDiff = flowers[minIndex] - flowers[minIndex + 1];
+    const gardenCount = n - lastUniqueIndex;
+    suffixCost[minIndex] = suffixCost[minIndex + 1] + flowerDiff * gardenCount;
+
+    if (suffixCost[minIndex] > newFlowers) break;
+
+    if (flowers[minIndex] !== flowers[minIndex - 1]) {
+      lastUniqueIndex = minIndex;
+    }
+  }
+
+  ++minIndex;
+
+  const remainingFlowersForMin = newFlowers - suffixCost[minIndex];
+  const gardenCountForMin = n - minIndex;
+  let minFlowerValue = Math.min(
+    target - 1,
+    flowers[minIndex] + Math.floor(remainingFlowersForMin / gardenCountForMin)
+  );
+
+  let result = 0;
+
+  for (let i = 0; i < n; ++i) {
+    if (flowers[i] >= target) {
+      continue;
+    }
+
+    const currentBeauty = i * full + minFlowerValue * partial;
+    result = Math.max(result, currentBeauty);
+
+    const flowersNeeded = target - flowers[i];
+    if (flowersNeeded > newFlowers) break;
+
+    newFlowers -= flowersNeeded;
+    flowers[i] = target;
+
+    while (
+      minIndex <= i || newFlowers < suffixCost[minIndex]
+      || (minIndex > 0 && flowers[minIndex] === flowers[minIndex - 1])
+    ) {
+      ++minIndex;
+    }
+
+    const updatedRemaining = newFlowers - suffixCost[minIndex];
+    const updatedGardenCount = n - minIndex;
+    minFlowerValue = Math.min(
+      target - 1,
+      flowers[minIndex] + Math.floor(updatedRemaining / updatedGardenCount)
+    );
+  }
+
+  if (flowers[n - 1] >= target) {
+    result = Math.max(result, n * full);
+  }
+
+  return result;
+};
diff --git a/solutions/2236-root-equals-sum-of-children.js b/solutions/2236-root-equals-sum-of-children.js
new file mode 100644
index 00000000..fa9a3cd9
--- /dev/null
+++ b/solutions/2236-root-equals-sum-of-children.js
@@ -0,0 +1,27 @@
+/**
+ * 2236. Root Equals Sum of Children
+ * https://leetcode.com/problems/root-equals-sum-of-children/
+ * Difficulty: Easy
+ *
+ * You are given the root of a binary tree that consists of exactly 3 nodes: the root, its
+ * left child, and its right child.
+ *
+ * Return true if the value of the root is equal to the sum of the values of its two children,
+ * or false otherwise.
+ */
+
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @return {boolean}
+ */
+var checkTree = function(root) {
+  return root.val === root.left.val + root.right.val;
+};
diff --git a/solutions/2239-find-closest-number-to-zero.js b/solutions/2239-find-closest-number-to-zero.js
new file mode 100644
index 00000000..56d07d8e
--- /dev/null
+++ b/solutions/2239-find-closest-number-to-zero.js
@@ -0,0 +1,27 @@
+/**
+ * 2239. Find Closest Number to Zero
+ * https://leetcode.com/problems/find-closest-number-to-zero/
+ * Difficulty: Easy
+ *
+ * Given an integer array nums of size n, return the number with the value closest to 0 in nums.
+ * If there are multiple answers, return the number with the largest value.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var findClosestNumber = function(nums) {
+  let result = nums[0];
+  let minDistance = Math.abs(nums[0]);
+
+  for (const num of nums) {
+    const distance = Math.abs(num);
+    if (distance < minDistance || (distance === minDistance && num > result)) {
+      minDistance = distance;
+      result = num;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2240-number-of-ways-to-buy-pens-and-pencils.js b/solutions/2240-number-of-ways-to-buy-pens-and-pencils.js
new file mode 100644
index 00000000..b55f1cff
--- /dev/null
+++ b/solutions/2240-number-of-ways-to-buy-pens-and-pencils.js
@@ -0,0 +1,31 @@
+/**
+ * 2240. Number of Ways to Buy Pens and Pencils
+ * https://leetcode.com/problems/number-of-ways-to-buy-pens-and-pencils/
+ * Difficulty: Medium
+ *
+ * You are given an integer total indicating the amount of money you have. You are also given
+ * two integers cost1 and cost2 indicating the price of a pen and pencil respectively. You can
+ * spend part or all of your money to buy multiple quantities (or none) of each kind of writing
+ * utensil.
+ *
+ * Return the number of distinct ways you can buy some number of pens and pencils.
+ */
+
+/**
+ * @param {number} total
+ * @param {number} cost1
+ * @param {number} cost2
+ * @return {number}
+ */
+var waysToBuyPensPencils = function(total, cost1, cost2) {
+  let result = 0;
+  const maxPens = Math.floor(total / cost1);
+
+  for (let pens = 0; pens <= maxPens; pens++) {
+    const remaining = total - pens * cost1;
+    const pencils = Math.floor(remaining / cost2);
+    result += pencils + 1;
+  }
+
+  return result;
+};
diff --git a/solutions/2241-design-an-atm-machine.js b/solutions/2241-design-an-atm-machine.js
new file mode 100644
index 00000000..4a5afa87
--- /dev/null
+++ b/solutions/2241-design-an-atm-machine.js
@@ -0,0 +1,64 @@
+/**
+ * 2241. Design an ATM Machine
+ * https://leetcode.com/problems/design-an-atm-machine/
+ * Difficulty: Medium
+ *
+ * There is an ATM machine that stores banknotes of 5 denominations: 20, 50, 100, 200, and
+ * 500 dollars. Initially the ATM is empty. The user can use the machine to deposit or
+ * withdraw any amount of money.
+ *
+ * When withdrawing, the machine prioritizes using banknotes of larger values.
+ * - For example, if you want to withdraw $300 and there are 2 $50 banknotes, 1 $100 banknote,
+ *   and 1 $200 banknote, then the machine will use the $100 and $200 banknotes.
+ * - However, if you try to withdraw $600 and there are 3 $200 banknotes and 1 $500 banknote,
+ *   then the withdraw request will be rejected because the machine will first try to use the
+ *   $500 banknote and then be unable to use banknotes to complete the remaining $100. Note
+ *   that the machine is not allowed to use the $200 banknotes instead of the $500 banknote.
+ *
+ * Implement the ATM class:
+ * - ATM() Initializes the ATM object.
+ * - void deposit(int[] banknotesCount) Deposits new banknotes in the order $20, $50, $100,
+ *   $200, and $500.
+ * - int[] withdraw(int amount) Returns an array of length 5 of the number of banknotes that
+ *   will be handed to the user in the order $20, $50, $100, $200, and $500, and update the
+ *   number of banknotes in the ATM after withdrawing. Returns [-1] if it is not possible (do
+ *   not withdraw any banknotes in this case).
+ */
+
+var ATM = function() {
+  this.denominations = [20, 50, 100, 200, 500];
+  this.notes = [0, 0, 0, 0, 0];
+};
+
+/**
+ * @param {number[]} banknotesCount
+ * @return {void}
+ */
+ATM.prototype.deposit = function(banknotesCount) {
+  for (let i = 0; i < 5; i++) {
+    this.notes[i] += banknotesCount[i];
+  }
+};
+
+/**
+ * @param {number} amount
+ * @return {number[]}
+ */
+ATM.prototype.withdraw = function(amount) {
+  const result = [0, 0, 0, 0, 0];
+  let remaining = amount;
+
+  for (let i = 4; i >= 0; i--) {
+    const count = Math.min(Math.floor(remaining / this.denominations[i]), this.notes[i]);
+    result[i] = count;
+    remaining -= count * this.denominations[i];
+  }
+
+  if (remaining !== 0) return [-1];
+
+  for (let i = 0; i < 5; i++) {
+    this.notes[i] -= result[i];
+  }
+
+  return result;
+};
diff --git a/solutions/2242-maximum-score-of-a-node-sequence.js b/solutions/2242-maximum-score-of-a-node-sequence.js
new file mode 100644
index 00000000..b64c16ce
--- /dev/null
+++ b/solutions/2242-maximum-score-of-a-node-sequence.js
@@ -0,0 +1,55 @@
+/**
+ * 2242. Maximum Score of a Node Sequence
+ * https://leetcode.com/problems/maximum-score-of-a-node-sequence/
+ * Difficulty: Hard
+ *
+ * There is an undirected graph with n nodes, numbered from 0 to n - 1.
+ *
+ * You are given a 0-indexed integer array scores of length n where scores[i] denotes the score
+ * of node i. You are also given a 2D integer array edges where edges[i] = [ai, bi] denotes that
+ * there exists an undirected edge connecting nodes ai and bi.
+ *
+ * A node sequence is valid if it meets the following conditions:
+ * - There is an edge connecting every pair of adjacent nodes in the sequence.
+ * - No node appears more than once in the sequence.
+ *
+ * The score of a node sequence is defined as the sum of the scores of the nodes in the sequence.
+ *
+ * Return the maximum score of a valid node sequence with a length of 4. If no such sequence exists,
+ * return -1.
+ */
+
+/**
+ * @param {number[]} scores
+ * @param {number[][]} edges
+ * @return {number}
+ */
+var maximumScore = function(scores, edges) {
+  const n = scores.length;
+  const graph = new Array(n).fill().map(() => []);
+
+  for (const [a, b] of edges) {
+    graph[a].push(b);
+    graph[b].push(a);
+  }
+
+  for (let i = 0; i < n; i++) {
+    graph[i].sort((a, b) => scores[b] - scores[a]);
+    if (graph[i].length > 3) {
+      graph[i] = graph[i].slice(0, 3);
+    }
+  }
+
+  let result = -1;
+  for (const [a, b] of edges) {
+    for (const c of graph[a]) {
+      if (c === b) continue;
+      for (const d of graph[b]) {
+        if (d === a || d === c) continue;
+        result = Math.max(result, scores[a] + scores[b] + scores[c] + scores[d]);
+      }
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2243-calculate-digit-sum-of-a-string.js b/solutions/2243-calculate-digit-sum-of-a-string.js
new file mode 100644
index 00000000..293438ed
--- /dev/null
+++ b/solutions/2243-calculate-digit-sum-of-a-string.js
@@ -0,0 +1,36 @@
+/**
+ * 2243. Calculate Digit Sum of a String
+ * https://leetcode.com/problems/calculate-digit-sum-of-a-string/
+ * Difficulty: Easy
+ *
+ * You are given a string s consisting of digits and an integer k.
+ *
+ * A round can be completed if the length of s is greater than k. In one round, do the following:
+ * 1. Divide s into consecutive groups of size k such that the first k characters are in the first
+ *    group, the next k characters are in the second group, and so on. Note that the size of the
+ *    last group can be smaller than k.
+ * 2. Replace each group of s with a string representing the sum of all its digits. For example,
+ *    "346" is replaced with "13" because 3 + 4 + 6 = 13.
+ * 3. Merge consecutive groups together to form a new string. If the length of the string is greater
+ *    than k, repeat from step 1.
+ *
+ * Return s after all rounds have been completed.
+ */
+
+/**
+ * @param {string} s
+ * @param {number} k
+ * @return {string}
+ */
+var digitSum = function(s, k) {
+  while (s.length > k) {
+    let next = '';
+    for (let i = 0; i < s.length; i += k) {
+      const group = s.slice(i, i + k);
+      const sum = group.split('').reduce((acc, digit) => acc + Number(digit), 0);
+      next += sum;
+    }
+    s = next;
+  }
+  return s;
+};
diff --git a/solutions/2245-maximum-trailing-zeros-in-a-cornered-path.js b/solutions/2245-maximum-trailing-zeros-in-a-cornered-path.js
new file mode 100644
index 00000000..7f18ed32
--- /dev/null
+++ b/solutions/2245-maximum-trailing-zeros-in-a-cornered-path.js
@@ -0,0 +1,112 @@
+/**
+ * 2245. Maximum Trailing Zeros in a Cornered Path
+ * https://leetcode.com/problems/maximum-trailing-zeros-in-a-cornered-path/
+ * Difficulty: Medium
+ *
+ * You are given a 2D integer array grid of size m x n, where each cell contains a positive integer.
+ *
+ * A cornered path is defined as a set of adjacent cells with at most one turn. More specifically,
+ * the path should exclusively move either horizontally or vertically up to the turn (if there is
+ * one), without returning to a previously visited cell. After the turn, the path will then move
+ * exclusively in the alternate direction: move vertically if it moved horizontally, and vice versa,
+ * also without returning to a previously visited cell.
+ *
+ * The product of a path is defined as the product of all the values in the path.
+ *
+ * Return the maximum number of trailing zeros in the product of a cornered path found in grid.
+ *
+ * Note:
+ * - Horizontal movement means moving in either the left or right direction.
+ * - Vertical movement means moving in either the up or down direction.
+ */
+
+/**
+ * @param {number[][]} grid
+ * @return {number}
+ */
+var maxTrailingZeros = function(grid) {
+  const m = grid.length;
+  const n = grid[0].length;
+
+  function countFactors(num) {
+    let count2 = 0;
+    let count5 = 0;
+
+    while (num % 2 === 0) {
+      count2++;
+      num = Math.floor(num / 2);
+    }
+
+    while (num % 5 === 0) {
+      count5++;
+      num = Math.floor(num / 5);
+    }
+
+    return [count2, count5];
+  }
+
+  const factors = new Array(m).fill().map(() => new Array(n).fill().map(() => [0, 0]));
+  for (let i = 0; i < m; i++) {
+    for (let j = 0; j < n; j++) {
+      factors[i][j] = countFactors(grid[i][j]);
+    }
+  }
+
+  const rowPrefix = new Array(m).fill().map(() => new Array(n + 1).fill().map(() => [0, 0]));
+  for (let i = 0; i < m; i++) {
+    for (let j = 0; j < n; j++) {
+      rowPrefix[i][j + 1][0] = rowPrefix[i][j][0] + factors[i][j][0];
+      rowPrefix[i][j + 1][1] = rowPrefix[i][j][1] + factors[i][j][1];
+    }
+  }
+
+  const colPrefix = new Array(m + 1).fill().map(() => new Array(n).fill().map(() => [0, 0]));
+  for (let j = 0; j < n; j++) {
+    for (let i = 0; i < m; i++) {
+      colPrefix[i + 1][j][0] = colPrefix[i][j][0] + factors[i][j][0];
+      colPrefix[i + 1][j][1] = colPrefix[i][j][1] + factors[i][j][1];
+    }
+  }
+
+  let maxZeros = 0;
+  for (let i = 0; i < m; i++) {
+    for (let j = 0; j < n; j++) {
+      const [count2, count5] = factors[i][j];
+
+      const leftUp = [
+        rowPrefix[i][j][0] + colPrefix[i][j][0],
+        rowPrefix[i][j][1] + colPrefix[i][j][1]
+      ];
+
+      const leftDown = [
+        rowPrefix[i][j][0] + (colPrefix[m][j][0] - colPrefix[i + 1][j][0]),
+        rowPrefix[i][j][1] + (colPrefix[m][j][1] - colPrefix[i + 1][j][1])
+      ];
+
+      const rightUp = [
+        (rowPrefix[i][n][0] - rowPrefix[i][j + 1][0]) + colPrefix[i][j][0],
+        (rowPrefix[i][n][1] - rowPrefix[i][j + 1][1]) + colPrefix[i][j][1]
+      ];
+
+      const rightDown = [
+        (rowPrefix[i][n][0] - rowPrefix[i][j + 1][0])
+          + (colPrefix[m][j][0] - colPrefix[i + 1][j][0]),
+        (rowPrefix[i][n][1] - rowPrefix[i][j + 1][1])
+          + (colPrefix[m][j][1] - colPrefix[i + 1][j][1])
+      ];
+
+      const paths = [
+        [leftUp[0] + count2, leftUp[1] + count5],
+        [leftDown[0] + count2, leftDown[1] + count5],
+        [rightUp[0] + count2, rightUp[1] + count5],
+        [rightDown[0] + count2, rightDown[1] + count5]
+      ];
+
+      for (const [path2, path5] of paths) {
+        maxZeros = Math.max(maxZeros, Math.min(path2, path5));
+      }
+    }
+  }
+
+  return maxZeros;
+};
diff --git a/solutions/2246-longest-path-with-different-adjacent-characters.js b/solutions/2246-longest-path-with-different-adjacent-characters.js
new file mode 100644
index 00000000..4eb8cb53
--- /dev/null
+++ b/solutions/2246-longest-path-with-different-adjacent-characters.js
@@ -0,0 +1,54 @@
+/**
+ * 2246. Longest Path With Different Adjacent Characters
+ * https://leetcode.com/problems/longest-path-with-different-adjacent-characters/
+ * Difficulty: Hard
+ *
+ * You are given a tree (i.e. a connected, undirected graph that has no cycles) rooted at node
+ * 0 consisting of n nodes numbered from 0 to n - 1. The tree is represented by a 0-indexed array
+ * parent of size n, where parent[i] is the parent of node i. Since node 0 is the root,
+ * parent[0] == -1.
+ *
+ * You are also given a string s of length n, where s[i] is the character assigned to node i.
+ *
+ * Return the length of the longest path in the tree such that no pair of adjacent nodes on the path
+ * have the same character assigned to them.
+ */
+
+/**
+ * @param {number[]} parent
+ * @param {string} s
+ * @return {number}
+ */
+var longestPath = function(parent, s) {
+  const n = parent.length;
+  const children = Array.from({ length: n }, () => []);
+  let result = 1;
+
+  for (let i = 1; i < n; i++) {
+    children[parent[i]].push(i);
+  }
+
+  findLongest(0);
+
+  return result;
+
+  function findLongest(node) {
+    let longest = 0;
+    let secondLongest = 0;
+
+    for (const child of children[node]) {
+      const length = findLongest(child);
+      if (s[child] !== s[node]) {
+        if (length > longest) {
+          secondLongest = longest;
+          longest = length;
+        } else if (length > secondLongest) {
+          secondLongest = length;
+        }
+      }
+    }
+
+    result = Math.max(result, longest + secondLongest + 1);
+    return longest + 1;
+  }
+};
diff --git a/solutions/2248-intersection-of-multiple-arrays.js b/solutions/2248-intersection-of-multiple-arrays.js
new file mode 100644
index 00000000..fa162656
--- /dev/null
+++ b/solutions/2248-intersection-of-multiple-arrays.js
@@ -0,0 +1,31 @@
+/**
+ * 2248. Intersection of Multiple Arrays
+ * https://leetcode.com/problems/intersection-of-multiple-arrays/
+ * Difficulty: Easy
+ *
+ * Given a 2D integer array nums where nums[i] is a non-empty array of distinct positive integers,
+ * return the list of integers that are present in each array of nums sorted in ascending order.
+ */
+
+/**
+ * @param {number[][]} nums
+ * @return {number[]}
+ */
+var intersection = function(nums) {
+  const count = new Map();
+
+  for (const array of nums) {
+    for (const num of array) {
+      count.set(num, (count.get(num) || 0) + 1);
+    }
+  }
+
+  const result = [];
+  for (const [num, freq] of count) {
+    if (freq === nums.length) {
+      result.push(num);
+    }
+  }
+
+  return result.sort((a, b) => a - b);
+};
diff --git a/solutions/2249-count-lattice-points-inside-a-circle.js b/solutions/2249-count-lattice-points-inside-a-circle.js
new file mode 100644
index 00000000..2c762806
--- /dev/null
+++ b/solutions/2249-count-lattice-points-inside-a-circle.js
@@ -0,0 +1,33 @@
+/**
+ * 2249. Count Lattice Points Inside a Circle
+ * https://leetcode.com/problems/count-lattice-points-inside-a-circle/
+ * Difficulty: Medium
+ *
+ * Given a 2D integer array circles where circles[i] = [xi, yi, ri] represents the center (xi, yi)
+ * and radius ri of the ith circle drawn on a grid, return the number of lattice points that are
+ * present inside at least one circle.
+ *
+ * Note:
+ * - A lattice point is a point with integer coordinates.
+ * - Points that lie on the circumference of a circle are also considered to be inside it.
+ */
+
+/**
+ * @param {number[][]} circles
+ * @return {number}
+ */
+var countLatticePoints = function(circles) {
+  const set = new Set();
+
+  for (const [x, y, r] of circles) {
+    for (let i = x - r; i <= x + r; i++) {
+      for (let j = y - r; j <= y + r; j++) {
+        if ((x - i) ** 2 + (y - j) ** 2 <= r ** 2) {
+          set.add(`${i},${j}`);
+        }
+      }
+    }
+  }
+
+  return set.size;
+};
diff --git a/solutions/2250-count-number-of-rectangles-containing-each-point.js b/solutions/2250-count-number-of-rectangles-containing-each-point.js
new file mode 100644
index 00000000..79ed186c
--- /dev/null
+++ b/solutions/2250-count-number-of-rectangles-containing-each-point.js
@@ -0,0 +1,62 @@
+/**
+ * 2250. Count Number of Rectangles Containing Each Point
+ * https://leetcode.com/problems/count-number-of-rectangles-containing-each-point/
+ * Difficulty: Medium
+ *
+ * You are given a 2D integer array rectangles where rectangles[i] = [li, hi] indicates that ith
+ * rectangle has a length of li and a height of hi. You are also given a 2D integer array points
+ * where points[j] = [xj, yj] is a point with coordinates (xj, yj).
+ *
+ * The ith rectangle has its bottom-left corner point at the coordinates (0, 0) and its top-right
+ * corner point at (li, hi).
+ *
+ * Return an integer array count of length points.length where count[j] is the number of rectangles
+ * that contain the jth point.
+ *
+ * The ith rectangle contains the jth point if 0 <= xj <= li and 0 <= yj <= hi. Note that points
+ * that lie on the edges of a rectangle are also considered to be contained by that rectangle.
+ */
+
+/**
+ * @param {number[][]} rectangles
+ * @param {number[][]} points
+ * @return {number[]}
+ */
+var countRectangles = function(rectangles, points) {
+  const rectByHeight = new Array(101).fill().map(() => []);
+
+  for (const [length, height] of rectangles) {
+    rectByHeight[height].push(length);
+  }
+
+  for (let h = 0; h <= 100; h++) {
+    rectByHeight[h].sort((a, b) => a - b);
+  }
+
+  const result = new Array(points.length).fill(0);
+  for (let i = 0; i < points.length; i++) {
+    const [x, y] = points[i];
+
+    for (let h = y; h <= 100; h++) {
+      const lengths = rectByHeight[h];
+
+      let left = 0;
+      let right = lengths.length - 1;
+      let insertPos = lengths.length;
+
+      while (left <= right) {
+        const mid = Math.floor((left + right) / 2);
+        if (lengths[mid] >= x) {
+          insertPos = mid;
+          right = mid - 1;
+        } else {
+          left = mid + 1;
+        }
+      }
+
+      result[i] += lengths.length - insertPos;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2251-number-of-flowers-in-full-bloom.js b/solutions/2251-number-of-flowers-in-full-bloom.js
new file mode 100644
index 00000000..c9ae393f
--- /dev/null
+++ b/solutions/2251-number-of-flowers-in-full-bloom.js
@@ -0,0 +1,43 @@
+/**
+ * 2251. Number of Flowers in Full Bloom
+ * https://leetcode.com/problems/number-of-flowers-in-full-bloom/
+ * Difficulty: Hard
+ *
+ * You are given a 0-indexed 2D integer array flowers, where flowers[i] = [starti, endi] means
+ * the ith flower will be in full bloom from starti to endi (inclusive). You are also given a
+ * 0-indexed integer array people of size n, where people[i] is the time that the ith person
+ * will arrive to see the flowers.
+ *
+ * Return an integer array answer of size n, where answer[i] is the number of flowers that are
+ * in full bloom when the ith person arrives.
+ */
+
+/**
+ * @param {number[][]} flowers
+ * @param {number[]} people
+ * @return {number[]}
+ */
+var fullBloomFlowers = function(flowers, people) {
+  const events = [];
+  for (const [start, end] of flowers) {
+    events.push([start, 1]);
+    events.push([end + 1, -1]);
+  }
+  events.sort((a, b) => a[0] - b[0] || a[1] - b[1]);
+
+  const result = new Array(people.length).fill(0);
+  const peopleWithIndex = people.map((time, index) => [time, index]);
+  peopleWithIndex.sort((a, b) => a[0] - b[0]);
+
+  let bloomCount = 0;
+  let eventIndex = 0;
+  for (const [time, personIndex] of peopleWithIndex) {
+    while (eventIndex < events.length && events[eventIndex][0] <= time) {
+      bloomCount += events[eventIndex][1];
+      eventIndex++;
+    }
+    result[personIndex] = bloomCount;
+  }
+
+  return result;
+};
diff --git a/solutions/2255-count-prefixes-of-a-given-string.js b/solutions/2255-count-prefixes-of-a-given-string.js
new file mode 100644
index 00000000..25f3ecbf
--- /dev/null
+++ b/solutions/2255-count-prefixes-of-a-given-string.js
@@ -0,0 +1,30 @@
+/**
+ * 2255. Count Prefixes of a Given String
+ * https://leetcode.com/problems/count-prefixes-of-a-given-string/
+ * Difficulty: Easy
+ *
+ * You are given a string array words and a string s, where words[i] and s comprise only of
+ * lowercase English letters.
+ *
+ * Return the number of strings in words that are a prefix of s.
+ *
+ * A prefix of a string is a substring that occurs at the beginning of the string. A substring
+ * is a contiguous sequence of characters within a string.
+ */
+
+/**
+ * @param {string[]} words
+ * @param {string} s
+ * @return {number}
+ */
+var countPrefixes = function(words, s) {
+  let result = 0;
+
+  for (const word of words) {
+    if (word.length <= s.length && s.startsWith(word)) {
+      result++;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2256-minimum-average-difference.js b/solutions/2256-minimum-average-difference.js
new file mode 100644
index 00000000..8c7ebdde
--- /dev/null
+++ b/solutions/2256-minimum-average-difference.js
@@ -0,0 +1,46 @@
+/**
+ * 2256. Minimum Average Difference
+ * https://leetcode.com/problems/minimum-average-difference/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed integer array nums of length n.
+ *
+ * The average difference of the index i is the absolute difference between the average of the
+ * first i + 1 elements of nums and the average of the last n - i - 1 elements. Both averages
+ * should be rounded down to the nearest integer.
+ *
+ * Return the index with the minimum average difference. If there are multiple such indices,
+ * return the smallest one.
+ *
+ * Note:
+ * - The absolute difference of two numbers is the absolute value of their difference.
+ * - The average of n elements is the sum of the n elements divided (integer division) by n.
+ * - The average of 0 elements is considered to be 0.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var minimumAverageDifference = function(nums) {
+  const n = nums.length;
+  let leftSum = 0;
+  let rightSum = nums.reduce((sum, num) => sum + num, 0);
+  let minDiff = Infinity;
+  let result = 0;
+
+  for (let i = 0; i < n; i++) {
+    leftSum += nums[i];
+    rightSum -= nums[i];
+    const leftAvg = Math.floor(leftSum / (i + 1));
+    const rightAvg = i === n - 1 ? 0 : Math.floor(rightSum / (n - i - 1));
+    const diff = Math.abs(leftAvg - rightAvg);
+
+    if (diff < minDiff) {
+      minDiff = diff;
+      result = i;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2257-count-unguarded-cells-in-the-grid.js b/solutions/2257-count-unguarded-cells-in-the-grid.js
new file mode 100644
index 00000000..87cbde16
--- /dev/null
+++ b/solutions/2257-count-unguarded-cells-in-the-grid.js
@@ -0,0 +1,59 @@
+/**
+ * 2257. Count Unguarded Cells in the Grid
+ * https://leetcode.com/problems/count-unguarded-cells-in-the-grid/
+ * Difficulty: Medium
+ *
+ * You are given two integers m and n representing a 0-indexed m x n grid. You are also given two
+ * 2D integer arrays guards and walls where guards[i] = [rowi, coli] and walls[j] = [rowj, colj]
+ * represent the positions of the ith guard and jth wall respectively.
+ *
+ * A guard can see every cell in the four cardinal directions (north, east, south, or west) starting
+ * from their position unless obstructed by a wall or another guard. A cell is guarded if there is
+ * at least one guard that can see it.
+ *
+ * Return the number of unoccupied cells that are not guarded.
+ */
+
+/**
+ * @param {number} m
+ * @param {number} n
+ * @param {number[][]} guards
+ * @param {number[][]} walls
+ * @return {number}
+ */
+var countUnguarded = function(m, n, guards, walls) {
+  const grid = Array.from({ length: m }, () => Array(n).fill(0));
+
+  for (const [row, col] of walls) {
+    grid[row][col] = 1;
+  }
+  for (const [row, col] of guards) {
+    grid[row][col] = 2;
+  }
+
+  for (const [row, col] of guards) {
+    for (let i = row - 1; i >= 0 && grid[i][col] !== 1 && grid[i][col] !== 2; i--) {
+      grid[i][col] = 3;
+    }
+    for (let i = row + 1; i < m && grid[i][col] !== 1 && grid[i][col] !== 2; i++) {
+      grid[i][col] = 3;
+    }
+    for (let j = col - 1; j >= 0 && grid[row][j] !== 1 && grid[row][j] !== 2; j--) {
+      grid[row][j] = 3;
+    }
+    for (let j = col + 1; j < n && grid[row][j] !== 1 && grid[row][j] !== 2; j++) {
+      grid[row][j] = 3;
+    }
+  }
+
+  let result = 0;
+  for (let i = 0; i < m; i++) {
+    for (let j = 0; j < n; j++) {
+      if (grid[i][j] === 0) {
+        result++;
+      }
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2258-escape-the-spreading-fire.js b/solutions/2258-escape-the-spreading-fire.js
new file mode 100644
index 00000000..7e26720a
--- /dev/null
+++ b/solutions/2258-escape-the-spreading-fire.js
@@ -0,0 +1,118 @@
+/**
+ * 2258. Escape the Spreading Fire
+ * https://leetcode.com/problems/escape-the-spreading-fire/
+ * Difficulty: Hard
+ *
+ * You are given a 0-indexed 2D integer array grid of size m x n which represents a field.
+ * Each cell has one of three values:
+ * - 0 represents grass,
+ * - 1 represents fire,
+ * - 2 represents a wall that you and fire cannot pass through.
+ *
+ * You are situated in the top-left cell, (0, 0), and you want to travel to the safehouse at the
+ * bottom-right cell, (m - 1, n - 1). Every minute, you may move to an adjacent grass cell. After
+ * your move, every fire cell will spread to all adjacent cells that are not walls.
+ *
+ * Return the maximum number of minutes that you can stay in your initial position before moving
+ * while still safely reaching the safehouse. If this is impossible, return -1. If you can always
+ * reach the safehouse regardless of the minutes stayed, return 109.
+ *
+ * Note that even if the fire spreads to the safehouse immediately after you have reached it, it
+ * will be counted as safely reaching the safehouse.
+ *
+ * A cell is adjacent to another cell if the former is directly north, east, south, or west of the
+ * latter (i.e., their sides are touching).
+ */
+
+/**
+ * @param {number[][]} grid
+ * @return {number}
+ */
+var maximumMinutes = function(grid) {
+  const m = grid.length;
+  const n = grid[0].length;
+  const directions = [[0, 1], [1, 0], [0, -1], [-1, 0]];
+  const MAX_ANSWER = 1000000000;
+
+  function isValid(x, y) {
+    return x >= 0 && x < m && y >= 0 && y < n && grid[x][y] === 0;
+  }
+
+  function calculateFireTime() {
+    const fireTime = new Array(m).fill().map(() => new Array(n).fill(Infinity));
+    const queue = [];
+
+    for (let i = 0; i < m; i++) {
+      for (let j = 0; j < n; j++) {
+        if (grid[i][j] === 1) {
+          queue.push([i, j, 0]);
+          fireTime[i][j] = 0;
+        }
+      }
+    }
+
+    while (queue.length > 0) {
+      const [x, y, time] = queue.shift();
+
+      for (const [dx, dy] of directions) {
+        const nx = x + dx;
+        const ny = y + dy;
+
+        if (isValid(nx, ny) && fireTime[nx][ny] === Infinity) {
+          fireTime[nx][ny] = time + 1;
+          queue.push([nx, ny, time + 1]);
+        }
+      }
+    }
+
+    return fireTime;
+  }
+
+  function canReachSafehouse(delay) {
+    const fireTime = calculateFireTime();
+    const visited = new Array(m).fill().map(() => new Array(n).fill(false));
+    const queue = [[0, 0, delay]];
+    visited[0][0] = true;
+
+    while (queue.length > 0) {
+      const [x, y, time] = queue.shift();
+
+      for (const [dx, dy] of directions) {
+        const nx = x + dx;
+        const ny = y + dy;
+
+        if (!isValid(nx, ny) || visited[nx][ny]) continue;
+
+        if (nx === m - 1 && ny === n - 1) {
+          if (time + 1 <= fireTime[nx][ny] || fireTime[nx][ny] === Infinity) {
+            return true;
+          }
+        }
+
+        if (time + 1 < fireTime[nx][ny]) {
+          visited[nx][ny] = true;
+          queue.push([nx, ny, time + 1]);
+        }
+      }
+    }
+
+    return false;
+  }
+
+  let left = 0;
+  let right = MAX_ANSWER;
+  let result = -1;
+
+  while (left <= right) {
+    const mid = Math.floor((left + right) / 2);
+
+    if (canReachSafehouse(mid)) {
+      result = mid;
+      left = mid + 1;
+    } else {
+      right = mid - 1;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2259-remove-digit-from-number-to-maximize-result.js b/solutions/2259-remove-digit-from-number-to-maximize-result.js
new file mode 100644
index 00000000..d9ee4397
--- /dev/null
+++ b/solutions/2259-remove-digit-from-number-to-maximize-result.js
@@ -0,0 +1,31 @@
+/**
+ * 2259. Remove Digit From Number to Maximize Result
+ * https://leetcode.com/problems/remove-digit-from-number-to-maximize-result/
+ * Difficulty: Easy
+ *
+ * You are given a string number representing a positive integer and a character digit.
+ *
+ * Return the resulting string after removing exactly one occurrence of digit from number such that
+ * the value of the resulting string in decimal form is maximized. The test cases are generated such
+ * that digit occurs at least once in number.
+ */
+
+/**
+ * @param {string} number
+ * @param {character} digit
+ * @return {string}
+ */
+var removeDigit = function(number, digit) {
+  let result = '';
+
+  for (let i = 0; i < number.length; i++) {
+    if (number[i] === digit) {
+      const candidate = number.slice(0, i) + number.slice(i + 1);
+      if (candidate > result) {
+        result = candidate;
+      }
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2260-minimum-consecutive-cards-to-pick-up.js b/solutions/2260-minimum-consecutive-cards-to-pick-up.js
new file mode 100644
index 00000000..4463e10c
--- /dev/null
+++ b/solutions/2260-minimum-consecutive-cards-to-pick-up.js
@@ -0,0 +1,29 @@
+/**
+ * 2260. Minimum Consecutive Cards to Pick Up
+ * https://leetcode.com/problems/minimum-consecutive-cards-to-pick-up/
+ * Difficulty: Medium
+ *
+ * You are given an integer array cards where cards[i] represents the value of the ith card. A pair
+ * of cards are matching if the cards have the same value.
+ *
+ * Return the minimum number of consecutive cards you have to pick up to have a pair of matching
+ * cards among the picked cards. If it is impossible to have matching cards, return -1.
+ */
+
+/**
+ * @param {number[]} cards
+ * @return {number}
+ */
+var minimumCardPickup = function(cards) {
+  const map = new Map();
+  let minLength = Infinity;
+
+  for (let i = 0; i < cards.length; i++) {
+    if (map.has(cards[i])) {
+      minLength = Math.min(minLength, i - map.get(cards[i]) + 1);
+    }
+    map.set(cards[i], i);
+  }
+
+  return minLength === Infinity ? -1 : minLength;
+};
diff --git a/solutions/2261-k-divisible-elements-subarrays.js b/solutions/2261-k-divisible-elements-subarrays.js
new file mode 100644
index 00000000..72d781d1
--- /dev/null
+++ b/solutions/2261-k-divisible-elements-subarrays.js
@@ -0,0 +1,41 @@
+/**
+ * 2261. K Divisible Elements Subarrays
+ * https://leetcode.com/problems/k-divisible-elements-subarrays/
+ * Difficulty: Medium
+ *
+ * Given an integer array nums and two integers k and p, return the number of distinct subarrays,
+ * which have at most k elements that are divisible by p.
+ *
+ * Two arrays nums1 and nums2 are said to be distinct if:
+ * - They are of different lengths, or
+ * - There exists at least one index i where nums1[i] != nums2[i].
+ *
+ * A subarray is defined as a non-empty contiguous sequence of elements in an array.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @param {number} p
+ * @return {number}
+ */
+var countDistinct = function(nums, k, p) {
+  const n = nums.length;
+  const seen = new Set();
+
+  for (let start = 0; start < n; start++) {
+    const subarray = [];
+    let divisibleCount = 0;
+
+    for (let end = start; end < n; end++) {
+      subarray.push(nums[end]);
+      if (nums[end] % p === 0) divisibleCount++;
+
+      if (divisibleCount <= k) {
+        seen.add(subarray.join(','));
+      }
+    }
+  }
+
+  return seen.size;
+};
diff --git a/solutions/2262-total-appeal-of-a-string.js b/solutions/2262-total-appeal-of-a-string.js
new file mode 100644
index 00000000..20c2e2c3
--- /dev/null
+++ b/solutions/2262-total-appeal-of-a-string.js
@@ -0,0 +1,31 @@
+/**
+ * 2262. Total Appeal of A String
+ * https://leetcode.com/problems/total-appeal-of-a-string/
+ * Difficulty: Hard
+ *
+ * The appeal of a string is the number of distinct characters found in the string.
+ * - For example, the appeal of "abbca" is 3 because it has 3 distinct characters:
+ *   'a', 'b', and 'c'.
+ *
+ * Given a string s, return the total appeal of all of its substrings.
+ *
+ * A substring is a contiguous sequence of characters within a string.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var appealSum = function(s) {
+  const n = s.length;
+  const lastSeen = new Array(26).fill(-1);
+  let result = 0;
+
+  for (let i = 0; i < n; i++) {
+    const charIndex = s.charCodeAt(i) - 97;
+    result += (i - lastSeen[charIndex]) * (n - i);
+    lastSeen[charIndex] = i;
+  }
+
+  return result;
+};
diff --git a/solutions/2264-largest-3-same-digit-number-in-string.js b/solutions/2264-largest-3-same-digit-number-in-string.js
new file mode 100644
index 00000000..ef1df3a9
--- /dev/null
+++ b/solutions/2264-largest-3-same-digit-number-in-string.js
@@ -0,0 +1,35 @@
+/**
+ * 2264. Largest 3-Same-Digit Number in String
+ * https://leetcode.com/problems/largest-3-same-digit-number-in-string/
+ * Difficulty: Easy
+ *
+ * You are given a string num representing a large integer. An integer is good if it meets
+ * the following conditions:
+ * - It is a substring of num with length 3.
+ * - It consists of only one unique digit.
+ *
+ * Return the maximum good integer as a string or an empty string "" if no such integer exists.
+ *
+ * Note:
+ * - A substring is a contiguous sequence of characters within a string.
+ * - There may be leading zeroes in num or a good integer.
+ */
+
+/**
+ * @param {string} num
+ * @return {string}
+ */
+var largestGoodInteger = function(num) {
+  let result = '';
+
+  for (let i = 0; i <= num.length - 3; i++) {
+    const substring = num.slice(i, i + 3);
+    if (substring[0] === substring[1] && substring[1] === substring[2]) {
+      if (substring > result) {
+        result = substring;
+      }
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2265-count-nodes-equal-to-average-of-subtree.js b/solutions/2265-count-nodes-equal-to-average-of-subtree.js
new file mode 100644
index 00000000..2e3f4cb5
--- /dev/null
+++ b/solutions/2265-count-nodes-equal-to-average-of-subtree.js
@@ -0,0 +1,46 @@
+/**
+ * 2265. Count Nodes Equal to Average of Subtree
+ * https://leetcode.com/problems/count-nodes-equal-to-average-of-subtree/
+ * Difficulty: Medium
+ *
+ * Given the root of a binary tree, return the number of nodes where the value of the node is
+ * equal to the average of the values in its subtree.
+ *
+ * Note:
+ * - The average of n elements is the sum of the n elements divided by n and rounded down to
+ *   the nearest integer.
+ * - A subtree of root is a tree consisting of root and all of its descendants.
+ */
+
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @return {number}
+ */
+var averageOfSubtree = function(root) {
+  let count = 0;
+  traverse(root);
+  return count;
+
+  function traverse(node) {
+    if (!node) return [0, 0];
+
+    const [leftSum, leftCount] = traverse(node.left);
+    const [rightSum, rightCount] = traverse(node.right);
+    const totalSum = leftSum + rightSum + node.val;
+    const totalCount = leftCount + rightCount + 1;
+
+    if (Math.floor(totalSum / totalCount) === node.val) {
+      count++;
+    }
+
+    return [totalSum, totalCount];
+  }
+};
diff --git a/solutions/2266-count-number-of-texts.js b/solutions/2266-count-number-of-texts.js
new file mode 100644
index 00000000..d2f9cf84
--- /dev/null
+++ b/solutions/2266-count-number-of-texts.js
@@ -0,0 +1,57 @@
+/**
+ * 2266. Count Number of Texts
+ * https://leetcode.com/problems/count-number-of-texts/
+ * Difficulty: Medium
+ *
+ * Alice is texting Bob using her phone. The mapping of digits to letters is shown in the
+ * figure below.
+ *
+ * In order to add a letter, Alice has to press the key of the corresponding digit i times,
+ * where i is the position of the letter in the key.
+ * - For example, to add the letter 's', Alice has to press '7' four times. Similarly, to
+ *   add the letter 'k', Alice has to press '5' twice.
+ * - Note that the digits '0' and '1' do not map to any letters, so Alice does not use them.
+ *
+ * However, due to an error in transmission, Bob did not receive Alice's text message but
+ * received a string of pressed keys instead.
+ * - For example, when Alice sent the message "bob", Bob received the string "2266622".
+ *
+ * Given a string pressedKeys representing the string received by Bob, return the total
+ * number of possible text messages Alice could have sent.
+ *
+ * Since the answer may be very large, return it modulo 109 + 7.
+ */
+
+/**
+ * @param {string} pressedKeys
+ * @return {number}
+ */
+var countTexts = function(pressedKeys) {
+  const mod = 1e9 + 7;
+  const maxGroup = pressedKeys.length;
+  const dp = new Array(maxGroup + 1).fill(0);
+  dp[0] = 1;
+
+  const map = new Map([
+    ['2', 3], ['3', 3], ['4', 3], ['5', 3],
+    ['6', 3], ['7', 4], ['8', 3], ['9', 4]
+  ]);
+
+  for (let i = 1; i <= maxGroup; i++) {
+    dp[i] = dp[i - 1];
+    if (i >= 2 && pressedKeys[i - 1] === pressedKeys[i - 2]) {
+      dp[i] = (dp[i] + dp[i - 2]) % mod;
+    }
+    if (i >= 3 && pressedKeys[i - 1] === pressedKeys[i - 2]
+        && pressedKeys[i - 2] === pressedKeys[i - 3]) {
+      dp[i] = (dp[i] + dp[i - 3]) % mod;
+    }
+    if (i >= 4 && pressedKeys[i - 1] === pressedKeys[i - 2]
+        && pressedKeys[i - 2] === pressedKeys[i - 3]
+        && pressedKeys[i - 3] === pressedKeys[i - 4] && map.get(pressedKeys[i - 1]) === 4) {
+      dp[i] = (dp[i] + dp[i - 4]) % mod;
+    }
+  }
+
+  return dp[maxGroup];
+};
diff --git a/solutions/2267-check-if-there-is-a-valid-parentheses-string-path.js b/solutions/2267-check-if-there-is-a-valid-parentheses-string-path.js
new file mode 100644
index 00000000..8dc163ed
--- /dev/null
+++ b/solutions/2267-check-if-there-is-a-valid-parentheses-string-path.js
@@ -0,0 +1,57 @@
+/**
+ * 2267. Check if There Is a Valid Parentheses String Path
+ * https://leetcode.com/problems/check-if-there-is-a-valid-parentheses-string-path/
+ * Difficulty: Hard
+ *
+ * A parentheses string is a non-empty string consisting only of '(' and ')'. It is valid if any of
+ * the following conditions is true:
+ * - It is ().
+ * - It can be written as AB (A concatenated with B), where A and B are valid parentheses strings.
+ * - It can be written as (A), where A is a valid parentheses string.
+ *
+ * You are given an m x n matrix of parentheses grid. A valid parentheses string path in the grid
+ * is a path satisfying all of the following conditions:
+ * - The path starts from the upper left cell (0, 0).
+ * - The path ends at the bottom-right cell (m - 1, n - 1).
+ * - The path only ever moves down or right.
+ * - The resulting parentheses string formed by the path is valid.
+ *
+ * Return true if there exists a valid parentheses string path in the grid. Otherwise, return false.
+ */
+
+/**
+* @param {character[][]} grid
+* @return {boolean}
+*/
+var hasValidPath = function(grid) {
+  const m = grid.length;
+  const n = grid[0].length;
+
+  if ((m + n - 1) % 2 !== 0) {
+    return false;
+  }
+
+  if (grid[0][0] === ')' || grid[m - 1][n - 1] === '(') {
+    return false;
+  }
+
+  const visited = new Set();
+  return dfs(0, 0, 0);
+
+  function dfs(i, j, openCount) {
+    if (i >= m || j >= n || openCount < 0) {
+      return false;
+    }
+    openCount += grid[i][j] === '(' ? 1 : -1;
+    if (i === m - 1 && j === n - 1) {
+      return openCount === 0;
+    }
+    const key = `${i},${j},${openCount}`;
+    if (visited.has(key)) {
+      return false;
+    }
+    visited.add(key);
+
+    return dfs(i + 1, j, openCount) || dfs(i, j + 1, openCount);
+  }
+};
diff --git a/solutions/2269-find-the-k-beauty-of-a-number.js b/solutions/2269-find-the-k-beauty-of-a-number.js
new file mode 100644
index 00000000..3027e4d4
--- /dev/null
+++ b/solutions/2269-find-the-k-beauty-of-a-number.js
@@ -0,0 +1,37 @@
+/**
+ * 2269. Find the K-Beauty of a Number
+ * https://leetcode.com/problems/find-the-k-beauty-of-a-number/
+ * Difficulty: Easy
+ *
+ * The k-beauty of an integer num is defined as the number of substrings of num when it is
+ * read as a string that meet the following conditions:
+ * - It has a length of k.
+ * - It is a divisor of num.
+ *
+ * Given integers num and k, return the k-beauty of num.
+ *
+ * Note:
+ * - Leading zeros are allowed.
+ * - 0 is not a divisor of any value.
+ *
+ * A substring is a contiguous sequence of characters in a string.
+ */
+
+/**
+ * @param {number} num
+ * @param {number} k
+ * @return {number}
+ */
+var divisorSubstrings = function(num, k) {
+  const numStr = num.toString();
+  let result = 0;
+
+  for (let i = 0; i <= numStr.length - k; i++) {
+    const substring = parseInt(numStr.slice(i, i + k));
+    if (substring !== 0 && num % substring === 0) {
+      result++;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2271-maximum-white-tiles-covered-by-a-carpet.js b/solutions/2271-maximum-white-tiles-covered-by-a-carpet.js
new file mode 100644
index 00000000..0af443fc
--- /dev/null
+++ b/solutions/2271-maximum-white-tiles-covered-by-a-carpet.js
@@ -0,0 +1,50 @@
+/**
+ * 2271. Maximum White Tiles Covered by a Carpet
+ * https://leetcode.com/problems/maximum-white-tiles-covered-by-a-carpet/
+ * Difficulty: Medium
+ *
+ * You are given a 2D integer array tiles where tiles[i] = [li, ri] represents that every tile
+ * j in the range li <= j <= ri is colored white.
+ *
+ * You are also given an integer carpetLen, the length of a single carpet that can be placed
+ * anywhere.
+ *
+ * Return the maximum number of white tiles that can be covered by the carpet.
+ */
+
+/**
+ * @param {number[][]} tiles
+ * @param {number} carpetLen
+ * @return {number}
+ */
+var maximumWhiteTiles = function(tiles, carpetLen) {
+  tiles.sort((a, b) => a[0] - b[0]);
+
+  let result = 0;
+  let covered = 0;
+  let j = 0;
+
+  for (let i = 0; i < tiles.length; i++) {
+    const carpetEnd = tiles[i][0] + carpetLen - 1;
+
+    while (j < tiles.length && tiles[j][0] <= carpetEnd) {
+      if (tiles[j][1] <= carpetEnd) {
+        covered += tiles[j][1] - tiles[j][0] + 1;
+        j++;
+      } else {
+        covered += carpetEnd - tiles[j][0] + 1;
+        result = Math.max(result, covered);
+        covered -= carpetEnd - tiles[j][0] + 1;
+        break;
+      }
+    }
+
+    result = Math.max(result, covered);
+
+    if (j === tiles.length) break;
+
+    covered -= tiles[i][1] - tiles[i][0] + 1;
+  }
+
+  return result;
+};
diff --git a/solutions/2272-substring-with-largest-variance.js b/solutions/2272-substring-with-largest-variance.js
new file mode 100644
index 00000000..f11361c2
--- /dev/null
+++ b/solutions/2272-substring-with-largest-variance.js
@@ -0,0 +1,84 @@
+/**
+ * 2272. Substring With Largest Variance
+ * https://leetcode.com/problems/substring-with-largest-variance/
+ * Difficulty: Hard
+ *
+ * The variance of a string is defined as the largest difference between the number of occurrences
+ * of any 2 characters present in the string. Note the two characters may or may not be the same.
+ *
+ * Given a string s consisting of lowercase English letters only, return the largest variance
+ * possible among all substrings of s.
+ *
+ * A substring is a contiguous sequence of characters within a string.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var largestVariance = function(s) {
+  const map = new Set(s);
+  let result = 0;
+
+  for (const charA of map) {
+    for (const charB of map) {
+      if (charA === charB) continue;
+
+      let countA = 0;
+      let countB = 0;
+      let hasA = false;
+      let hasB = false;
+
+      for (const char of s) {
+        if (char === charA) {
+          countA++;
+          hasA = true;
+        }
+        if (char === charB) {
+          countB++;
+          hasB = true;
+        }
+
+        if (hasA && hasB) {
+          result = Math.max(result, Math.abs(countA - countB));
+        }
+
+        if (countA < countB) {
+          countA = 0;
+          countB = 0;
+          hasA = false;
+          hasB = false;
+        }
+      }
+
+      countA = 0;
+      countB = 0;
+      hasA = false;
+      hasB = false;
+
+      for (let i = s.length - 1; i >= 0; i--) {
+        if (s[i] === charA) {
+          countA++;
+          hasA = true;
+        }
+        if (s[i] === charB) {
+          countB++;
+          hasB = true;
+        }
+
+        if (hasA && hasB) {
+          result = Math.max(result, Math.abs(countA - countB));
+        }
+
+        if (countA < countB) {
+          countA = 0;
+          countB = 0;
+          hasA = false;
+          hasB = false;
+        }
+      }
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2273-find-resultant-array-after-removing-anagrams.js b/solutions/2273-find-resultant-array-after-removing-anagrams.js
new file mode 100644
index 00000000..8e0ef746
--- /dev/null
+++ b/solutions/2273-find-resultant-array-after-removing-anagrams.js
@@ -0,0 +1,38 @@
+/**
+ * 2273. Find Resultant Array After Removing Anagrams
+ * https://leetcode.com/problems/find-resultant-array-after-removing-anagrams/
+ * Difficulty: Easy
+ *
+ * You are given a 0-indexed string array words, where words[i] consists of lowercase
+ * English letters.
+ *
+ * In one operation, select any index i such that 0 < i < words.length and words[i - 1] and
+ * words[i] are anagrams, and delete words[i] from words. Keep performing this operation as
+ * long as you can select an index that satisfies the conditions.
+ *
+ * Return words after performing all operations. It can be shown that selecting the indices
+ * for each operation in any arbitrary order will lead to the same result.
+ *
+ * An Anagram is a word or phrase formed by rearranging the letters of a different word or
+ * phrase using all the original letters exactly once. For example, "dacb" is an anagram of
+ * "abdc".
+ */
+
+/**
+ * @param {string[]} words
+ * @return {string[]}
+ */
+var removeAnagrams = function(words) {
+  const result = [words[0]];
+
+  for (let i = 1; i < words.length; i++) {
+    const prevSorted = result[result.length - 1].split('').sort().join('');
+    const currSorted = words[i].split('').sort().join('');
+
+    if (prevSorted !== currSorted) {
+      result.push(words[i]);
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2274-maximum-consecutive-floors-without-special-floors.js b/solutions/2274-maximum-consecutive-floors-without-special-floors.js
new file mode 100644
index 00000000..6ccb00f4
--- /dev/null
+++ b/solutions/2274-maximum-consecutive-floors-without-special-floors.js
@@ -0,0 +1,31 @@
+/**
+ * 2274. Maximum Consecutive Floors Without Special Floors
+ * https://leetcode.com/problems/maximum-consecutive-floors-without-special-floors/
+ * Difficulty: Medium
+ *
+ * Alice manages a company and has rented some floors of a building as office space. Alice has
+ * decided some of these floors should be special floors, used for relaxation only.
+ *
+ * You are given two integers bottom and top, which denote that Alice has rented all the floors
+ * from bottom to top (inclusive). You are also given the integer array special, where special[i]
+ * denotes a special floor that Alice has designated for relaxation.
+ *
+ * Return the maximum number of consecutive floors without a special floor.
+ */
+
+/**
+ * @param {number} bottom
+ * @param {number} top
+ * @param {number[]} special
+ * @return {number}
+ */
+var maxConsecutive = function(bottom, top, special) {
+  special.sort((a, b) => a - b);
+  let result = Math.max(special[0] - bottom, top - special[special.length - 1]);
+
+  for (let i = 1; i < special.length; i++) {
+    result = Math.max(result, special[i] - special[i - 1] - 1);
+  }
+
+  return result;
+};
diff --git a/solutions/2275-largest-combination-with-bitwise-and-greater-than-zero.js b/solutions/2275-largest-combination-with-bitwise-and-greater-than-zero.js
new file mode 100644
index 00000000..e75f68d3
--- /dev/null
+++ b/solutions/2275-largest-combination-with-bitwise-and-greater-than-zero.js
@@ -0,0 +1,37 @@
+/**
+ * 2275. Largest Combination With Bitwise AND Greater Than Zero
+ * https://leetcode.com/problems/largest-combination-with-bitwise-and-greater-than-zero/
+ * Difficulty: Medium
+ *
+ * The bitwise AND of an array nums is the bitwise AND of all integers in nums.
+ * - For example, for nums = [1, 5, 3], the bitwise AND is equal to 1 & 5 & 3 = 1.
+ * - Also, for nums = [7], the bitwise AND is 7.
+ *
+ * You are given an array of positive integers candidates. Compute the bitwise AND for all
+ * possible combinations of elements in the candidates array.
+ *
+ * Return the size of the largest combination of candidates with a bitwise AND greater than 0.
+ */
+
+/**
+ * @param {number[]} candidates
+ * @return {number}
+ */
+var largestCombination = function(candidates) {
+  const bitCounts = new Array(32).fill(0);
+  let result = 0;
+
+  for (const num of candidates) {
+    for (let i = 0; i < 32; i++) {
+      if (num & (1 << i)) {
+        bitCounts[i]++;
+      }
+    }
+  }
+
+  for (const count of bitCounts) {
+    result = Math.max(result, count);
+  }
+
+  return result;
+};
diff --git a/solutions/2278-percentage-of-letter-in-string.js b/solutions/2278-percentage-of-letter-in-string.js
new file mode 100644
index 00000000..fb3dfdbb
--- /dev/null
+++ b/solutions/2278-percentage-of-letter-in-string.js
@@ -0,0 +1,17 @@
+/**
+ * 2278. Percentage of Letter in String
+ * https://leetcode.com/problems/percentage-of-letter-in-string/
+ * Difficulty: Easy
+ *
+ * Given a string s and a character letter, return the percentage of characters in s that equal
+ * letter rounded down to the nearest whole percent.
+ */
+
+/**
+ * @param {string} s
+ * @param {character} letter
+ * @return {number}
+ */
+var percentageLetter = function(s, letter) {
+  return Math.floor((s.split('').filter(char => char === letter).length / s.length) * 100);
+};
diff --git a/solutions/2279-maximum-bags-with-full-capacity-of-rocks.js b/solutions/2279-maximum-bags-with-full-capacity-of-rocks.js
new file mode 100644
index 00000000..77d36c1a
--- /dev/null
+++ b/solutions/2279-maximum-bags-with-full-capacity-of-rocks.js
@@ -0,0 +1,38 @@
+/**
+ * 2279. Maximum Bags With Full Capacity of Rocks
+ * https://leetcode.com/problems/maximum-bags-with-full-capacity-of-rocks/
+ * Difficulty: Medium
+ *
+ * You have n bags numbered from 0 to n - 1. You are given two 0-indexed integer arrays capacity
+ * and rocks. The ith bag can hold a maximum of capacity[i] rocks and currently contains rocks[i]
+ * rocks. You are also given an integer additionalRocks, the number of additional rocks you can
+ * place in any of the bags.
+ *
+ * Return the maximum number of bags that could have full capacity after placing the additional
+ * rocks in some bags.
+ */
+
+/**
+ * @param {number[]} capacity
+ * @param {number[]} rocks
+ * @param {number} additionalRocks
+ * @return {number}
+ */
+var maximumBags = function(capacity, rocks, additionalRocks) {
+  const remaining = capacity.map((cap, i) => cap - rocks[i]);
+  remaining.sort((a, b) => a - b);
+
+  let result = 0;
+  let rocksLeft = additionalRocks;
+
+  for (const needed of remaining) {
+    if (needed <= rocksLeft) {
+      result++;
+      rocksLeft -= needed;
+    } else {
+      break;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2280-minimum-lines-to-represent-a-line-chart.js b/solutions/2280-minimum-lines-to-represent-a-line-chart.js
new file mode 100644
index 00000000..ff401c32
--- /dev/null
+++ b/solutions/2280-minimum-lines-to-represent-a-line-chart.js
@@ -0,0 +1,38 @@
+/**
+ * 2280. Minimum Lines to Represent a Line Chart
+ * https://leetcode.com/problems/minimum-lines-to-represent-a-line-chart/
+ * Difficulty: Medium
+ *
+ * You are given a 2D integer array stockPrices where stockPrices[i] = [dayi, pricei] indicates
+ * the price of the stock on day dayi is pricei. A line chart is created from the array by
+ * plotting the points on an XY plane with the X-axis representing the day and the Y-axis
+ * representing the price and connecting adjacent points. One such example is shown below.
+ *
+ * Return the minimum number of lines needed to represent the line chart.
+ */
+
+/**
+ * @param {number[][]} stockPrices
+ * @return {number}
+ */
+var minimumLines = function(stockPrices) {
+  if (stockPrices.length <= 2) return stockPrices.length - 1;
+
+  stockPrices.sort((a, b) => a[0] - b[0]);
+
+  let lines = 1;
+  for (let i = 2; i < stockPrices.length; i++) {
+    const [x0, y0] = stockPrices[i - 2];
+    const [x1, y1] = stockPrices[i - 1];
+    const [x2, y2] = stockPrices[i];
+
+    const dx1 = BigInt(x1 - x0);
+    const dy1 = BigInt(y1 - y0);
+    const dx2 = BigInt(x2 - x1);
+    const dy2 = BigInt(y2 - y1);
+
+    if (dy1 * dx2 !== dy2 * dx1) lines++;
+  }
+
+  return lines;
+};
diff --git a/solutions/2283-check-if-number-has-equal-digit-count-and-digit-value.js b/solutions/2283-check-if-number-has-equal-digit-count-and-digit-value.js
new file mode 100644
index 00000000..ea342877
--- /dev/null
+++ b/solutions/2283-check-if-number-has-equal-digit-count-and-digit-value.js
@@ -0,0 +1,30 @@
+/**
+ * 2283. Check if Number Has Equal Digit Count and Digit Value
+ * https://leetcode.com/problems/check-if-number-has-equal-digit-count-and-digit-value/
+ * Difficulty: Easy
+ *
+ * You are given a 0-indexed string num of length n consisting of digits.
+ *
+ * Return true if for every index i in the range 0 <= i < n, the digit i occurs num[i] times in num,
+ * otherwise return false.
+ */
+
+/**
+ * @param {string} num
+ * @return {boolean}
+ */
+var digitCount = function(num) {
+  const frequency = new Array(10).fill(0);
+
+  for (const digit of num) {
+    frequency[digit]++;
+  }
+
+  for (let i = 0; i < num.length; i++) {
+    if (frequency[i] !== Number(num[i])) {
+      return false;
+    }
+  }
+
+  return true;
+};
diff --git a/solutions/2284-sender-with-largest-word-count.js b/solutions/2284-sender-with-largest-word-count.js
new file mode 100644
index 00000000..d4c02260
--- /dev/null
+++ b/solutions/2284-sender-with-largest-word-count.js
@@ -0,0 +1,44 @@
+/**
+ * 2284. Sender With Largest Word Count
+ * https://leetcode.com/problems/sender-with-largest-word-count/
+ * Difficulty: Medium
+ *
+ * You have a chat log of n messages. You are given two string arrays messages and senders
+ * where messages[i] is a message sent by senders[i].
+ *
+ * A message is list of words that are separated by a single space with no leading or trailing
+ * spaces. The word count of a sender is the total number of words sent by the sender. Note
+ * that a sender may send more than one message.
+ *
+ * Return the sender with the largest word count. If there is more than one sender with the
+ * largest word count, return the one with the lexicographically largest name.
+ *
+ * Note:
+ * - Uppercase letters come before lowercase letters in lexicographical order.
+ * - "Alice" and "alice" are distinct.
+ */
+
+/**
+ * @param {string[]} messages
+ * @param {string[]} senders
+ * @return {string}
+ */
+var largestWordCount = function(messages, senders) {
+  const map = new Map();
+  let maxWords = 0;
+  let result = '';
+
+  for (let i = 0; i < messages.length; i++) {
+    const wordCount = messages[i].split(' ').length;
+    const sender = senders[i];
+    const totalWords = (map.get(sender) || 0) + wordCount;
+    map.set(sender, totalWords);
+
+    if (totalWords > maxWords || (totalWords === maxWords && sender > result)) {
+      maxWords = totalWords;
+      result = sender;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2287-rearrange-characters-to-make-target-string.js b/solutions/2287-rearrange-characters-to-make-target-string.js
new file mode 100644
index 00000000..310eb5a2
--- /dev/null
+++ b/solutions/2287-rearrange-characters-to-make-target-string.js
@@ -0,0 +1,38 @@
+/**
+ * 2287. Rearrange Characters to Make Target String
+ * https://leetcode.com/problems/rearrange-characters-to-make-target-string/
+ * Difficulty: Easy
+ *
+ * You are given two 0-indexed strings s and target. You can take some letters from s and
+ * rearrange them to form new strings.
+ *
+ * Return the maximum number of copies of target that can be formed by taking letters from
+ * s and rearranging them.
+ */
+
+/**
+ * @param {string} s
+ * @param {string} target
+ * @return {number}
+ */
+var rearrangeCharacters = function(s, target) {
+  const sourceFreq = new Array(26).fill(0);
+  const targetFreq = new Array(26).fill(0);
+
+  for (const char of s) {
+    sourceFreq[char.charCodeAt(0) - 97]++;
+  }
+
+  for (const char of target) {
+    targetFreq[char.charCodeAt(0) - 97]++;
+  }
+
+  let result = Infinity;
+  for (let i = 0; i < 26; i++) {
+    if (targetFreq[i] > 0) {
+      result = Math.min(result, Math.floor(sourceFreq[i] / targetFreq[i]));
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2288-apply-discount-to-prices.js b/solutions/2288-apply-discount-to-prices.js
new file mode 100644
index 00000000..4645204e
--- /dev/null
+++ b/solutions/2288-apply-discount-to-prices.js
@@ -0,0 +1,37 @@
+/**
+ * 2288. Apply Discount to Prices
+ * https://leetcode.com/problems/apply-discount-to-prices/
+ * Difficulty: Medium
+ *
+ * A sentence is a string of single-space separated words where each word can contain digits,
+ * lowercase letters, and the dollar sign '$'. A word represents a price if it is a sequence
+ * of digits preceded by a dollar sign.
+ * - For example, "$100", "$23", and "$6" represent prices while "100", "$", and "$1e5" do not.
+ *
+ * You are given a string sentence representing a sentence and an integer discount. For each
+ * word representing a price, apply a discount of discount% on the price and update the word
+ * in the sentence. All updated prices should be represented with exactly two decimal places.
+ *
+ * Return a string representing the modified sentence.
+ *
+ * Note that all prices will contain at most 10 digits.
+ */
+
+/**
+ * @param {string} sentence
+ * @param {number} discount
+ * @return {string}
+ */
+var discountPrices = function(sentence, discount) {
+  const words = sentence.split(' ');
+  const discountFactor = 1 - discount / 100;
+
+  for (let i = 0; i < words.length; i++) {
+    if (words[i].startsWith('$') && /^[0-9]+$/.test(words[i].slice(1))) {
+      const price = parseInt(words[i].slice(1)) * discountFactor;
+      words[i] = `$${price.toFixed(2)}`;
+    }
+  }
+
+  return words.join(' ');
+};
diff --git a/solutions/2293-min-max-game.js b/solutions/2293-min-max-game.js
new file mode 100644
index 00000000..b6952f9c
--- /dev/null
+++ b/solutions/2293-min-max-game.js
@@ -0,0 +1,43 @@
+/**
+ * 2293. Min Max Game
+ * https://leetcode.com/problems/min-max-game/
+ * Difficulty: Easy
+ *
+ * You are given a 0-indexed integer array nums whose length is a power of 2.
+ *
+ * Apply the following algorithm on nums:
+ * 1. Let n be the length of nums. If n == 1, end the process. Otherwise, create a new 0-indexed
+ *    integer array newNums of length n / 2.
+ * 2. For every even index i where 0 <= i < n / 2, assign the value of newNums[i] as
+ *    min(nums[2 * i], nums[2 * i + 1]).
+ * 3. For every odd index i where 0 <= i < n / 2, assign the value of newNums[i] as
+ *    max(nums[2 * i], nums[2 * i + 1]).
+ * 4. Replace the array nums with newNums.
+ * 5. Repeat the entire process starting from step 1.
+ *
+ * Return the last number that remains in nums after applying the algorithm.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var minMaxGame = function(nums) {
+  let array = nums.slice();
+
+  while (array.length > 1) {
+    const newArray = new Array(array.length / 2);
+
+    for (let i = 0; i < newArray.length; i++) {
+      if (i % 2 === 0) {
+        newArray[i] = Math.min(array[2 * i], array[2 * i + 1]);
+      } else {
+        newArray[i] = Math.max(array[2 * i], array[2 * i + 1]);
+      }
+    }
+
+    array = newArray;
+  }
+
+  return array[0];
+};
diff --git a/solutions/2294-partition-array-such-that-maximum-difference-is-k.js b/solutions/2294-partition-array-such-that-maximum-difference-is-k.js
new file mode 100644
index 00000000..06cad3ae
--- /dev/null
+++ b/solutions/2294-partition-array-such-that-maximum-difference-is-k.js
@@ -0,0 +1,34 @@
+/**
+ * 2294. Partition Array Such That Maximum Difference Is K
+ * https://leetcode.com/problems/partition-array-such-that-maximum-difference-is-k/
+ * Difficulty: Medium
+ *
+ * You are given an integer array nums and an integer k. You may partition nums into one or more
+ * subsequences such that each element in nums appears in exactly one of the subsequences.
+ *
+ * Return the minimum number of subsequences needed such that the difference between the maximum
+ * and minimum values in each subsequence is at most k.
+ *
+ * A subsequence is a sequence that can be derived from another sequence by deleting some or no
+ * elements without changing the order of the remaining elements.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+var partitionArray = function(nums, k) {
+  nums.sort((a, b) => a - b);
+  let result = 1;
+  let minValue = nums[0];
+
+  for (let i = 1; i < nums.length; i++) {
+    if (nums[i] - minValue > k) {
+      result++;
+      minValue = nums[i];
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2295-replace-elements-in-an-array.js b/solutions/2295-replace-elements-in-an-array.js
new file mode 100644
index 00000000..de3bbad0
--- /dev/null
+++ b/solutions/2295-replace-elements-in-an-array.js
@@ -0,0 +1,36 @@
+/**
+ * 2295. Replace Elements in an Array
+ * https://leetcode.com/problems/replace-elements-in-an-array/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed array nums that consists of n distinct positive integers.
+ * Apply m operations to this array, where in the ith operation you replace the number
+ * operations[i][0] with operations[i][1].
+ *
+ * It is guaranteed that in the ith operation:
+ * - operations[i][0] exists in nums.
+ * - operations[i][1] does not exist in nums.
+ *
+ * Return the array obtained after applying all the operations.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number[][]} operations
+ * @return {number[]}
+ */
+var arrayChange = function(nums, operations) {
+  const map = new Map();
+  for (let i = 0; i < nums.length; i++) {
+    map.set(nums[i], i);
+  }
+
+  for (const [oldVal, newVal] of operations) {
+    const index = map.get(oldVal);
+    nums[index] = newVal;
+    map.set(newVal, index);
+    map.delete(oldVal);
+  }
+
+  return nums;
+};
diff --git a/solutions/2296-design-a-text-editor.js b/solutions/2296-design-a-text-editor.js
new file mode 100644
index 00000000..6020598d
--- /dev/null
+++ b/solutions/2296-design-a-text-editor.js
@@ -0,0 +1,78 @@
+/**
+ * 2296. Design a Text Editor
+ * https://leetcode.com/problems/design-a-text-editor/
+ * Difficulty: Hard
+ *
+ * Design a text editor with a cursor that can do the following:
+ * - Add text to where the cursor is.
+ * - Delete text from where the cursor is (simulating the backspace key).
+ * - Move the cursor either left or right.
+ *
+ * When deleting text, only characters to the left of the cursor will be deleted. The cursor
+ * will also remain within the actual text and cannot be moved beyond it. More formally,
+ * we have that 0 <= cursor.position <= currentText.length always holds.
+ *
+ * Implement the TextEditor class:
+ * - TextEditor() Initializes the object with empty text.
+ * - void addText(string text) Appends text to where the cursor is. The cursor ends to the
+ *   right of text.
+ * - int deleteText(int k) Deletes k characters to the left of the cursor. Returns the number
+ *   of characters actually deleted.
+ * - string cursorLeft(int k) Moves the cursor to the left k times. Returns the last
+ *   min(10, len) characters to the left of the cursor, where len is the number of characters
+ *   to the left of the cursor.
+ * - string cursorRight(int k) Moves the cursor to the right k times. Returns the last
+ *   min(10, len) characters to the left of the cursor, where len is the number of characters
+ *   to the left of the cursor.
+ */
+
+var TextEditor = function() {
+  this.left = [];
+  this.right = [];
+};
+
+/**
+ * @param {string} text
+ * @return {void}
+ */
+TextEditor.prototype.addText = function(text) {
+  for (const char of text) {
+    this.left.push(char);
+  }
+};
+
+/**
+ * @param {number} k
+ * @return {number}
+ */
+TextEditor.prototype.deleteText = function(k) {
+  const deleteCount = Math.min(k, this.left.length);
+  for (let i = 0; i < deleteCount; i++) {
+    this.left.pop();
+  }
+  return deleteCount;
+};
+
+/**
+ * @param {number} k
+ * @return {string}
+ */
+TextEditor.prototype.cursorLeft = function(k) {
+  for (let i = 0; i < k && this.left.length; i++) {
+    this.right.push(this.left.pop());
+  }
+  const start = Math.max(0, this.left.length - 10);
+  return this.left.slice(start).join('');
+};
+
+/**
+ * @param {number} k
+ * @return {string}
+ */
+TextEditor.prototype.cursorRight = function(k) {
+  for (let i = 0; i < k && this.right.length; i++) {
+    this.left.push(this.right.pop());
+  }
+  const start = Math.max(0, this.left.length - 10);
+  return this.left.slice(start).join('');
+};
diff --git a/solutions/2299-strong-password-checker-ii.js b/solutions/2299-strong-password-checker-ii.js
new file mode 100644
index 00000000..f96fac07
--- /dev/null
+++ b/solutions/2299-strong-password-checker-ii.js
@@ -0,0 +1,43 @@
+/**
+ * 2299. Strong Password Checker II
+ * https://leetcode.com/problems/strong-password-checker-ii/
+ * Difficulty: Easy
+ *
+ * A password is said to be strong if it satisfies all the following criteria:
+ * - It has at least 8 characters.
+ * - It contains at least one lowercase letter.
+ * - It contains at least one uppercase letter.
+ * - It contains at least one digit.
+ * - It contains at least one special character. The special characters are the characters
+ *   in the following string: "!@#$%^&*()-+".
+ * - It does not contain 2 of the same character in adjacent positions (i.e., "aab" violates
+ *   this condition, but "aba" does not).
+ *
+ * Given a string password, return true if it is a strong password. Otherwise, return false.
+ */
+
+/**
+ * @param {string} password
+ * @return {boolean}
+ */
+var strongPasswordCheckerII = function(password) {
+  if (password.length < 8) return false;
+
+  let hasLower = false;
+  let hasUpper = false;
+  let hasDigit = false;
+  let hasSpecial = false;
+  const set = new Set(['!', '@', '#', '$', '%', '^', '&', '*', '(', ')', '-', '+']);
+
+  for (let i = 0; i < password.length; i++) {
+    if (i > 0 && password[i] === password[i - 1]) return false;
+
+    const char = password[i];
+    if (/[a-z]/.test(char)) hasLower = true;
+    else if (/[A-Z]/.test(char)) hasUpper = true;
+    else if (/\d/.test(char)) hasDigit = true;
+    else if (set.has(char)) hasSpecial = true;
+  }
+
+  return hasLower && hasUpper && hasDigit && hasSpecial;
+};
diff --git a/solutions/2301-match-substring-after-replacement.js b/solutions/2301-match-substring-after-replacement.js
new file mode 100644
index 00000000..fc7e5a91
--- /dev/null
+++ b/solutions/2301-match-substring-after-replacement.js
@@ -0,0 +1,50 @@
+/**
+ * 2301. Match Substring After Replacement
+ * https://leetcode.com/problems/match-substring-after-replacement/
+ * Difficulty: Hard
+ *
+ * You are given two strings s and sub. You are also given a 2D character array mappings where
+ * mappings[i] = [oldi, newi] indicates that you may perform the following operation any number
+ * of times:
+ * - Replace a character oldi of sub with newi.
+ *
+ * Each character in sub cannot be replaced more than once.
+ *
+ * Return true if it is possible to make sub a substring of s by replacing zero or more characters
+ * according to mappings. Otherwise, return false.
+ *
+ * A substring is a contiguous non-empty sequence of characters within a string.
+ */
+
+/**
+ * @param {string} s
+ * @param {string} sub
+ * @param {character[][]} mappings
+ * @return {boolean}
+ */
+var matchReplacement = function(s, sub, mappings) {
+  const map = new Map();
+
+  for (const [oldChar, newChar] of mappings) {
+    if (!map.has(oldChar)) {
+      map.set(oldChar, new Set());
+    }
+    map.get(oldChar).add(newChar);
+  }
+
+  const subLen = sub.length;
+  for (let i = 0; i <= s.length - subLen; i++) {
+    let isValid = true;
+    for (let j = 0; j < subLen; j++) {
+      const sChar = s[i + j];
+      const subChar = sub[j];
+      if (sChar !== subChar && (!map.has(subChar) || !map.get(subChar).has(sChar))) {
+        isValid = false;
+        break;
+      }
+    }
+    if (isValid) return true;
+  }
+
+  return false;
+};
diff --git a/solutions/2303-calculate-amount-paid-in-taxes.js b/solutions/2303-calculate-amount-paid-in-taxes.js
new file mode 100644
index 00000000..6944f54b
--- /dev/null
+++ b/solutions/2303-calculate-amount-paid-in-taxes.js
@@ -0,0 +1,41 @@
+/**
+ * 2303. Calculate Amount Paid in Taxes
+ * https://leetcode.com/problems/calculate-amount-paid-in-taxes/
+ * Difficulty: Easy
+ *
+ * You are given a 0-indexed 2D integer array brackets where brackets[i] = [upperi, percenti]
+ * means that the ith tax bracket has an upper bound of upperi and is taxed at a rate of percenti.
+ * The brackets are sorted by upper bound (i.e. upperi-1 < upperi for 0 < i < brackets.length).
+ *
+ * Tax is calculated as follows:
+ * - The first upper0 dollars earned are taxed at a rate of percent0.
+ * - The next upper1 - upper0 dollars earned are taxed at a rate of percent1.
+ * - The next upper2 - upper1 dollars earned are taxed at a rate of percent2.
+ * - And so on.
+ *
+ * You are given an integer income representing the amount of money you earned. Return the
+ * amount of money that you have to pay in taxes. Answers within 10-5 of the actual answer
+ * will be accepted.
+ */
+
+/**
+ * @param {number[][]} brackets
+ * @param {number} income
+ * @return {number}
+ */
+var calculateTax = function(brackets, income) {
+  let result = 0;
+  let previousUpper = 0;
+
+  for (const [upper, percent] of brackets) {
+    if (income > previousUpper) {
+      const taxable = Math.min(income, upper) - previousUpper;
+      result += taxable * (percent / 100);
+      previousUpper = upper;
+    } else {
+      break;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2304-minimum-path-cost-in-a-grid.js b/solutions/2304-minimum-path-cost-in-a-grid.js
new file mode 100644
index 00000000..eb1f10c4
--- /dev/null
+++ b/solutions/2304-minimum-path-cost-in-a-grid.js
@@ -0,0 +1,48 @@
+/**
+ * 2304. Minimum Path Cost in a Grid
+ * https://leetcode.com/problems/minimum-path-cost-in-a-grid/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed m x n integer matrix grid consisting of distinct integers from
+ * 0 to m * n - 1. You can move in this matrix from a cell to any other cell in the next row.
+ * That is, if you are in cell (x, y) such that x < m - 1, you can move to any of the cells
+ * (x + 1, 0), (x + 1, 1), ..., (x + 1, n - 1). Note that it is not possible to move from
+ * cells in the last row.
+ *
+ * Each possible move has a cost given by a 0-indexed 2D array moveCost of size (m * n) x n,
+ * where moveCost[i][j] is the cost of moving from a cell with value i to a cell in column j
+ * of the next row. The cost of moving from cells in the last row of grid can be ignored.
+ *
+ * The cost of a path in grid is the sum of all values of cells visited plus the sum of costs
+ * of all the moves made. Return the minimum cost of a path that starts from any cell in the
+ * first row and ends at any cell in the last row.
+ */
+
+/**
+ * @param {number[][]} grid
+ * @param {number[][]} moveCost
+ * @return {number}
+ */
+var minPathCost = function(grid, moveCost) {
+  const m = grid.length;
+  const n = grid[0].length;
+  const dp = Array.from({ length: m }, () => new Array(n).fill(Infinity));
+
+  for (let j = 0; j < n; j++) {
+    dp[0][j] = grid[0][j];
+  }
+
+  for (let i = 0; i < m - 1; i++) {
+    for (let j = 0; j < n; j++) {
+      const value = grid[i][j];
+      for (let k = 0; k < n; k++) {
+        dp[i + 1][k] = Math.min(
+          dp[i + 1][k],
+          dp[i][j] + grid[i + 1][k] + moveCost[value][k]
+        );
+      }
+    }
+  }
+
+  return Math.min(...dp[m - 1]);
+};
diff --git a/solutions/2305-fair-distribution-of-cookies.js b/solutions/2305-fair-distribution-of-cookies.js
new file mode 100644
index 00000000..dffcb4a4
--- /dev/null
+++ b/solutions/2305-fair-distribution-of-cookies.js
@@ -0,0 +1,46 @@
+/**
+ * 2305. Fair Distribution of Cookies
+ * https://leetcode.com/problems/fair-distribution-of-cookies/
+ * Difficulty: Medium
+ *
+ * You are given an integer array cookies, where cookies[i] denotes the number of cookies in the
+ * ith bag. You are also given an integer k that denotes the number of children to distribute
+ * all the bags of cookies to. All the cookies in the same bag must go to the same child and
+ * cannot be split up.
+ *
+ * The unfairness of a distribution is defined as the maximum total cookies obtained by a single
+ * child in the distribution.
+ *
+ * Return the minimum unfairness of all distributions.
+ */
+
+/**
+ * @param {number[]} cookies
+ * @param {number} k
+ * @return {number}
+ */
+var distributeCookies = function(cookies, k) {
+  const distribution = new Array(k).fill(0);
+  let result = Infinity;
+
+  distribute(0);
+
+  return result;
+
+  function distribute(index) {
+    if (index === cookies.length) {
+      result = Math.min(result, Math.max(...distribution));
+      return;
+    }
+
+    for (let i = 0; i < k; i++) {
+      distribution[i] += cookies[index];
+      if (distribution[i] < result) {
+        distribute(index + 1);
+      }
+      distribution[i] -= cookies[index];
+
+      if (distribution[i] === 0) break;
+    }
+  }
+};
diff --git a/solutions/2306-naming-a-company.js b/solutions/2306-naming-a-company.js
new file mode 100644
index 00000000..bef81dfe
--- /dev/null
+++ b/solutions/2306-naming-a-company.js
@@ -0,0 +1,37 @@
+/**
+ * 2306. Naming a Company
+ * https://leetcode.com/problems/naming-a-company/
+ * Difficulty: Hard
+ *
+ * You are given an array of strings ideas that represents a list of names to be used in the
+ * process of naming a company. The process of naming a company is as follows:
+ * 1. Choose 2 distinct names from ideas, call them ideaA and ideaB.
+ * 2. Swap the first letters of ideaA and ideaB with each other.
+ * 3. If both of the new names are not found in the original ideas, then the name ideaA ideaB
+ *    (the concatenation of ideaA and ideaB, separated by a space) is a valid company name.
+ * 4. Otherwise, it is not a valid name.
+ *
+ * Return the number of distinct valid names for the company.
+ */
+
+/**
+ * @param {string[]} ideas
+ * @return {number}
+ */
+var distinctNames = function(ideas) {
+  const groups = Array.from({ length: 26 }, () => new Set());
+  let result = 0;
+
+  for (const idea of ideas) {
+    groups[idea.charCodeAt(0) - 97].add(idea.slice(1));
+  }
+
+  for (let i = 0; i < 25; i++) {
+    for (let j = i + 1; j < 26; j++) {
+      const mutualCount = [...groups[i]].filter(suffix => groups[j].has(suffix)).length;
+      result += 2 * (groups[i].size - mutualCount) * (groups[j].size - mutualCount);
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2309-greatest-english-letter-in-upper-and-lower-case.js b/solutions/2309-greatest-english-letter-in-upper-and-lower-case.js
new file mode 100644
index 00000000..b86b8f8b
--- /dev/null
+++ b/solutions/2309-greatest-english-letter-in-upper-and-lower-case.js
@@ -0,0 +1,38 @@
+/**
+ * 2309. Greatest English Letter in Upper and Lower Case
+ * https://leetcode.com/problems/greatest-english-letter-in-upper-and-lower-case/
+ * Difficulty: Easy
+ *
+ * Given a string of English letters s, return the greatest English letter which occurs as both
+ * a lowercase and uppercase letter in s. The returned letter should be in uppercase. If no such
+ * letter exists, return an empty string.
+ *
+ * An English letter b is greater than another letter a if b appears after a in the English
+ * alphabet.
+ */
+
+/**
+ * @param {string} s
+ * @return {string}
+ */
+var greatestLetter = function(s) {
+  const lowerSet = new Set();
+  const upperSet = new Set();
+
+  for (const char of s) {
+    if (char >= 'a' && char <= 'z') {
+      lowerSet.add(char);
+    } else {
+      upperSet.add(char.toLowerCase());
+    }
+  }
+
+  let maxLetter = '';
+  for (const char of lowerSet) {
+    if (upperSet.has(char) && char > maxLetter) {
+      maxLetter = char;
+    }
+  }
+
+  return maxLetter.toUpperCase();
+};
diff --git a/solutions/2312-selling-pieces-of-wood.js b/solutions/2312-selling-pieces-of-wood.js
new file mode 100644
index 00000000..6edab3a0
--- /dev/null
+++ b/solutions/2312-selling-pieces-of-wood.js
@@ -0,0 +1,46 @@
+/**
+ * 2312. Selling Pieces of Wood
+ * https://leetcode.com/problems/selling-pieces-of-wood/
+ * Difficulty: Hard
+ *
+ * You are given two integers m and n that represent the height and width of a rectangular piece
+ * of wood. You are also given a 2D integer array prices, where prices[i] = [hi, wi, pricei]
+ * indicates you can sell a rectangular piece of wood of height hi and width wi for pricei dollars.
+ *
+ * To cut a piece of wood, you must make a vertical or horizontal cut across the entire height
+ * or width of the piece to split it into two smaller pieces. After cutting a piece of wood into
+ * some number of smaller pieces, you can sell pieces according to prices. You may sell multiple
+ * pieces of the same shape, and you do not have to sell all the shapes. The grain of the wood
+ * makes a difference, so you cannot rotate a piece to swap its height and width.
+ *
+ * Return the maximum money you can earn after cutting an m x n piece of wood.
+ *
+ * Note that you can cut the piece of wood as many times as you want.
+ */
+
+/**
+ * @param {number} m
+ * @param {number} n
+ * @param {number[][]} prices
+ * @return {number}
+ */
+var sellingWood = function(m, n, prices) {
+  const dp = Array.from({ length: m + 1 }, () => new Array(n + 1).fill(0));
+
+  for (const [h, w, price] of prices) {
+    dp[h][w] = price;
+  }
+
+  for (let i = 1; i <= m; i++) {
+    for (let j = 1; j <= n; j++) {
+      for (let k = 1; k < i; k++) {
+        dp[i][j] = Math.max(dp[i][j], dp[k][j] + dp[i - k][j]);
+      }
+      for (let k = 1; k < j; k++) {
+        dp[i][j] = Math.max(dp[i][j], dp[i][k] + dp[i][j - k]);
+      }
+    }
+  }
+
+  return dp[m][n];
+};
diff --git a/solutions/2315-count-asterisks.js b/solutions/2315-count-asterisks.js
new file mode 100644
index 00000000..650055eb
--- /dev/null
+++ b/solutions/2315-count-asterisks.js
@@ -0,0 +1,32 @@
+/**
+ * 2315. Count Asterisks
+ * https://leetcode.com/problems/count-asterisks/
+ * Difficulty: Easy
+ *
+ * You are given a string s, where every two consecutive vertical bars '|' are grouped into
+ * a pair. In other words, the 1st and 2nd '|' make a pair, the 3rd and 4th '|' make a pair,
+ * and so forth.
+ *
+ * Return the number of '*' in s, excluding the '*' between each pair of '|'.
+ *
+ * Note that each '|' will belong to exactly one pair.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var countAsterisks = function(s) {
+  let isInsidePair = false;
+  let result = 0;
+
+  for (const char of s) {
+    if (char === '|') {
+      isInsidePair = !isInsidePair;
+    } else if (char === '*' && !isInsidePair) {
+      result++;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2316-count-unreachable-pairs-of-nodes-in-an-undirected-graph.js b/solutions/2316-count-unreachable-pairs-of-nodes-in-an-undirected-graph.js
new file mode 100644
index 00000000..4de0c07b
--- /dev/null
+++ b/solutions/2316-count-unreachable-pairs-of-nodes-in-an-undirected-graph.js
@@ -0,0 +1,48 @@
+/**
+ * 2316. Count Unreachable Pairs of Nodes in an Undirected Graph
+ * https://leetcode.com/problems/count-unreachable-pairs-of-nodes-in-an-undirected-graph/
+ * Difficulty: Medium
+ *
+ * You are given an integer n. There is an undirected graph with n nodes, numbered from 0
+ * to n - 1. You are given a 2D integer array edges where edges[i] = [ai, bi] denotes that
+ * there exists an undirected edge connecting nodes ai and bi.
+ *
+ * Return the number of pairs of different nodes that are unreachable from each other.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} edges
+ * @return {number}
+ */
+var countPairs = function(n, edges) {
+  const graph = Array.from({ length: n }, () => []);
+  for (const [a, b] of edges) {
+    graph[a].push(b);
+    graph[b].push(a);
+  }
+
+  const visited = new Array(n).fill(false);
+  let result = 0;
+
+  let totalNodes = 0;
+  for (let node = 0; node < n; node++) {
+    if (!visited[node]) {
+      const componentSize = exploreComponent(node);
+      result += totalNodes * componentSize;
+      totalNodes += componentSize;
+    }
+  }
+
+  return result;
+
+  function exploreComponent(node) {
+    if (visited[node]) return 0;
+    visited[node] = true;
+    let size = 1;
+    for (const neighbor of graph[node]) {
+      size += exploreComponent(neighbor);
+    }
+    return size;
+  }
+};
diff --git a/solutions/2317-maximum-xor-after-operations.js b/solutions/2317-maximum-xor-after-operations.js
new file mode 100644
index 00000000..7049675b
--- /dev/null
+++ b/solutions/2317-maximum-xor-after-operations.js
@@ -0,0 +1,27 @@
+/**
+ * 2317. Maximum XOR After Operations
+ * https://leetcode.com/problems/maximum-xor-after-operations/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed integer array nums. In one operation, select any non-negative
+ * integer x and an index i, then update nums[i] to be equal to nums[i] AND (nums[i] XOR x).
+ *
+ * Note that AND is the bitwise AND operation and XOR is the bitwise XOR operation.
+ *
+ * Return the maximum possible bitwise XOR of all elements of nums after applying the operation
+ * any number of times.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var maximumXOR = function(nums) {
+  let maxXor = 0;
+
+  for (const num of nums) {
+    maxXor |= num;
+  }
+
+  return maxXor;
+};
diff --git a/solutions/2318-number-of-distinct-roll-sequences.js b/solutions/2318-number-of-distinct-roll-sequences.js
new file mode 100644
index 00000000..5f67f5c0
--- /dev/null
+++ b/solutions/2318-number-of-distinct-roll-sequences.js
@@ -0,0 +1,55 @@
+/**
+ * 2318. Number of Distinct Roll Sequences
+ * https://leetcode.com/problems/number-of-distinct-roll-sequences/
+ * Difficulty: Hard
+ *
+ * You are given an integer n. You roll a fair 6-sided dice n times. Determine the total number
+ * of distinct sequences of rolls possible such that the following conditions are satisfied:
+ * 1. The greatest common divisor of any adjacent values in the sequence is equal to 1.
+ * 2. There is at least a gap of 2 rolls between equal valued rolls. More formally, if the
+ *    value of the ith roll is equal to the value of the jth roll, then abs(i - j) > 2.
+ *
+ * Return the total number of distinct sequences possible. Since the answer may be very large,
+ * return it modulo 109 + 7.
+ *
+ * Two sequences are considered distinct if at least one element is different.
+ */
+
+/**
+ * @param {number} n
+ * @return {number}
+ */
+var distinctSequences = function(n) {
+  const MOD = 1e9 + 7;
+  const gcd = (a, b) => b === 0 ? a : gcd(b, a % b);
+
+  const dp = new Array(n + 1)
+    .fill()
+    .map(() => new Array(7).fill().map(() => new Array(7).fill(0)));
+
+  for (let i = 1; i <= 6; i++) {
+    dp[1][i][0] = 1;
+  }
+
+  for (let len = 2; len <= n; len++) {
+    for (let curr = 1; curr <= 6; curr++) {
+      for (let prev = 0; prev <= 6; prev++) {
+        for (let prevPrev = 0; prevPrev <= 6; prevPrev++) {
+          if (dp[len - 1][prev][prevPrev] === 0) continue;
+          if (curr === prev || curr === prevPrev) continue;
+          if (prev !== 0 && gcd(curr, prev) !== 1) continue;
+          dp[len][curr][prev] = (dp[len][curr][prev] + dp[len - 1][prev][prevPrev]) % MOD;
+        }
+      }
+    }
+  }
+
+  let result = 0;
+  for (let curr = 1; curr <= 6; curr++) {
+    for (let prev = 0; prev <= 6; prev++) {
+      result = (result + dp[n][curr][prev]) % MOD;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2319-check-if-matrix-is-x-matrix.js b/solutions/2319-check-if-matrix-is-x-matrix.js
new file mode 100644
index 00000000..ee45ca42
--- /dev/null
+++ b/solutions/2319-check-if-matrix-is-x-matrix.js
@@ -0,0 +1,32 @@
+/**
+ * 2319. Check if Matrix Is X-Matrix
+ * https://leetcode.com/problems/check-if-matrix-is-x-matrix/
+ * Difficulty: Easy
+ *
+ * A square matrix is said to be an X-Matrix if both of the following conditions hold:
+ * 1. All the elements in the diagonals of the matrix are non-zero.
+ * 2. All other elements are 0.
+ *
+ * Given a 2D integer array grid of size n x n representing a square matrix, return true if grid
+ * is an X-Matrix. Otherwise, return false.
+ */
+
+/**
+ * @param {number[][]} grid
+ * @return {boolean}
+ */
+var checkXMatrix = function(grid) {
+  const size = grid.length;
+
+  for (let row = 0; row < size; row++) {
+    for (let col = 0; col < size; col++) {
+      const isDiagonal = row === col || row + col === size - 1;
+      const isNonZero = grid[row][col] !== 0;
+
+      if (isDiagonal && !isNonZero) return false;
+      if (!isDiagonal && isNonZero) return false;
+    }
+  }
+
+  return true;
+};
diff --git a/solutions/2320-count-number-of-ways-to-place-houses.js b/solutions/2320-count-number-of-ways-to-place-houses.js
new file mode 100644
index 00000000..39c28f72
--- /dev/null
+++ b/solutions/2320-count-number-of-ways-to-place-houses.js
@@ -0,0 +1,34 @@
+/**
+ * 2320. Count Number of Ways to Place Houses
+ * https://leetcode.com/problems/count-number-of-ways-to-place-houses/
+ * Difficulty: Medium
+ *
+ * There is a street with n * 2 plots, where there are n plots on each side of the street. The plots
+ * on each side are numbered from 1 to n. On each plot, a house can be placed.
+ *
+ * Return the number of ways houses can be placed such that no two houses are adjacent to each other
+ * on the same side of the street. Since the answer may be very large, return it modulo 109 + 7.
+ *
+ * Note that if a house is placed on the ith plot on one side of the street, a house can also be
+ * placed on the ith plot on the other side of the street.
+ */
+
+/**
+ * @param {number} n
+ * @return {number}
+ */
+var countHousePlacements = function(n) {
+  const MOD = 1e9 + 7;
+  let empty = 1n;
+  let house = 1n;
+
+  for (let i = 2; i <= n; i++) {
+    const nextEmpty = (empty + house) % BigInt(MOD);
+    const nextHouse = empty;
+    empty = nextEmpty;
+    house = nextHouse;
+  }
+
+  const sideWays = (empty + house) % BigInt(MOD);
+  return Number((sideWays * sideWays) % BigInt(MOD));
+};
diff --git a/solutions/2321-maximum-score-of-spliced-array.js b/solutions/2321-maximum-score-of-spliced-array.js
new file mode 100644
index 00000000..3ecf5068
--- /dev/null
+++ b/solutions/2321-maximum-score-of-spliced-array.js
@@ -0,0 +1,52 @@
+/**
+ * 2321. Maximum Score Of Spliced Array
+ * https://leetcode.com/problems/maximum-score-of-spliced-array/
+ * Difficulty: Hard
+ *
+ * You are given two 0-indexed integer arrays nums1 and nums2, both of length n.
+ *
+ * You can choose two integers left and right where 0 <= left <= right < n and swap the subarray
+ * nums1[left...right] with the subarray nums2[left...right].
+ * - For example, if nums1 = [1,2,3,4,5] and nums2 = [11,12,13,14,15] and you choose left = 1 and
+ *   right = 2, nums1 becomes [1,12,13,4,5] and nums2 becomes [11,2,3,14,15].
+ *
+ * You may choose to apply the mentioned operation once or not do anything.
+ *
+ * The score of the arrays is the maximum of sum(nums1) and sum(nums2), where sum(arr) is the sum
+ * of all the elements in the array arr.
+ *
+ * Return the maximum possible score.
+ *
+ * A subarray is a contiguous sequence of elements within an array. arr[left...right] denotes the
+ * subarray that contains the elements of nums between indices left and right (inclusive).
+ */
+
+/**
+ * @param {number[]} nums1
+ * @param {number[]} nums2
+ * @return {number}
+ */
+var maximumsSplicedArray = function(nums1, nums2) {
+  const length = nums1.length;
+  let sum1 = 0;
+  let sum2 = 0;
+
+  for (let i = 0; i < length; i++) {
+    sum1 += nums1[i];
+    sum2 += nums2[i];
+  }
+
+  let maxGain1 = 0;
+  let maxGain2 = 0;
+  let currGain1 = 0;
+  let currGain2 = 0;
+
+  for (let i = 0; i < length; i++) {
+    currGain1 = Math.max(0, currGain1 + nums2[i] - nums1[i]);
+    currGain2 = Math.max(0, currGain2 + nums1[i] - nums2[i]);
+    maxGain1 = Math.max(maxGain1, currGain1);
+    maxGain2 = Math.max(maxGain2, currGain2);
+  }
+
+  return Math.max(sum1 + maxGain1, sum2 + maxGain2);
+};
diff --git a/solutions/2322-minimum-score-after-removals-on-a-tree.js b/solutions/2322-minimum-score-after-removals-on-a-tree.js
new file mode 100644
index 00000000..4b0b6cd0
--- /dev/null
+++ b/solutions/2322-minimum-score-after-removals-on-a-tree.js
@@ -0,0 +1,112 @@
+/**
+ * 2322. Minimum Score After Removals on a Tree
+ * https://leetcode.com/problems/minimum-score-after-removals-on-a-tree/
+ * Difficulty: Hard
+ *
+ * There is an undirected connected tree with n nodes labeled from 0 to n - 1 and n - 1 edges.
+ *
+ * You are given a 0-indexed integer array nums of length n where nums[i] represents the value
+ * of the ith node. You are also given a 2D integer array edges of length n - 1 where
+ * edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree.
+ *
+ * Remove two distinct edges of the tree to form three connected components. For a pair of
+ * removed edges, the following steps are defined:
+ * 1. Get the XOR of all the values of the nodes for each of the three components respectively.
+ * 2. The difference between the largest XOR value and the smallest XOR value is the score of
+ *    the pair.
+ * 3. For example, say the three components have the node values: [4,5,7], [1,9], and [3,3,3].
+ *    The three XOR values are 4 ^ 5 ^ 7 = 6, 1 ^ 9 = 8, and 3 ^ 3 ^ 3 = 3. The largest XOR
+ *    value is 8 and the smallest XOR value is 3. The score is then 8 - 3 = 5.
+ *
+ * Return the minimum score of any possible pair of edge removals on the given tree.
+ */
+
+/**
+* @param {number[]} nums
+* @param {number[][]} edges
+* @return {number}
+*/
+var minimumScore = function(nums, edges) {
+  const n = nums.length;
+  const graph = Array.from({ length: n }, () => []);
+
+  for (const [a, b] of edges) {
+    graph[a].push(b);
+    graph[b].push(a);
+  }
+
+  const totalXor = nums.reduce((acc, num) => acc ^ num, 0);
+
+  const subtreeXor = new Array(n).fill(0);
+
+  const tin = new Array(n);
+  const tout = new Array(n);
+  let timer = 0;
+
+  function dfs(node, parent) {
+    tin[node] = timer++;
+    subtreeXor[node] = nums[node];
+
+    for (const neighbor of graph[node]) {
+      if (neighbor !== parent) {
+        dfs(neighbor, node);
+        subtreeXor[node] ^= subtreeXor[neighbor];
+      }
+    }
+
+    tout[node] = timer++;
+  }
+
+  dfs(0, -1);
+
+  function isAncestor(a, b) {
+    return tin[a] <= tin[b] && tout[a] >= tout[b];
+  }
+
+  let result = Infinity;
+  for (let i = 0; i < n - 1; i++) {
+    for (let j = i + 1; j < n - 1; j++) {
+      const edge1 = edges[i];
+      const edge2 = edges[j];
+
+      let node1;
+      if (isAncestor(edge1[0], edge1[1])) {
+        node1 = edge1[1];
+      } else {
+        node1 = edge1[0];
+      }
+      const subtree1 = subtreeXor[node1];
+
+      let node2;
+      if (isAncestor(edge2[0], edge2[1])) {
+        node2 = edge2[1];
+      } else {
+        node2 = edge2[0];
+      }
+      const subtree2 = subtreeXor[node2];
+
+      let component1;
+      let component2;
+      let component3;
+      if (isAncestor(node1, node2)) {
+        component1 = subtree2;
+        component2 = subtree1 ^ component1;
+        component3 = totalXor ^ subtree1;
+      } else if (isAncestor(node2, node1)) {
+        component1 = subtree1;
+        component2 = subtree2 ^ component1;
+        component3 = totalXor ^ subtree2;
+      } else {
+        component1 = subtree1;
+        component2 = subtree2;
+        component3 = totalXor ^ component1 ^ component2;
+      }
+
+      const maxXor = Math.max(component1, component2, component3);
+      const minXor = Math.min(component1, component2, component3);
+      result = Math.min(result, maxXor - minXor);
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2325-decode-the-message.js b/solutions/2325-decode-the-message.js
new file mode 100644
index 00000000..bde90e23
--- /dev/null
+++ b/solutions/2325-decode-the-message.js
@@ -0,0 +1,41 @@
+/**
+ * 2325. Decode the Message
+ * https://leetcode.com/problems/decode-the-message/
+ * Difficulty: Easy
+ *
+ * You are given the strings key and message, which represent a cipher key and a secret message,
+ * respectively. The steps to decode message are as follows:
+ * 1. Use the first appearance of all 26 lowercase English letters in key as the order of the
+ *    substitution table.
+ * 2. Align the substitution table with the regular English alphabet.
+ * 3. Each letter in message is then substituted using the table.
+ * 4. Spaces ' ' are transformed to themselves.
+ * 5. For example, given key = "happy boy" (actual key would have at least one instance of each
+ *    letter in the alphabet), we have the partial substitution table of ('h' -> 'a', 'a' -> 'b',
+ *    'p' -> 'c', 'y' -> 'd', 'b' -> 'e', 'o' -> 'f').
+ *
+ * Return the decoded message.
+ */
+
+/**
+ * @param {string} key
+ * @param {string} message
+ * @return {string}
+ */
+var decodeMessage = function(key, message) {
+  const substitution = new Map();
+  let alphabetIndex = 0;
+
+  for (const char of key) {
+    if (char !== ' ' && !substitution.has(char)) {
+      substitution.set(char, String.fromCharCode(97 + alphabetIndex++));
+    }
+  }
+
+  let result = '';
+  for (const char of message) {
+    result += char === ' ' ? ' ' : substitution.get(char);
+  }
+
+  return result;
+};
diff --git a/solutions/2326-spiral-matrix-iv.js b/solutions/2326-spiral-matrix-iv.js
new file mode 100644
index 00000000..1fa98050
--- /dev/null
+++ b/solutions/2326-spiral-matrix-iv.js
@@ -0,0 +1,65 @@
+/**
+ * 2326. Spiral Matrix IV
+ * https://leetcode.com/problems/spiral-matrix-iv/
+ * Difficulty: Medium
+ *
+ * You are given two integers m and n, which represent the dimensions of a matrix.
+ *
+ * You are also given the head of a linked list of integers.
+ *
+ * Generate an m x n matrix that contains the integers in the linked list presented in spiral
+ * order (clockwise), starting from the top-left of the matrix. If there are remaining empty
+ * spaces, fill them with -1.
+ *
+ * Return the generated matrix.
+ */
+
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * @param {number} m
+ * @param {number} n
+ * @param {ListNode} head
+ * @return {number[][]}
+ */
+var spiralMatrix = function(m, n, head) {
+  const matrix = new Array(m).fill().map(() => new Array(n).fill(-1));
+  let top = 0;
+  let bottom = m - 1;
+  let left = 0;
+  let right = n - 1;
+  let current = head;
+
+  while (top <= bottom && left <= right && current) {
+    for (let col = left; col <= right && current; col++) {
+      matrix[top][col] = current.val;
+      current = current.next;
+    }
+    top++;
+
+    for (let row = top; row <= bottom && current; row++) {
+      matrix[row][right] = current.val;
+      current = current.next;
+    }
+    right--;
+
+    for (let col = right; col >= left && current; col--) {
+      matrix[bottom][col] = current.val;
+      current = current.next;
+    }
+    bottom--;
+
+    for (let row = bottom; row >= top && current; row--) {
+      matrix[row][left] = current.val;
+      current = current.next;
+    }
+    left++;
+  }
+
+  return matrix;
+};
diff --git a/solutions/2328-number-of-increasing-paths-in-a-grid.js b/solutions/2328-number-of-increasing-paths-in-a-grid.js
new file mode 100644
index 00000000..f239c80c
--- /dev/null
+++ b/solutions/2328-number-of-increasing-paths-in-a-grid.js
@@ -0,0 +1,53 @@
+/**
+ * 2328. Number of Increasing Paths in a Grid
+ * https://leetcode.com/problems/number-of-increasing-paths-in-a-grid/
+ * Difficulty: Hard
+ *
+ * You are given an m x n integer matrix grid, where you can move from a cell to any adjacent
+ * cell in all 4 directions.
+ *
+ * Return the number of strictly increasing paths in the grid such that you can start from any
+ * cell and end at any cell. Since the answer may be very large, return it modulo 109 + 7.
+ *
+ * Two paths are considered different if they do not have exactly the same sequence of visited
+ * cells.
+ */
+
+/**
+ * @param {number[][]} grid
+ * @return {number}
+ */
+var countPaths = function(grid) {
+  const MOD = 1e9 + 7;
+  const rows = grid.length;
+  const cols = grid[0].length;
+  const cache = new Array(rows).fill().map(() => new Array(cols).fill(0));
+  const directions = [[0, 1], [1, 0], [0, -1], [-1, 0]];
+
+  function explore(row, col) {
+    if (cache[row][col]) return cache[row][col];
+
+    let paths = 1;
+    for (const [dr, dc] of directions) {
+      const newRow = row + dr;
+      const newCol = col + dc;
+      if (
+        newRow >= 0 && newRow < rows && newCol >= 0
+        && newCol < cols && grid[newRow][newCol] > grid[row][col]
+      ) {
+        paths = (paths + explore(newRow, newCol)) % MOD;
+      }
+    }
+
+    return cache[row][col] = paths;
+  }
+
+  let total = 0;
+  for (let i = 0; i < rows; i++) {
+    for (let j = 0; j < cols; j++) {
+      total = (total + explore(i, j)) % MOD;
+    }
+  }
+
+  return total;
+};
diff --git a/solutions/2331-evaluate-boolean-binary-tree.js b/solutions/2331-evaluate-boolean-binary-tree.js
new file mode 100644
index 00000000..d6f4d851
--- /dev/null
+++ b/solutions/2331-evaluate-boolean-binary-tree.js
@@ -0,0 +1,42 @@
+/**
+ * 2331. Evaluate Boolean Binary Tree
+ * https://leetcode.com/problems/evaluate-boolean-binary-tree/
+ * Difficulty: Easy
+ *
+ * You are given the root of a full binary tree with the following properties:
+ * - Leaf nodes have either the value 0 or 1, where 0 represents False and 1 represents True.
+ * - Non-leaf nodes have either the value 2 or 3, where 2 represents the boolean OR and 3 represents
+ *   the boolean AND.
+ *
+ * The evaluation of a node is as follows:
+ * - If the node is a leaf node, the evaluation is the value of the node, i.e. True or False.
+ * - Otherwise, evaluate the node's two children and apply the boolean operation of its value with
+ *   the children's evaluations.
+ *
+ * Return the boolean result of evaluating the root node.
+ *
+ * A full binary tree is a binary tree where each node has either 0 or 2 children.
+ *
+ * A leaf node is a node that has zero children.
+ */
+
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @return {boolean}
+ */
+var evaluateTree = function(root) {
+  if (!root.left && !root.right) return root.val === 1;
+
+  const leftResult = evaluateTree(root.left);
+  const rightResult = evaluateTree(root.right);
+
+  return root.val === 2 ? leftResult || rightResult : leftResult && rightResult;
+};
diff --git a/solutions/2334-subarray-with-elements-greater-than-varying-threshold.js b/solutions/2334-subarray-with-elements-greater-than-varying-threshold.js
new file mode 100644
index 00000000..789a5ff3
--- /dev/null
+++ b/solutions/2334-subarray-with-elements-greater-than-varying-threshold.js
@@ -0,0 +1,43 @@
+/**
+ * 2334. Subarray With Elements Greater Than Varying Threshold
+ * https://leetcode.com/problems/subarray-with-elements-greater-than-varying-threshold/
+ * Difficulty: Hard
+ *
+ * You are given an integer array nums and an integer threshold.
+ *
+ * Find any subarray of nums of length k such that every element in the subarray is greater
+ * than threshold / k.
+ *
+ * Return the size of any such subarray. If there is no such subarray, return -1.
+ *
+ * A subarray is a contiguous non-empty sequence of elements within an array.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} threshold
+ * @return {number}
+ */
+var validSubarraySize = function(nums, threshold) {
+  const n = nums.length;
+  const stack = [];
+  const nextSmaller = new Array(n).fill(n);
+  const prevSmaller = new Array(n).fill(-1);
+
+  for (let i = 0; i < n; i++) {
+    while (stack.length && nums[stack[stack.length - 1]] >= nums[i]) {
+      nextSmaller[stack.pop()] = i;
+    }
+    if (stack.length) prevSmaller[i] = stack[stack.length - 1];
+    stack.push(i);
+  }
+
+  for (let i = 0; i < n; i++) {
+    const k = nextSmaller[i] - prevSmaller[i] - 1;
+    if (k > 0 && nums[i] > threshold / k) {
+      return k;
+    }
+  }
+
+  return -1;
+};
diff --git a/solutions/2337-move-pieces-to-obtain-a-string.js b/solutions/2337-move-pieces-to-obtain-a-string.js
new file mode 100644
index 00000000..64f29a9a
--- /dev/null
+++ b/solutions/2337-move-pieces-to-obtain-a-string.js
@@ -0,0 +1,41 @@
+/**
+ * 2337. Move Pieces to Obtain a String
+ * https://leetcode.com/problems/move-pieces-to-obtain-a-string/
+ * Difficulty: Medium
+ *
+ * You are given two strings start and target, both of length n. Each string consists only of
+ * the characters 'L', 'R', and '_' where:
+ * - The characters 'L' and 'R' represent pieces, where a piece 'L' can move to the left only if
+ *   there is a blank space directly to its left, and a piece 'R' can move to the right only if
+ *   there is a blank space directly to its right.
+ * - The character '_' represents a blank space that can be occupied by any of the 'L' or 'R'
+ *   pieces.
+ *
+ * Return true if it is possible to obtain the string target by moving the pieces of the string
+ * start any number of times. Otherwise, return false.
+ */
+
+/**
+ * @param {string} start
+ * @param {string} target
+ * @return {boolean}
+ */
+var canChange = function(start, target) {
+  const pieces = [];
+  const targetPieces = [];
+
+  for (let i = 0; i < start.length; i++) {
+    if (start[i] !== '_') pieces.push([start[i], i]);
+    if (target[i] !== '_') targetPieces.push([target[i], i]);
+  }
+
+  if (pieces.length !== targetPieces.length) return false;
+
+  for (let i = 0; i < pieces.length; i++) {
+    if (pieces[i][0] !== targetPieces[i][0]) return false;
+    if (pieces[i][0] === 'L' && pieces[i][1] < targetPieces[i][1]) return false;
+    if (pieces[i][0] === 'R' && pieces[i][1] > targetPieces[i][1]) return false;
+  }
+
+  return true;
+};
diff --git a/solutions/2341-maximum-number-of-pairs-in-array.js b/solutions/2341-maximum-number-of-pairs-in-array.js
new file mode 100644
index 00000000..2cdcce2c
--- /dev/null
+++ b/solutions/2341-maximum-number-of-pairs-in-array.js
@@ -0,0 +1,35 @@
+/**
+ * 2341. Maximum Number of Pairs in Array
+ * https://leetcode.com/problems/maximum-number-of-pairs-in-array/
+ * Difficulty: Easy
+ *
+ * You are given a 0-indexed integer array nums. In one operation, you may do the following:
+ * - Choose two integers in nums that are equal.
+ * - Remove both integers from nums, forming a pair.
+ *
+ * The operation is done on nums as many times as possible.
+ *
+ * Return a 0-indexed integer array answer of size 2 where answer[0] is the number of pairs that
+ * are formed and answer[1] is the number of leftover integers in nums after doing the operation
+ * as many times as possible.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var numberOfPairs = function(nums) {
+  const frequency = new Map();
+  let pairs = 0;
+
+  for (const num of nums) {
+    frequency.set(num, (frequency.get(num) || 0) + 1);
+    if (frequency.get(num) === 2) {
+      pairs++;
+      frequency.set(num, 0);
+    }
+  }
+
+  const leftovers = nums.length - pairs * 2;
+  return [pairs, leftovers];
+};
diff --git a/solutions/2347-best-poker-hand.js b/solutions/2347-best-poker-hand.js
new file mode 100644
index 00000000..586dcda2
--- /dev/null
+++ b/solutions/2347-best-poker-hand.js
@@ -0,0 +1,41 @@
+/**
+ * 2347. Best Poker Hand
+ * https://leetcode.com/problems/best-poker-hand/
+ * Difficulty: Easy
+ *
+ * You are given an integer array ranks and a character array suits. You have 5 cards where the
+ * ith card has a rank of ranks[i] and a suit of suits[i].
+ *
+ * The following are the types of poker hands you can make from best to worst:
+ * 1. "Flush": Five cards of the same suit.
+ * 2. "Three of a Kind": Three cards of the same rank.
+ * 3. "Pair": Two cards of the same rank.
+ * 4. "High Card": Any single card.
+ *
+ * Return a string representing the best type of poker hand you can make with the given cards.
+ *
+ * Note that the return values are case-sensitive.
+ */
+
+/**
+ * @param {number[]} ranks
+ * @param {character[]} suits
+ * @return {string}
+ */
+var bestHand = function(ranks, suits) {
+  const suitCount = new Map();
+  const rankCount = new Map();
+
+  for (let i = 0; i < 5; i++) {
+    suitCount.set(suits[i], (suitCount.get(suits[i]) || 0) + 1);
+    rankCount.set(ranks[i], (rankCount.get(ranks[i]) || 0) + 1);
+  }
+
+  if (suitCount.size === 1) return 'Flush';
+
+  const maxRankCount = Math.max(...rankCount.values());
+  if (maxRankCount >= 3) return 'Three of a Kind';
+  if (maxRankCount === 2) return 'Pair';
+
+  return 'High Card';
+};
diff --git a/solutions/2348-number-of-zero-filled-subarrays.js b/solutions/2348-number-of-zero-filled-subarrays.js
new file mode 100644
index 00000000..fa93ec19
--- /dev/null
+++ b/solutions/2348-number-of-zero-filled-subarrays.js
@@ -0,0 +1,29 @@
+/**
+ * 2348. Number of Zero-Filled Subarrays
+ * https://leetcode.com/problems/number-of-zero-filled-subarrays/
+ * Difficulty: Medium
+ *
+ * Given an integer array nums, return the number of subarrays filled with 0.
+ *
+ * A subarray is a contiguous non-empty sequence of elements within an array.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var zeroFilledSubarray = function(nums) {
+  let result = 0;
+  let zeroCount = 0;
+
+  for (const num of nums) {
+    if (num === 0) {
+      zeroCount++;
+      result += zeroCount;
+    } else {
+      zeroCount = 0;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2350-shortest-impossible-sequence-of-rolls.js b/solutions/2350-shortest-impossible-sequence-of-rolls.js
new file mode 100644
index 00000000..ea27a604
--- /dev/null
+++ b/solutions/2350-shortest-impossible-sequence-of-rolls.js
@@ -0,0 +1,32 @@
+/**
+ * 2350. Shortest Impossible Sequence of Rolls
+ * https://leetcode.com/problems/shortest-impossible-sequence-of-rolls/
+ * Difficulty: Hard
+ *
+ * You are given an integer array rolls of length n and an integer k. You roll a k sided dice
+ * numbered from 1 to k, n times, where the result of the ith roll is rolls[i].
+ *
+ * Return the length of the shortest sequence of rolls so that there's no such subsequence in rolls.
+ *
+ * A sequence of rolls of length len is the result of rolling a k sided dice len times.
+ */
+
+/**
+ * @param {number[]} rolls
+ * @param {number} k
+ * @return {number}
+ */
+var shortestSequence = function(rolls, k) {
+  const set = new Set();
+  let sequences = 0;
+
+  for (const roll of rolls) {
+    set.add(roll);
+    if (set.size === k) {
+      sequences++;
+      set.clear();
+    }
+  }
+
+  return sequences + 1;
+};
diff --git a/solutions/2351-first-letter-to-appear-twice.js b/solutions/2351-first-letter-to-appear-twice.js
new file mode 100644
index 00000000..b5fff8f5
--- /dev/null
+++ b/solutions/2351-first-letter-to-appear-twice.js
@@ -0,0 +1,26 @@
+/**
+ * 2351. First Letter to Appear Twice
+ * https://leetcode.com/problems/first-letter-to-appear-twice/
+ * Difficulty: Easy
+ *
+ * Given a string s consisting of lowercase English letters, return the first letter
+ * to appear twice.
+ *
+ * Note:
+ * - A letter a appears twice before another letter b if the second occurrence of a is before
+ *   the second occurrence of b.
+ * - s will contain at least one letter that appears twice.
+ */
+
+/**
+ * @param {string} s
+ * @return {character}
+ */
+var repeatedCharacter = function(s) {
+  const set = new Set();
+
+  for (const char of s) {
+    if (set.has(char)) return char;
+    set.add(char);
+  }
+};
diff --git a/solutions/2354-number-of-excellent-pairs.js b/solutions/2354-number-of-excellent-pairs.js
new file mode 100644
index 00000000..bf60424a
--- /dev/null
+++ b/solutions/2354-number-of-excellent-pairs.js
@@ -0,0 +1,46 @@
+/**
+ * 2354. Number of Excellent Pairs
+ * https://leetcode.com/problems/number-of-excellent-pairs/
+ * Difficulty: Hard
+ *
+ * You are given a 0-indexed positive integer array nums and a positive integer k.
+ *
+ * A pair of numbers (num1, num2) is called excellent if the following conditions are satisfied:
+ * - Both the numbers num1 and num2 exist in the array nums.
+ * - The sum of the number of set bits in num1 OR num2 and num1 AND num2 is greater than or equal
+ *   to k, where OR is the bitwise OR operation and AND is the bitwise AND operation.
+ *
+ * Return the number of distinct excellent pairs.
+ *
+ * Two pairs (a, b) and (c, d) are considered distinct if either a != c or b != d. For example,
+ * (1, 2) and (2, 1) are distinct.
+ *
+ * Note that a pair (num1, num2) such that num1 == num2 can also be excellent if you have at least
+ * one occurrence of num1 in the array.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+var countExcellentPairs = function(nums, k) {
+  const map = new Map();
+  const set = new Set(nums);
+
+  for (const num of set) {
+    const bits = num.toString(2).split('0').join('').length;
+    map.set(bits, (map.get(bits) || 0) + 1);
+  }
+
+  let result = 0;
+  for (const [i, countI] of map) {
+    for (const [j, countJ] of map) {
+      if (i + j >= k) {
+        result += countI * countJ;
+      }
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2358-maximum-number-of-groups-entering-a-competition.js b/solutions/2358-maximum-number-of-groups-entering-a-competition.js
new file mode 100644
index 00000000..2637927b
--- /dev/null
+++ b/solutions/2358-maximum-number-of-groups-entering-a-competition.js
@@ -0,0 +1,31 @@
+/**
+ * 2358. Maximum Number of Groups Entering a Competition
+ * https://leetcode.com/problems/maximum-number-of-groups-entering-a-competition/
+ * Difficulty: Medium
+ *
+ * You are given a positive integer array grades which represents the grades of students in
+ * a university. You would like to enter all these students into a competition in ordered
+ * non-empty groups, such that the ordering meets the following conditions:
+ * - The sum of the grades of students in the ith group is less than the sum of the grades of
+ *   students in the (i + 1)th group, for all groups (except the last).
+ * - The total number of students in the ith group is less than the total number of students
+ *   in the (i + 1)th group, for all groups (except the last).
+ *
+ * Return the maximum number of groups that can be formed.
+ */
+
+/**
+ * @param {number[]} grades
+ * @return {number}
+ */
+var maximumGroups = function(grades) {
+  let result = 0;
+  let studentsUsed = 0;
+
+  while (studentsUsed + result + 1 <= grades.length) {
+    result++;
+    studentsUsed += result;
+  }
+
+  return result;
+};
diff --git a/solutions/2359-find-closest-node-to-given-two-nodes.js b/solutions/2359-find-closest-node-to-given-two-nodes.js
new file mode 100644
index 00000000..c048153f
--- /dev/null
+++ b/solutions/2359-find-closest-node-to-given-two-nodes.js
@@ -0,0 +1,63 @@
+/**
+ * 2359. Find Closest Node to Given Two Nodes
+ * https://leetcode.com/problems/find-closest-node-to-given-two-nodes/
+ * Difficulty: Medium
+ *
+ * You are given a directed graph of n nodes numbered from 0 to n - 1, where each node has at
+ * most one outgoing edge.
+ *
+ * The graph is represented with a given 0-indexed array edges of size n, indicating that there
+ * is a directed edge from node i to node edges[i]. If there is no outgoing edge from i, then
+ * edges[i] == -1.
+ *
+ * You are also given two integers node1 and node2.
+ *
+ * Return the index of the node that can be reached from both node1 and node2, such that the maximum
+ * between the distance from node1 to that node, and from node2 to that node is minimized. If there
+ * are multiple answers, return the node with the smallest index, and if no possible answer exists,
+ * return -1.
+ *
+ * Note that edges may contain cycles.
+ */
+
+/**
+ * @param {number[]} edges
+ * @param {number} node1
+ * @param {number} node2
+ * @return {number}
+ */
+var closestMeetingNode = function(edges, node1, node2) {
+  const n = edges.length;
+  const dist1 = new Array(n).fill(Infinity);
+  const dist2 = new Array(n).fill(Infinity);
+
+  computeDistances(node1, dist1);
+  computeDistances(node2, dist2);
+
+  let minDist = Infinity;
+  let result = -1;
+
+  for (let i = 0; i < n; i++) {
+    if (dist1[i] !== Infinity && dist2[i] !== Infinity) {
+      const maxDist = Math.max(dist1[i], dist2[i]);
+      if (maxDist < minDist) {
+        minDist = maxDist;
+        result = i;
+      }
+    }
+  }
+
+  return result;
+
+  function computeDistances(start, distances) {
+    let current = start;
+    let steps = 0;
+    const visited = new Set();
+
+    while (current !== -1 && !visited.has(current)) {
+      distances[current] = steps++;
+      visited.add(current);
+      current = edges[current];
+    }
+  }
+};
diff --git a/solutions/2360-longest-cycle-in-a-graph.js b/solutions/2360-longest-cycle-in-a-graph.js
new file mode 100644
index 00000000..6da96750
--- /dev/null
+++ b/solutions/2360-longest-cycle-in-a-graph.js
@@ -0,0 +1,53 @@
+/**
+ * 2360. Longest Cycle in a Graph
+ * https://leetcode.com/problems/longest-cycle-in-a-graph/
+ * Difficulty: Hard
+ *
+ * You are given a directed graph of n nodes numbered from 0 to n - 1, where each node has at
+ * most one outgoing edge.
+ *
+ * The graph is represented with a given 0-indexed array edges of size n, indicating that
+ * there is a directed edge from node i to node edges[i]. If there is no outgoing edge from
+ * node i, then edges[i] == -1.
+ *
+ * Return the length of the longest cycle in the graph. If no cycle exists, return -1.
+ *
+ * A cycle is a path that starts and ends at the same node.
+ */
+
+/**
+ * @param {number[]} edges
+ * @return {number}
+ */
+var longestCycle = function(edges) {
+  const n = edges.length;
+  const visited = new Array(n).fill(false);
+  let result = -1;
+
+  for (let i = 0; i < n; i++) {
+    if (!visited[i]) {
+      findCycle(i, 0, []);
+    }
+  }
+
+  return result;
+
+  function findCycle(node, steps, path) {
+    if (visited[node]) {
+      const cycleStart = path.indexOf(node);
+      if (cycleStart !== -1) {
+        result = Math.max(result, steps - cycleStart);
+      }
+      return;
+    }
+
+    visited[node] = true;
+    path.push(node);
+
+    if (edges[node] !== -1) {
+      findCycle(edges[node], steps + 1, path);
+    }
+
+    path.pop();
+  }
+};
diff --git a/solutions/2363-merge-similar-items.js b/solutions/2363-merge-similar-items.js
new file mode 100644
index 00000000..da16c478
--- /dev/null
+++ b/solutions/2363-merge-similar-items.js
@@ -0,0 +1,40 @@
+/**
+ * 2363. Merge Similar Items
+ * https://leetcode.com/problems/merge-similar-items/
+ * Difficulty: Easy
+ *
+ * You are given two 2D integer arrays, items1 and items2, representing two sets of items.
+ * Each array items has the following properties:
+ * - items[i] = [valuei, weighti] where valuei represents the value and weighti represents
+ *   the weight of the ith item.
+ * - The value of each item in items is unique.
+ *
+ * Return a 2D integer array ret where ret[i] = [valuei, weighti], with weighti being the sum of
+ * weights of all items with value valuei.
+ *
+ * Note: ret should be returned in ascending order by value.
+ */
+
+/**
+ * @param {number[][]} items1
+ * @param {number[][]} items2
+ * @return {number[][]}
+ */
+var mergeSimilarItems = function(items1, items2) {
+  const map = new Map();
+
+  for (const [value, weight] of items1) {
+    map.set(value, (map.get(value) || 0) + weight);
+  }
+
+  for (const [value, weight] of items2) {
+    map.set(value, (map.get(value) || 0) + weight);
+  }
+
+  const merged = [];
+  for (const [value, weight] of map) {
+    merged.push([value, weight]);
+  }
+
+  return merged.sort((a, b) => a[0] - b[0]);
+};
diff --git a/solutions/2365-task-scheduler-ii.js b/solutions/2365-task-scheduler-ii.js
new file mode 100644
index 00000000..166fdfdf
--- /dev/null
+++ b/solutions/2365-task-scheduler-ii.js
@@ -0,0 +1,41 @@
+/**
+ * 2365. Task Scheduler II
+ * https://leetcode.com/problems/task-scheduler-ii/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed array of positive integers tasks, representing tasks that need
+ * to be completed in order, where tasks[i] represents the type of the ith task.
+ *
+ * You are also given a positive integer space, which represents the minimum number of days
+ * that must pass after the completion of a task before another task of the same type can be
+ * performed.
+ *
+ * Each day, until all tasks have been completed, you must either:
+ * - Complete the next task from tasks, or
+ * - Take a break.
+ *
+ * Return the minimum number of days needed to complete all tasks.
+ */
+
+/**
+ * @param {number[]} tasks
+ * @param {number} space
+ * @return {number}
+ */
+var taskSchedulerII = function(tasks, space) {
+  const map = new Map();
+  let result = 0;
+
+  for (const task of tasks) {
+    result++;
+    if (map.has(task)) {
+      const lastDay = map.get(task);
+      if (result - lastDay <= space) {
+        result = lastDay + space + 1;
+      }
+    }
+    map.set(task, result);
+  }
+
+  return result;
+};
diff --git a/solutions/2366-minimum-replacements-to-sort-the-array.js b/solutions/2366-minimum-replacements-to-sort-the-array.js
new file mode 100644
index 00000000..422d21f6
--- /dev/null
+++ b/solutions/2366-minimum-replacements-to-sort-the-array.js
@@ -0,0 +1,33 @@
+/**
+ * 2366. Minimum Replacements to Sort the Array
+ * https://leetcode.com/problems/minimum-replacements-to-sort-the-array/
+ * Difficulty: Hard
+ *
+ * You are given a 0-indexed integer array nums. In one operation you can replace any element
+ * of the array with any two elements that sum to it.
+ * - For example, consider nums = [5,6,7]. In one operation, we can replace nums[1] with 2 and 4
+ *   and convert nums to [5,2,4,7].
+ *
+ * Return the minimum number of operations to make an array that is sorted in non-decreasing order.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var minimumReplacement = function(nums) {
+  let result = 0;
+  let prev = nums[nums.length - 1];
+
+  for (let i = nums.length - 2; i >= 0; i--) {
+    if (nums[i] > prev) {
+      const splits = Math.ceil(nums[i] / prev);
+      result += splits - 1;
+      prev = Math.floor(nums[i] / splits);
+    } else {
+      prev = nums[i];
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2367-number-of-arithmetic-triplets.js b/solutions/2367-number-of-arithmetic-triplets.js
new file mode 100644
index 00000000..21c02a86
--- /dev/null
+++ b/solutions/2367-number-of-arithmetic-triplets.js
@@ -0,0 +1,41 @@
+/**
+ * 2367. Number of Arithmetic Triplets
+ * https://leetcode.com/problems/number-of-arithmetic-triplets/
+ * Difficulty: Easy
+ *
+ * You are given a 0-indexed, strictly increasing integer array nums and a positive integer diff.
+ * A triplet (i, j, k) is an arithmetic triplet if the following conditions are met:
+ * - i < j < k,
+ * - nums[j] - nums[i] == diff, and
+ * - nums[k] - nums[j] == diff.
+ *
+ * Return the number of unique arithmetic triplets.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} diff
+ * @return {number}
+ */
+var arithmeticTriplets = function(nums, diff) {
+  let result = 0;
+
+  for (let j = 1; j < nums.length - 1; j++) {
+    let i = j - 1;
+    let k = j + 1;
+
+    while (i >= 0 && nums[j] - nums[i] <= diff) {
+      if (nums[j] - nums[i] === diff) {
+        while (k < nums.length && nums[k] - nums[j] <= diff) {
+          if (nums[k] - nums[j] === diff) {
+            result++;
+          }
+          k++;
+        }
+      }
+      i--;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2368-reachable-nodes-with-restrictions.js b/solutions/2368-reachable-nodes-with-restrictions.js
new file mode 100644
index 00000000..1ecd0821
--- /dev/null
+++ b/solutions/2368-reachable-nodes-with-restrictions.js
@@ -0,0 +1,43 @@
+/**
+ * 2368. Reachable Nodes With Restrictions
+ * https://leetcode.com/problems/reachable-nodes-with-restrictions/
+ * Difficulty: Medium
+ *
+ * There is an undirected tree with n nodes labeled from 0 to n - 1 and n - 1 edges.
+ *
+ * You are given a 2D integer array edges of length n - 1 where edges[i] = [ai, bi] indicates
+ * that there is an edge between nodes ai and bi in the tree. You are also given an integer
+ * array restricted which represents restricted nodes.
+ *
+ * Return the maximum number of nodes you can reach from node 0 without visiting a restricted node.
+ *
+ * Note that node 0 will not be a restricted node.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} edges
+ * @param {number[]} restricted
+ * @return {number}
+ */
+var reachableNodes = function(n, edges, restricted) {
+  const graph = Array(n).fill().map(() => []);
+  const restrictedSet = new Set(restricted);
+  const visited = new Set();
+
+  for (const [u, v] of edges) {
+    graph[u].push(v);
+    graph[v].push(u);
+  }
+
+  explore(0);
+  return visited.size;
+
+  function explore(node) {
+    if (visited.has(node) || restrictedSet.has(node)) return;
+    visited.add(node);
+    for (const neighbor of graph[node]) {
+      explore(neighbor);
+    }
+  }
+};
diff --git a/solutions/2369-check-if-there-is-a-valid-partition-for-the-array.js b/solutions/2369-check-if-there-is-a-valid-partition-for-the-array.js
new file mode 100644
index 00000000..be0e9844
--- /dev/null
+++ b/solutions/2369-check-if-there-is-a-valid-partition-for-the-array.js
@@ -0,0 +1,44 @@
+/**
+ * 2369. Check if There is a Valid Partition For The Array
+ * https://leetcode.com/problems/check-if-there-is-a-valid-partition-for-the-array/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed integer array nums. You have to partition the array into one or
+ * more contiguous subarrays.
+ *
+ * We call a partition of the array valid if each of the obtained subarrays satisfies one of
+ * the following conditions:
+ * 1. The subarray consists of exactly 2, equal elements. For example, the subarray [2,2] is good.
+ * 2. The subarray consists of exactly 3, equal elements. For example, the subarray [4,4,4] is good.
+ * 3. The subarray consists of exactly 3 consecutive increasing elements, that is, the difference
+ *    between adjacent elements is 1. For example, the subarray [3,4,5] is good, but the subarray
+ *    [1,3,5] is not.
+ *
+ * Return true if the array has at least one valid partition. Otherwise, return false.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {boolean}
+ */
+var validPartition = function(nums) {
+  const n = nums.length;
+  const dp = new Array(n + 1).fill(false);
+  dp[0] = true;
+
+  for (let i = 2; i <= n; i++) {
+    if (nums[i - 1] === nums[i - 2]) {
+      dp[i] = dp[i] || dp[i - 2];
+    }
+    if (i >= 3) {
+      if (nums[i - 1] === nums[i - 2] && nums[i - 2] === nums[i - 3]) {
+        dp[i] = dp[i] || dp[i - 3];
+      }
+      if (nums[i - 1] === nums[i - 2] + 1 && nums[i - 2] === nums[i - 3] + 1) {
+        dp[i] = dp[i] || dp[i - 3];
+      }
+    }
+  }
+
+  return dp[n];
+};
diff --git a/solutions/2370-longest-ideal-subsequence.js b/solutions/2370-longest-ideal-subsequence.js
new file mode 100644
index 00000000..c07e27f9
--- /dev/null
+++ b/solutions/2370-longest-ideal-subsequence.js
@@ -0,0 +1,43 @@
+/**
+ * 2370. Longest Ideal Subsequence
+ * https://leetcode.com/problems/longest-ideal-subsequence/
+ * Difficulty: Medium
+ *
+ * You are given a string s consisting of lowercase letters and an integer k. We call a string
+ * t ideal if the following conditions are satisfied:
+ * - t is a subsequence of the string s.
+ * - The absolute difference in the alphabet order of every two adjacent letters in t is less
+ *   than or equal to k.
+ *
+ * Return the length of the longest ideal string.
+ *
+ * A subsequence is a string that can be derived from another string by deleting some or no
+ * characters without changing the order of the remaining characters.
+ *
+ * Note that the alphabet order is not cyclic. For example, the absolute difference in the
+ * alphabet order of 'a' and 'z' is 25, not 1.
+ */
+
+/**
+ * @param {string} s
+ * @param {number} k
+ * @return {number}
+ */
+var longestIdealString = function(s, k) {
+  const dp = new Array(26).fill(0);
+  let result = 0;
+
+  for (const char of s) {
+    const idx = char.charCodeAt(0) - 97;
+    let currentMax = 0;
+
+    for (let i = Math.max(0, idx - k); i <= Math.min(25, idx + k); i++) {
+      currentMax = Math.max(currentMax, dp[i]);
+    }
+
+    dp[idx] = currentMax + 1;
+    result = Math.max(result, dp[idx]);
+  }
+
+  return result;
+};
diff --git a/solutions/2373-largest-local-values-in-a-matrix.js b/solutions/2373-largest-local-values-in-a-matrix.js
new file mode 100644
index 00000000..46a65c45
--- /dev/null
+++ b/solutions/2373-largest-local-values-in-a-matrix.js
@@ -0,0 +1,38 @@
+/**
+ * 2373. Largest Local Values in a Matrix
+ * https://leetcode.com/problems/largest-local-values-in-a-matrix/
+ * Difficulty: Easy
+ *
+ * You are given an n x n integer matrix grid.
+ *
+ * Generate an integer matrix maxLocal of size (n - 2) x (n - 2) such that:
+ * - maxLocal[i][j] is equal to the largest value of the 3 x 3 matrix in grid centered around
+ *   row i + 1 and column j + 1.
+ *
+ * In other words, we want to find the largest value in every contiguous 3 x 3 matrix in grid.
+ *
+ * Return the generated matrix.
+ */
+
+/**
+ * @param {number[][]} grid
+ * @return {number[][]}
+ */
+var largestLocal = function(grid) {
+  const n = grid.length;
+  const result = new Array(n - 2).fill().map(() => new Array(n - 2).fill(0));
+
+  for (let i = 0; i < n - 2; i++) {
+    for (let j = 0; j < n - 2; j++) {
+      let maxVal = 0;
+      for (let r = i; r < i + 3; r++) {
+        for (let c = j; c < j + 3; c++) {
+          maxVal = Math.max(maxVal, grid[r][c]);
+        }
+      }
+      result[i][j] = maxVal;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2374-node-with-highest-edge-score.js b/solutions/2374-node-with-highest-edge-score.js
new file mode 100644
index 00000000..b8597422
--- /dev/null
+++ b/solutions/2374-node-with-highest-edge-score.js
@@ -0,0 +1,39 @@
+/**
+ * 2374. Node With Highest Edge Score
+ * https://leetcode.com/problems/node-with-highest-edge-score/
+ * Difficulty: Medium
+ *
+ * You are given a directed graph with n nodes labeled from 0 to n - 1, where each node has
+ * exactly one outgoing edge.
+ *
+ * The graph is represented by a given 0-indexed integer array edges of length n, where edges[i]
+ * indicates that there is a directed edge from node i to node edges[i].
+ *
+ * The edge score of a node i is defined as the sum of the labels of all the nodes that have an
+ * edge pointing to i.
+ *
+ * Return the node with the highest edge score. If multiple nodes have the same edge score, return
+ * the node with the smallest index.
+ */
+
+/**
+ * @param {number[]} edges
+ * @return {number}
+ */
+var edgeScore = function(edges) {
+  const scores = new Array(edges.length).fill(0);
+
+  for (let i = 0; i < edges.length; i++) {
+    scores[edges[i]] += i;
+  }
+  let maxScore = 0;
+  let result = 0;
+  for (let i = 0; i < scores.length; i++) {
+    if (scores[i] > maxScore) {
+      maxScore = scores[i];
+      result = i;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2380-time-needed-to-rearrange-a-binary-string.js b/solutions/2380-time-needed-to-rearrange-a-binary-string.js
new file mode 100644
index 00000000..d85d2234
--- /dev/null
+++ b/solutions/2380-time-needed-to-rearrange-a-binary-string.js
@@ -0,0 +1,32 @@
+/**
+ * 2380. Time Needed to Rearrange a Binary String
+ * https://leetcode.com/problems/time-needed-to-rearrange-a-binary-string/
+ * Difficulty: Medium
+ *
+ * You are given a binary string s. In one second, all occurrences of "01" are simultaneously
+ * replaced with "10". This process repeats until no occurrences of "01" exist.
+ *
+ * Return the number of seconds needed to complete this process.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var secondsToRemoveOccurrences = function(s) {
+  const binary = s.split('');
+  let result = 0;
+
+  while (binary.join('').includes('01')) {
+    for (let i = 0; i < binary.length - 1; i++) {
+      if (binary[i] === '0' && binary[i + 1] === '1') {
+        binary[i] = '1';
+        binary[i + 1] = '0';
+        i++;
+      }
+    }
+    result++;
+  }
+
+  return result;
+};
diff --git a/solutions/2382-maximum-segment-sum-after-removals.js b/solutions/2382-maximum-segment-sum-after-removals.js
new file mode 100644
index 00000000..78065c5e
--- /dev/null
+++ b/solutions/2382-maximum-segment-sum-after-removals.js
@@ -0,0 +1,84 @@
+/**
+ * 2382. Maximum Segment Sum After Removals
+ * https://leetcode.com/problems/maximum-segment-sum-after-removals/
+ * Difficulty: Hard
+ *
+ * You are given two 0-indexed integer arrays nums and removeQueries, both of length n. For
+ * the ith query, the element in nums at the index removeQueries[i] is removed, splitting
+ * nums into different segments.
+ *
+ * A segment is a contiguous sequence of positive integers in nums. A segment sum is the sum
+ * of every element in a segment.
+ *
+ * Return an integer array answer, of length n, where answer[i] is the maximum segment sum after
+ * applying the ith removal.
+ *
+ * Note: The same index will not be removed more than once.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number[]} removeQueries
+ * @return {number[]}
+ */
+var maximumSegmentSum = function(nums, removeQueries) {
+  const n = nums.length;
+  const result = new Array(n);
+  const prefixSum = new Array(n + 1).fill(0);
+  const parents = new Array(n).fill(-1);
+  const sizes = new Array(n).fill(0);
+  const sums = new Array(n).fill(0);
+  let maxSum = 0;
+
+  for (let i = 0; i < n; i++) {
+    prefixSum[i + 1] = prefixSum[i] + nums[i];
+  }
+
+  for (let i = n - 1; i >= 0; i--) {
+    result[i] = maxSum;
+    const index = removeQueries[i];
+
+    parents[index] = index;
+    sizes[index] = 1;
+    sums[index] = nums[index];
+
+    if (index > 0 && parents[index - 1] !== -1) {
+      const leftRoot = findRoot(parents, index - 1);
+      const segmentSum = sums[leftRoot] + sums[index];
+
+      parents[index] = leftRoot;
+      sizes[leftRoot] += sizes[index];
+      sums[leftRoot] = segmentSum;
+    }
+
+    if (index < n - 1 && parents[index + 1] !== -1) {
+      const root = findRoot(parents, index);
+      const rightRoot = findRoot(parents, index + 1);
+
+      if (root !== rightRoot) {
+        const segmentSum = sums[root] + sums[rightRoot];
+
+        if (sizes[root] < sizes[rightRoot]) {
+          parents[root] = rightRoot;
+          sizes[rightRoot] += sizes[root];
+          sums[rightRoot] = segmentSum;
+        } else {
+          parents[rightRoot] = root;
+          sizes[root] += sizes[rightRoot];
+          sums[root] = segmentSum;
+        }
+      }
+    }
+
+    maxSum = Math.max(maxSum, sums[findRoot(parents, index)]);
+  }
+
+  return result;
+};
+
+function findRoot(parents, x) {
+  if (parents[x] !== x) {
+    parents[x] = findRoot(parents, parents[x]);
+  }
+  return parents[x];
+}
diff --git a/solutions/2383-minimum-hours-of-training-to-win-a-competition.js b/solutions/2383-minimum-hours-of-training-to-win-a-competition.js
new file mode 100644
index 00000000..0699fdcc
--- /dev/null
+++ b/solutions/2383-minimum-hours-of-training-to-win-a-competition.js
@@ -0,0 +1,55 @@
+/**
+ * 2383. Minimum Hours of Training to Win a Competition
+ * https://leetcode.com/problems/minimum-hours-of-training-to-win-a-competition/
+ * Difficulty: Easy
+ *
+ * You are entering a competition, and are given two positive integers initialEnergy and
+ * initialExperience denoting your initial energy and initial experience respectively.
+ *
+ * You are also given two 0-indexed integer arrays energy and experience, both of length n.
+ *
+ * You will face n opponents in order. The energy and experience of the ith opponent is
+ * denoted by energy[i] and experience[i] respectively. When you face an opponent, you need
+ * to have both strictly greater experience and energy to defeat them and move to the next
+ * opponent if available.
+ *
+ * Defeating the ith opponent increases your experience by experience[i], but decreases your
+ * energy by energy[i].
+ *
+ * Before starting the competition, you can train for some number of hours. After each hour
+ * of training, you can either choose to increase your initial experience by one, or increase
+ * your initial energy by one.
+ *
+ * Return the minimum number of training hours required to defeat all n opponents.
+ */
+
+/**
+ * @param {number} initialEnergy
+ * @param {number} initialExperience
+ * @param {number[]} energy
+ * @param {number[]} experience
+ * @return {number}
+ */
+var minNumberOfHours = function(initialEnergy, initialExperience, energy, experience) {
+  let energyNeeded = 0;
+  let experienceNeeded = 0;
+  let currentEnergy = initialEnergy;
+  let currentExperience = initialExperience;
+
+  for (let i = 0; i < energy.length; i++) {
+    if (currentEnergy <= energy[i]) {
+      const deficit = energy[i] - currentEnergy + 1;
+      energyNeeded += deficit;
+      currentEnergy += deficit;
+    }
+    if (currentExperience <= experience[i]) {
+      const deficit = experience[i] - currentExperience + 1;
+      experienceNeeded += deficit;
+      currentExperience += deficit;
+    }
+    currentEnergy -= energy[i];
+    currentExperience += experience[i];
+  }
+
+  return energyNeeded + experienceNeeded;
+};
diff --git a/solutions/2385-amount-of-time-for-binary-tree-to-be-infected.js b/solutions/2385-amount-of-time-for-binary-tree-to-be-infected.js
new file mode 100644
index 00000000..00bd2ddd
--- /dev/null
+++ b/solutions/2385-amount-of-time-for-binary-tree-to-be-infected.js
@@ -0,0 +1,70 @@
+/**
+ * 2385. Amount of Time for Binary Tree to Be Infected
+ * https://leetcode.com/problems/amount-of-time-for-binary-tree-to-be-infected/
+ * Difficulty: Medium
+ *
+ * You are given the root of a binary tree with unique values, and an integer start.
+ * At minute 0, an infection starts from the node with value start.
+ *
+ * Each minute, a node becomes infected if:
+ * - The node is currently uninfected.
+ * - The node is adjacent to an infected node.
+ *
+ * Return the number of minutes needed for the entire tree to be infected.
+ */
+
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @param {number} start
+ * @return {number}
+ */
+var amountOfTime = function(root, start) {
+  const graph = new Map();
+  const queue = [start];
+  const visited = new Set([start]);
+  let result = -1;
+
+  buildGraph(root, null);
+
+  while (queue.length) {
+    result++;
+    const levelSize = queue.length;
+    for (let i = 0; i < levelSize; i++) {
+      const current = queue.shift();
+      for (const neighbor of graph.get(current) || []) {
+        if (!visited.has(neighbor)) {
+          visited.add(neighbor);
+          queue.push(neighbor);
+        }
+      }
+    }
+  }
+
+  return result;
+
+  function buildGraph(node, parent) {
+    if (!node) return;
+    if (!graph.has(node.val)) graph.set(node.val, []);
+    if (parent) graph.get(node.val).push(parent.val);
+    if (node.left) {
+      graph.get(node.val).push(node.left.val);
+      graph.set(node.left.val, graph.get(node.left.val) || []);
+      graph.get(node.left.val).push(node.val);
+    }
+    if (node.right) {
+      graph.get(node.val).push(node.right.val);
+      graph.set(node.right.val, graph.get(node.right.val) || []);
+      graph.get(node.right.val).push(node.val);
+    }
+    buildGraph(node.left, node);
+    buildGraph(node.right, node);
+  }
+};
diff --git a/solutions/2389-longest-subsequence-with-limited-sum.js b/solutions/2389-longest-subsequence-with-limited-sum.js
new file mode 100644
index 00000000..a498b7c8
--- /dev/null
+++ b/solutions/2389-longest-subsequence-with-limited-sum.js
@@ -0,0 +1,43 @@
+/**
+ * 2389. Longest Subsequence With Limited Sum
+ * https://leetcode.com/problems/longest-subsequence-with-limited-sum/
+ * Difficulty: Easy
+ *
+ * You are given an integer array nums of length n, and an integer array queries of length m.
+ *
+ * Return an array answer of length m where answer[i] is the maximum size of a subsequence that
+ * you can take from nums such that the sum of its elements is less than or equal to queries[i].
+ *
+ * A subsequence is an array that can be derived from another array by deleting some or no elements
+ * without changing the order of the remaining elements.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number[]} queries
+ * @return {number[]}
+ */
+var answerQueries = function(nums, queries) {
+  nums.sort((a, b) => a - b);
+  const prefixSums = [0];
+  for (const num of nums) {
+    prefixSums.push(prefixSums.at(-1) + num);
+  }
+
+  const result = new Array(queries.length);
+  for (let i = 0; i < queries.length; i++) {
+    let left = 0;
+    let right = nums.length;
+    while (left < right) {
+      const mid = Math.floor((left + right + 1) / 2);
+      if (prefixSums[mid] <= queries[i]) {
+        left = mid;
+      } else {
+        right = mid - 1;
+      }
+    }
+    result[i] = left;
+  }
+
+  return result;
+};
diff --git a/solutions/2391-minimum-amount-of-time-to-collect-garbage.js b/solutions/2391-minimum-amount-of-time-to-collect-garbage.js
new file mode 100644
index 00000000..bbddfc01
--- /dev/null
+++ b/solutions/2391-minimum-amount-of-time-to-collect-garbage.js
@@ -0,0 +1,50 @@
+/**
+ * 2391. Minimum Amount of Time to Collect Garbage
+ * https://leetcode.com/problems/minimum-amount-of-time-to-collect-garbage/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed array of strings garbage where garbage[i] represents the assortment
+ * of garbage at the ith house. garbage[i] consists only of the characters 'M', 'P' and 'G'
+ * representing one unit of metal, paper and glass garbage respectively. Picking up one unit of
+ * any type of garbage takes 1 minute.
+ *
+ * You are also given a 0-indexed integer array travel where travel[i] is the number of minutes
+ * needed to go from house i to house i + 1.
+ *
+ * There are three garbage trucks in the city, each responsible for picking up one type of garbage.
+ * Each garbage truck starts at house 0 and must visit each house in order; however, they do not
+ * need to visit every house.
+ *
+ * Only one garbage truck may be used at any given moment. While one truck is driving or picking up
+ * garbage, the other two trucks cannot do anything.
+ *
+ * Return the minimum number of minutes needed to pick up all the garbage.
+ */
+
+/**
+ * @param {string[]} garbage
+ * @param {number[]} travel
+ * @return {number}
+ */
+var garbageCollection = function(garbage, travel) {
+  let result = 0;
+  const lastHouse = {'M': 0, 'P': 0, 'G': 0};
+
+  for (let i = 0; i < garbage.length; i++) {
+    result += garbage[i].length;
+    for (const type of garbage[i]) {
+      lastHouse[type] = i;
+    }
+  }
+
+  const prefixTravel = [0];
+  for (const time of travel) {
+    prefixTravel.push(prefixTravel.at(-1) + time);
+  }
+
+  Object.keys(lastHouse).forEach(type => {
+    result += prefixTravel[lastHouse[type]];
+  });
+
+  return result;
+};
diff --git a/solutions/2392-build-a-matrix-with-conditions.js b/solutions/2392-build-a-matrix-with-conditions.js
new file mode 100644
index 00000000..b55cdd0f
--- /dev/null
+++ b/solutions/2392-build-a-matrix-with-conditions.js
@@ -0,0 +1,76 @@
+/**
+ * 2392. Build a Matrix With Conditions
+ * https://leetcode.com/problems/build-a-matrix-with-conditions/
+ * Difficulty: Hard
+ *
+ * You are given a positive integer k. You are also given:
+ * - a 2D integer array rowConditions of size n where rowConditions[i] = [abovei, belowi], and
+ * - a 2D integer array colConditions of size m where colConditions[i] = [lefti, righti].
+ *
+ * The two arrays contain integers from 1 to k.
+ *
+ * You have to build a k x k matrix that contains each of the numbers from 1 to k exactly once.
+ * The remaining cells should have the value 0.
+ *
+ * The matrix should also satisfy the following conditions:
+ * - The number abovei should appear in a row that is strictly above the row at which the number
+ *   belowi appears for all i from 0 to n - 1.
+ * - The number lefti should appear in a column that is strictly left of the column at which the
+ *   number righti appears for all i from 0 to m - 1.
+ *
+ * Return any matrix that satisfies the conditions. If no answer exists, return an empty matrix.
+ */
+
+/**
+ * @param {number} k
+ * @param {number[][]} rowConditions
+ * @param {number[][]} colConditions
+ * @return {number[][]}
+ */
+var buildMatrix = function(k, rowConditions, colConditions) {
+  const rowOrder = helper(rowConditions, k);
+  if (!rowOrder.length) return [];
+
+  const colOrder = helper(colConditions, k);
+  if (!colOrder.length) return [];
+
+  const matrix = Array.from({ length: k }, () => new Array(k).fill(0));
+  const rowIndex = new Array(k + 1).fill(0);
+  const colIndex = new Array(k + 1).fill(0);
+
+  for (let i = 0; i < k; i++) {
+    rowIndex[rowOrder[i]] = i;
+    colIndex[colOrder[i]] = i;
+  }
+
+  for (let num = 1; num <= k; num++) {
+    matrix[rowIndex[num]][colIndex[num]] = num;
+  }
+
+  return matrix;
+
+  function helper(edges, size) {
+    const graph = Array.from({ length: size + 1 }, () => []);
+    const inDegree = new Array(size + 1).fill(0);
+    for (const [u, v] of edges) {
+      graph[u].push(v);
+      inDegree[v]++;
+    }
+
+    const queue = [];
+    for (let i = 1; i <= size; i++) {
+      if (inDegree[i] === 0) queue.push(i);
+    }
+
+    const order = [];
+    while (queue.length) {
+      const node = queue.shift();
+      order.push(node);
+      for (const next of graph[node]) {
+        if (--inDegree[next] === 0) queue.push(next);
+      }
+    }
+
+    return order.length === size ? order : [];
+  }
+};
diff --git a/solutions/2395-find-subarrays-with-equal-sum.js b/solutions/2395-find-subarrays-with-equal-sum.js
new file mode 100644
index 00000000..9b4ec520
--- /dev/null
+++ b/solutions/2395-find-subarrays-with-equal-sum.js
@@ -0,0 +1,28 @@
+/**
+ * 2395. Find Subarrays With Equal Sum
+ * https://leetcode.com/problems/find-subarrays-with-equal-sum/
+ * Difficulty: Easy
+ *
+ * Given a 0-indexed integer array nums, determine whether there exist two subarrays of length
+ * 2 with equal sum. Note that the two subarrays must begin at different indices.
+ *
+ * Return true if these subarrays exist, and false otherwise.
+ *
+ * A subarray is a contiguous non-empty sequence of elements within an array.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {boolean}
+ */
+var findSubarrays = function(nums) {
+  const set = new Set();
+
+  for (let i = 0; i < nums.length - 1; i++) {
+    const sum = nums[i] + nums[i + 1];
+    if (set.has(sum)) return true;
+    set.add(sum);
+  }
+
+  return false;
+};
diff --git a/solutions/2397-maximum-rows-covered-by-columns.js b/solutions/2397-maximum-rows-covered-by-columns.js
new file mode 100644
index 00000000..0fa68a2b
--- /dev/null
+++ b/solutions/2397-maximum-rows-covered-by-columns.js
@@ -0,0 +1,61 @@
+/**
+ * 2397. Maximum Rows Covered by Columns
+ * https://leetcode.com/problems/maximum-rows-covered-by-columns/
+ * Difficulty: Medium
+ *
+ * You are given an m x n binary matrix matrix and an integer numSelect.
+ *
+ * Your goal is to select exactly numSelect distinct columns from matrix such that you cover
+ * as many rows as possible.
+ *
+ * A row is considered covered if all the 1's in that row are also part of a column that you
+ * have selected. If a row does not have any 1s, it is also considered covered.
+ *
+ * More formally, let us consider selected = {c1, c2, ...., cnumSelect} as the set of columns
+ * selected by you. A row i is covered by selected if:
+ * - For each cell where matrix[i][j] == 1, the column j is in selected.
+ * - Or, no cell in row i has a value of 1.
+ *
+ * Return the maximum number of rows that can be covered by a set of numSelect columns.
+ */
+
+/**
+ * @param {number[][]} matrix
+ * @param {number} numSelect
+ * @return {number}
+ */
+var maximumRows = function(matrix, numSelect) {
+  const rows = matrix.length;
+  const cols = matrix[0].length;
+  let result = 0;
+
+  backtrack(0, 0, 0);
+
+  return result;
+
+  function countCovered(selected) {
+    let covered = 0;
+    for (let i = 0; i < rows; i++) {
+      let isCovered = true;
+      for (let j = 0; j < cols; j++) {
+        if (matrix[i][j] === 1 && !(selected & (1 << j))) {
+          isCovered = false;
+          break;
+        }
+      }
+      if (isCovered) covered++;
+    }
+    return covered;
+  }
+
+  function backtrack(index, selected, count) {
+    if (count === numSelect) {
+      result = Math.max(result, countCovered(selected));
+      return;
+    }
+    if (index >= cols || cols - index + count < numSelect) return;
+
+    backtrack(index + 1, selected | (1 << index), count + 1);
+    backtrack(index + 1, selected, count);
+  }
+};
diff --git a/solutions/2399-check-distances-between-same-letters.js b/solutions/2399-check-distances-between-same-letters.js
new file mode 100644
index 00000000..7f977d1e
--- /dev/null
+++ b/solutions/2399-check-distances-between-same-letters.js
@@ -0,0 +1,44 @@
+/**
+ * 2399. Check Distances Between Same Letters
+ * https://leetcode.com/problems/check-distances-between-same-letters/
+ * Difficulty: Easy
+ *
+ * You are given a 0-indexed string s consisting of only lowercase English letters, where each
+ * letter in s appears exactly twice. You are also given a 0-indexed integer array distance of
+ * length 26.
+ *
+ * Each letter in the alphabet is numbered from 0 to 25 (i.e. 'a' -> 0, 'b' -> 1,
+ * 'c' -> 2, ... , 'z' -> 25).
+ *
+ * In a well-spaced string, the number of letters between the two occurrences of the ith letter
+ * is distance[i]. If the ith letter does not appear in s, then distance[i] can be ignored.
+ *
+ * Return true if s is a well-spaced string, otherwise return false.
+ */
+
+/**
+ * @param {string} s
+ * @param {number[]} distance
+ * @return {boolean}
+ */
+var checkDistances = function(s, distance) {
+  const map = new Map();
+
+  for (let i = 0; i < s.length; i++) {
+    const char = s[i];
+    if (map.has(char)) {
+      map.get(char).push(i);
+    } else {
+      map.set(char, [i]);
+    }
+  }
+
+  for (const [char, positions] of map) {
+    const letterIndex = char.charCodeAt(0) - 'a'.charCodeAt(0);
+    if (positions[1] - positions[0] - 1 !== distance[letterIndex]) {
+      return false;
+    }
+  }
+
+  return true;
+};
diff --git a/solutions/2404-most-frequent-even-element.js b/solutions/2404-most-frequent-even-element.js
new file mode 100644
index 00000000..e5fcd0c6
--- /dev/null
+++ b/solutions/2404-most-frequent-even-element.js
@@ -0,0 +1,33 @@
+/**
+ * 2404. Most Frequent Even Element
+ * https://leetcode.com/problems/most-frequent-even-element/
+ * Difficulty: Easy
+ *
+ * Given an integer array nums, return the most frequent even element.
+ *
+ * If there is a tie, return the smallest one. If there is no such element, return -1.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var mostFrequentEven = function(nums) {
+  const map = new Map();
+  let maxCount = 0;
+  let result = -1;
+
+  for (const num of nums) {
+    if (num % 2 === 0) {
+      const count = (map.get(num) || 0) + 1;
+      map.set(num, count);
+      if (count > maxCount || (count === maxCount && num < result)) {
+        maxCount = count;
+        result = num;
+      }
+    }
+  }
+
+  return result;
+};
+
diff --git a/solutions/2405-optimal-partition-of-string.js b/solutions/2405-optimal-partition-of-string.js
new file mode 100644
index 00000000..eb2849bc
--- /dev/null
+++ b/solutions/2405-optimal-partition-of-string.js
@@ -0,0 +1,31 @@
+/**
+ * 2405. Optimal Partition of String
+ * https://leetcode.com/problems/optimal-partition-of-string/
+ * Difficulty: Medium
+ *
+ * Given a string s, partition the string into one or more substrings such that the characters
+ * in each substring are unique. That is, no letter appears in a single substring more than once.
+ *
+ * Return the minimum number of substrings in such a partition.
+ *
+ * Note that each character should belong to exactly one substring in a partition.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var partitionString = function(s) {
+  const set = new Set();
+  let result = 1;
+
+  for (const char of s) {
+    if (set.has(char)) {
+      set.clear();
+      result++;
+    }
+    set.add(char);
+  }
+
+  return result;
+};
diff --git a/solutions/2409-count-days-spent-together.js b/solutions/2409-count-days-spent-together.js
new file mode 100644
index 00000000..1047ab0b
--- /dev/null
+++ b/solutions/2409-count-days-spent-together.js
@@ -0,0 +1,48 @@
+/**
+ * 2409. Count Days Spent Together
+ * https://leetcode.com/problems/count-days-spent-together/
+ * Difficulty: Easy
+ *
+ * Alice and Bob are traveling to Rome for separate business meetings.
+ *
+ * You are given 4 strings arriveAlice, leaveAlice, arriveBob, and leaveBob. Alice will be in
+ * the city from the dates arriveAlice to leaveAlice (inclusive), while Bob will be in the city
+ * from the dates arriveBob to leaveBob (inclusive). Each will be a 5-character string in the
+ * format "MM-DD", corresponding to the month and day of the date.
+ *
+ * Return the total number of days that Alice and Bob are in Rome together.
+ *
+ * You can assume that all dates occur in the same calendar year, which is not a leap year. Note
+ * that the number of days per month can be represented as: [31, 28, 31, 30, 31, 30, 31, 31, 30,
+ * 31, 30, 31].
+ */
+
+/**
+ * @param {string} arriveAlice
+ * @param {string} leaveAlice
+ * @param {string} arriveBob
+ * @param {string} leaveBob
+ * @return {number}
+ */
+var countDaysTogether = function(arriveAlice, leaveAlice, arriveBob, leaveBob) {
+  const daysInMonth = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
+
+  function toDays(date) {
+    const [month, day] = date.split('-').map(Number);
+    let totalDays = 0;
+    for (let i = 0; i < month - 1; i++) {
+      totalDays += daysInMonth[i];
+    }
+    return totalDays + day;
+  }
+
+  const aliceStart = toDays(arriveAlice);
+  const aliceEnd = toDays(leaveAlice);
+  const bobStart = toDays(arriveBob);
+  const bobEnd = toDays(leaveBob);
+
+  const overlapStart = Math.max(aliceStart, bobStart);
+  const overlapEnd = Math.min(aliceEnd, bobEnd);
+
+  return Math.max(0, overlapEnd - overlapStart + 1);
+};
diff --git a/solutions/2410-maximum-matching-of-players-with-trainers.js b/solutions/2410-maximum-matching-of-players-with-trainers.js
new file mode 100644
index 00000000..885db445
--- /dev/null
+++ b/solutions/2410-maximum-matching-of-players-with-trainers.js
@@ -0,0 +1,35 @@
+/**
+ * 2410. Maximum Matching of Players With Trainers
+ * https://leetcode.com/problems/maximum-matching-of-players-with-trainers/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed integer array players, where players[i] represents the ability of
+ * the ith player. You are also given a 0-indexed integer array trainers, where trainers[j]
+ * represents the training capacity of the jth trainer.
+ *
+ * The ith player can match with the jth trainer if the player's ability is less than or equal
+ * to the trainer's training capacity. Additionally, the ith player can be matched with at most
+ * one trainer, and the jth trainer can be matched with at most one player.
+ *
+ * Return the maximum number of matchings between players and trainers that satisfy these
+ * conditions.
+ */
+
+/**
+ * @param {number[]} players
+ * @param {number[]} trainers
+ * @return {number}
+ */
+var matchPlayersAndTrainers = function(players, trainers) {
+  players.sort((a, b) => a - b);
+  trainers.sort((a, b) => a - b);
+
+  let result = 0;
+  for (let j = 0; result < players.length && j < trainers.length; j++) {
+    if (trainers[j] >= players[result]) {
+      result++;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2411-smallest-subarrays-with-maximum-bitwise-or.js b/solutions/2411-smallest-subarrays-with-maximum-bitwise-or.js
new file mode 100644
index 00000000..784cf5d2
--- /dev/null
+++ b/solutions/2411-smallest-subarrays-with-maximum-bitwise-or.js
@@ -0,0 +1,46 @@
+/**
+ * 2411. Smallest Subarrays With Maximum Bitwise OR
+ * https://leetcode.com/problems/smallest-subarrays-with-maximum-bitwise-or/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed array nums of length n, consisting of non-negative integers. For
+ * each index i from 0 to n - 1, you must determine the size of the minimum sized non-empty
+ * subarray of nums starting at i (inclusive) that has the maximum possible bitwise OR.
+ * - In other words, let Bij be the bitwise OR of the subarray nums[i...j]. You need to find
+ *   the smallest subarray starting at i, such that bitwise OR of this subarray is equal to
+ *   max(Bik) where i <= k <= n - 1.
+ *
+ * The bitwise OR of an array is the bitwise OR of all the numbers in it.
+ *
+ * Return an integer array answer of size n where answer[i] is the length of the minimum sized
+ * subarray starting at i with maximum bitwise OR.
+ *
+ * A subarray is a contiguous non-empty sequence of elements within an array.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var smallestSubarrays = function(nums) {
+  const n = nums.length;
+  const result = new Array(n).fill(1);
+  const bitPositions = new Array(32).fill(0);
+
+  for (let i = n - 1; i >= 0; i--) {
+    for (let bit = 0; bit < 32; bit++) {
+      if (nums[i] & (1 << bit)) {
+        bitPositions[bit] = i;
+      }
+    }
+
+    let maxIndex = i;
+    for (let bit = 0; bit < 32; bit++) {
+      maxIndex = Math.max(maxIndex, bitPositions[bit]);
+    }
+
+    result[i] = maxIndex - i + 1;
+  }
+
+  return result;
+};
diff --git a/solutions/2412-minimum-money-required-before-transactions.js b/solutions/2412-minimum-money-required-before-transactions.js
new file mode 100644
index 00000000..240ca6ca
--- /dev/null
+++ b/solutions/2412-minimum-money-required-before-transactions.js
@@ -0,0 +1,36 @@
+/**
+ * 2412. Minimum Money Required Before Transactions
+ * https://leetcode.com/problems/minimum-money-required-before-transactions/
+ * Difficulty: Hard
+ *
+ * You are given a 0-indexed 2D integer array transactions, where
+ * transactions[i] = [costi, cashbacki].
+ *
+ * The array describes transactions, where each transaction must be completed exactly once
+ * in some order. At any given moment, you have a certain amount of money. In order to
+ * complete transaction i, money >= costi must hold true. After performing a transaction,
+ * money becomes money - costi + cashbacki.
+ *
+ * Return the minimum amount of money required before any transaction so that all of the
+ * transactions can be completed regardless of the order of the transactions.
+ */
+
+/**
+ * @param {number[][]} transactions
+ * @return {number}
+ */
+var minimumMoney = function(transactions) {
+  let totalLoss = 0;
+  let maxCost = 0;
+
+  for (const [cost, cashback] of transactions) {
+    if (cost > cashback) {
+      totalLoss += cost - cashback;
+      maxCost = Math.max(maxCost, cashback);
+    } else {
+      maxCost = Math.max(maxCost, cost);
+    }
+  }
+
+  return totalLoss + maxCost;
+};
diff --git a/solutions/2414-length-of-the-longest-alphabetical-continuous-substring.js b/solutions/2414-length-of-the-longest-alphabetical-continuous-substring.js
new file mode 100644
index 00000000..afb44c82
--- /dev/null
+++ b/solutions/2414-length-of-the-longest-alphabetical-continuous-substring.js
@@ -0,0 +1,32 @@
+/**
+ * 2414. Length of the Longest Alphabetical Continuous Substring
+ * https://leetcode.com/problems/length-of-the-longest-alphabetical-continuous-substring/
+ * Difficulty: Medium
+ *
+ * An alphabetical continuous string is a string consisting of consecutive letters in the alphabet.
+ * In other words, it is any substring of the string "abcdefghijklmnopqrstuvwxyz".
+ * - For example, "abc" is an alphabetical continuous string, while "acb" and "za" are not.
+ *
+ * Given a string s consisting of lowercase letters only, return the length of the longest
+ * alphabetical continuous substring.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var longestContinuousSubstring = function(s) {
+  let result = 1;
+  let currentLength = 1;
+
+  for (let i = 1; i < s.length; i++) {
+    if (s.charCodeAt(i) - s.charCodeAt(i - 1) === 1) {
+      currentLength++;
+      result = Math.max(result, currentLength);
+    } else {
+      currentLength = 1;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2415-reverse-odd-levels-of-binary-tree.js b/solutions/2415-reverse-odd-levels-of-binary-tree.js
new file mode 100644
index 00000000..9c70383b
--- /dev/null
+++ b/solutions/2415-reverse-odd-levels-of-binary-tree.js
@@ -0,0 +1,51 @@
+/**
+ * 2415. Reverse Odd Levels of Binary Tree
+ * https://leetcode.com/problems/reverse-odd-levels-of-binary-tree/
+ * Difficulty: Medium
+ *
+ * Given the root of a perfect binary tree, reverse the node values at each odd level of the tree.
+ * - For example, suppose the node values at level 3 are [2,1,3,4,7,11,29,18], then it should
+ *   become [18,29,11,7,4,3,1,2].
+ *
+ * Return the root of the reversed tree.
+ *
+ * A binary tree is perfect if all parent nodes have two children and all leaves are on the same
+ * level.
+ *
+ * The level of a node is the number of edges along the path between it and the root node.
+ */
+
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @return {TreeNode}
+ */
+var reverseOddLevels = function(root) {
+  reverseLevel([root], 0);
+  return root;
+
+  function reverseLevel(nodes, level) {
+    if (!nodes.length) return;
+
+    if (level % 2 === 1) {
+      for (let i = 0, j = nodes.length - 1; i < j; i++, j--) {
+        [nodes[i].val, nodes[j].val] = [nodes[j].val, nodes[i].val];
+      }
+    }
+
+    const nextLevel = [];
+    for (const node of nodes) {
+      if (node.left) nextLevel.push(node.left);
+      if (node.right) nextLevel.push(node.right);
+    }
+
+    reverseLevel(nextLevel, level + 1);
+  }
+};
diff --git a/solutions/2416-sum-of-prefix-scores-of-strings.js b/solutions/2416-sum-of-prefix-scores-of-strings.js
new file mode 100644
index 00000000..45d3a184
--- /dev/null
+++ b/solutions/2416-sum-of-prefix-scores-of-strings.js
@@ -0,0 +1,56 @@
+/**
+ * 2416. Sum of Prefix Scores of Strings
+ * https://leetcode.com/problems/sum-of-prefix-scores-of-strings/
+ * Difficulty: Hard
+ *
+ * You are given an array words of size n consisting of non-empty strings.
+ *
+ * We define the score of a string term as the number of strings words[i] such that term is
+ * a prefix of words[i].
+ * - For example, if words = ["a", "ab", "abc", "cab"], then the score of "ab" is 2, since
+ *   "ab" is a prefix of both "ab" and "abc".
+ *
+ * Return an array answer of size n where answer[i] is the sum of scores of every non-empty
+ * prefix of words[i].
+ *
+ * Note that a string is considered as a prefix of itself.
+ */
+
+/**
+ * @param {string[]} words
+ * @return {number[]}
+ */
+var sumPrefixScores = function(words) {
+  class TrieNode {
+    constructor() {
+      this.children = new Map();
+      this.count = 0;
+    }
+  }
+
+  const root = new TrieNode();
+
+  for (const word of words) {
+    let node = root;
+    for (const char of word) {
+      if (!node.children.has(char)) {
+        node.children.set(char, new TrieNode());
+      }
+      node = node.children.get(char);
+      node.count++;
+    }
+  }
+
+  const result = [];
+  for (const word of words) {
+    let node = root;
+    let score = 0;
+    for (const char of word) {
+      node = node.children.get(char);
+      score += node.count;
+    }
+    result.push(score);
+  }
+
+  return result;
+};
diff --git a/solutions/2418-sort-the-people.js b/solutions/2418-sort-the-people.js
new file mode 100644
index 00000000..c0477a6a
--- /dev/null
+++ b/solutions/2418-sort-the-people.js
@@ -0,0 +1,23 @@
+/**
+ * 2418. Sort the People
+ * https://leetcode.com/problems/sort-the-people/
+ * Difficulty: Easy
+ *
+ * You are given an array of strings names, and an array heights that consists of distinct
+ * positive integers. Both arrays are of length n.
+ *
+ * For each index i, names[i] and heights[i] denote the name and height of the ith person.
+ *
+ * Return names sorted in descending order by the people's heights.
+ */
+
+/**
+ * @param {string[]} names
+ * @param {number[]} heights
+ * @return {string[]}
+ */
+var sortPeople = function(names, heights) {
+  const people = names.map((name, index) => ({name, height: heights[index]}));
+  people.sort((a, b) => b.height - a.height);
+  return people.map(person => person.name);
+};
diff --git a/solutions/2419-longest-subarray-with-maximum-bitwise-and.js b/solutions/2419-longest-subarray-with-maximum-bitwise-and.js
new file mode 100644
index 00000000..c6a26711
--- /dev/null
+++ b/solutions/2419-longest-subarray-with-maximum-bitwise-and.js
@@ -0,0 +1,38 @@
+/**
+ * 2419. Longest Subarray With Maximum Bitwise AND
+ * https://leetcode.com/problems/longest-subarray-with-maximum-bitwise-and/
+ * Difficulty: Medium
+ *
+ * You are given an integer array nums of size n.
+ *
+ * Consider a non-empty subarray from nums that has the maximum possible bitwise AND.
+ * - In other words, let k be the maximum value of the bitwise AND of any subarray of nums.
+ *   Then, only subarrays with a bitwise AND equal to k should be considered.
+ *
+ * Return the length of the longest such subarray.
+ *
+ * The bitwise AND of an array is the bitwise AND of all the numbers in it.
+ *
+ * A subarray is a contiguous sequence of elements within an array.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var longestSubarray = function(nums) {
+  const maxValue = Math.max(...nums);
+  let result = 0;
+  let currentLength = 0;
+
+  for (const num of nums) {
+    if (num === maxValue) {
+      currentLength++;
+      result = Math.max(result, currentLength);
+    } else {
+      currentLength = 0;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2421-number-of-good-paths.js b/solutions/2421-number-of-good-paths.js
new file mode 100644
index 00000000..c0f7d606
--- /dev/null
+++ b/solutions/2421-number-of-good-paths.js
@@ -0,0 +1,86 @@
+/**
+ * 2421. Number of Good Paths
+ * https://leetcode.com/problems/number-of-good-paths/
+ * Difficulty: Hard
+ *
+ * There is a tree (i.e. a connected, undirected graph with no cycles) consisting of n nodes
+ * numbered from 0 to n - 1 and exactly n - 1 edges.
+ *
+ * You are given a 0-indexed integer array vals of length n where vals[i] denotes the value
+ * of the ith node. You are also given a 2D integer array edges where edges[i] = [ai, bi]
+ * denotes that there exists an undirected edge connecting nodes ai and bi.
+ *
+ * A good path is a simple path that satisfies the following conditions:
+ * 1. The starting node and the ending node have the same value.
+ * 2. All nodes between the starting node and the ending node have values less than or equal
+ *    to the starting node (i.e. the starting node's value should be the maximum value along
+ *    the path).
+ *
+ * Return the number of distinct good paths.
+ *
+ * Note that a path and its reverse are counted as the same path. For example, 0 -> 1 is
+ * considered to be the same as 1 -> 0. A single node is also considered as a valid path.
+ */
+
+/**
+ * @param {number[]} vals
+ * @param {number[][]} edges
+ * @return {number}
+ */
+var numberOfGoodPaths = function(vals, edges) {
+  const n = vals.length;
+  const graph = Array.from({length: n}, () => []);
+  for (const [u, v] of edges) {
+    graph[u].push(v);
+    graph[v].push(u);
+  }
+
+  const parent = new Array(n).fill(-1);
+  const rank = new Array(n).fill(0);
+  function find(x) {
+    if (parent[x] === -1) return x;
+    return parent[x] = find(parent[x]);
+  }
+
+  function union(x, y) {
+    let px = find(x);
+    let py = find(y);
+    if (px === py) return;
+    if (rank[px] < rank[py]) [px, py] = [py, px];
+    parent[py] = px;
+    if (rank[px] === rank[py]) rank[px]++;
+  }
+
+  const valueGroups = new Map();
+  for (let i = 0; i < n; i++) {
+    if (!valueGroups.has(vals[i])) {
+      valueGroups.set(vals[i], []);
+    }
+    valueGroups.get(vals[i]).push(i);
+  }
+
+  let result = 0;
+  for (const value of [...valueGroups.keys()].sort((a, b) => a - b)) {
+    const nodes = valueGroups.get(value);
+
+    for (const node of nodes) {
+      for (const neighbor of graph[node]) {
+        if (vals[neighbor] <= value) {
+          union(node, neighbor);
+        }
+      }
+    }
+
+    const groupCount = new Map();
+    for (const node of nodes) {
+      const root = find(node);
+      groupCount.set(root, (groupCount.get(root) || 0) + 1);
+    }
+
+    for (const count of groupCount.values()) {
+      result += (count * (count + 1)) / 2;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2426-number-of-pairs-satisfying-inequality.js b/solutions/2426-number-of-pairs-satisfying-inequality.js
new file mode 100644
index 00000000..e24957fd
--- /dev/null
+++ b/solutions/2426-number-of-pairs-satisfying-inequality.js
@@ -0,0 +1,66 @@
+/**
+ * 2426. Number of Pairs Satisfying Inequality
+ * https://leetcode.com/problems/number-of-pairs-satisfying-inequality/
+ * Difficulty: Hard
+ *
+ * You are given two 0-indexed integer arrays nums1 and nums2, each of size n, and an
+ * integer diff. Find the number of pairs (i, j) such that:
+ * - 0 <= i < j <= n - 1 and
+ * - nums1[i] - nums1[j] <= nums2[i] - nums2[j] + diff.
+ *
+ * Return the number of pairs that satisfy the conditions.
+ */
+
+/**
+ * @param {number[]} nums1
+ * @param {number[]} nums2
+ * @param {number} diff
+ * @return {number}
+ */
+var numberOfPairs = function(nums1, nums2, diff) {
+  const n = nums1.length;
+  const differences = new Array(n);
+  for (let i = 0; i < n; i++) {
+    differences[i] = nums1[i] - nums2[i];
+  }
+  let result = 0;
+  mergeSort(0, n);
+  return result;
+
+  function mergeSort(left, right) {
+    if (right - left <= 1) return;
+
+    const mid = Math.floor((left + right) / 2);
+    mergeSort(left, mid);
+    mergeSort(mid, right);
+
+    let i = left;
+    let j = mid;
+    while (i < mid && j < right) {
+      if (differences[i] <= differences[j] + diff) {
+        result += right - j;
+        i++;
+      } else {
+        j++;
+      }
+    }
+
+    const sorted = new Array(right - left);
+    let k = 0;
+    i = left;
+    j = mid;
+    while (i < mid && j < right) {
+      if (differences[i] <= differences[j]) {
+        sorted[k++] = differences[i++];
+      } else {
+        sorted[k++] = differences[j++];
+      }
+    }
+    while (i < mid) sorted[k++] = differences[i++];
+    while (j < right) sorted[k++] = differences[j++];
+
+    for (let p = 0; p < sorted.length; p++) {
+      differences[left + p] = sorted[p];
+    }
+  }
+};
diff --git a/solutions/2428-maximum-sum-of-an-hourglass.js b/solutions/2428-maximum-sum-of-an-hourglass.js
new file mode 100644
index 00000000..f6cb5a39
--- /dev/null
+++ b/solutions/2428-maximum-sum-of-an-hourglass.js
@@ -0,0 +1,33 @@
+/**
+ * 2428. Maximum Sum of an Hourglass
+ * https://leetcode.com/problems/maximum-sum-of-an-hourglass/
+ * Difficulty: Medium
+ *
+ * You are given an m x n integer matrix grid.
+ *
+ * We define an hourglass as a part of the matrix with the following form.
+ *
+ * Return the maximum sum of the elements of an hourglass.
+ *
+ * Note that an hourglass cannot be rotated and must be entirely contained within the matrix.
+ */
+
+/**
+ * @param {number[][]} grid
+ * @return {number}
+ */
+var maxSum = function(grid) {
+  let result = 0;
+  const rows = grid.length;
+  const cols = grid[0].length;
+
+  for (let i = 0; i <= rows - 3; i++) {
+    for (let j = 0; j <= cols - 3; j++) {
+      const hourglassSum = grid[i][j] + grid[i][j+1] + grid[i][j+2] + grid[i+1][j+1]
+        + grid[i+2][j] + grid[i+2][j+1] + grid[i+2][j+2];
+      result = Math.max(result, hourglassSum);
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2432-the-employee-that-worked-on-the-longest-task.js b/solutions/2432-the-employee-that-worked-on-the-longest-task.js
new file mode 100644
index 00000000..acab1a87
--- /dev/null
+++ b/solutions/2432-the-employee-that-worked-on-the-longest-task.js
@@ -0,0 +1,40 @@
+/**
+ * 2432. The Employee That Worked on the Longest Task
+ * https://leetcode.com/problems/the-employee-that-worked-on-the-longest-task/
+ * Difficulty: Easy
+ *
+ * There are n employees, each with a unique id from 0 to n - 1.
+ *
+ * You are given a 2D integer array logs where logs[i] = [idi, leaveTimei] where:
+ * - idi is the id of the employee that worked on the ith task, and
+ * - leaveTimei is the time at which the employee finished the ith task. All the values
+ *   leaveTimei are unique.
+ *
+ * Note that the ith task starts the moment right after the (i - 1)th task ends, and the 0th
+ * task starts at time 0.
+ *
+ * Return the id of the employee that worked the task with the longest time. If there is a tie
+ * between two or more employees, return the smallest id among them.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} logs
+ * @return {number}
+ */
+var hardestWorker = function(n, logs) {
+  let maxDuration = 0;
+  let result = n;
+  let startTime = 0;
+
+  for (const [employeeId, endTime] of logs) {
+    const duration = endTime - startTime;
+    if (duration > maxDuration || (duration === maxDuration && employeeId < result)) {
+      maxDuration = duration;
+      result = employeeId;
+    }
+    startTime = endTime;
+  }
+
+  return result;
+};
diff --git a/solutions/2433-find-the-original-array-of-prefix-xor.js b/solutions/2433-find-the-original-array-of-prefix-xor.js
new file mode 100644
index 00000000..0f41ebeb
--- /dev/null
+++ b/solutions/2433-find-the-original-array-of-prefix-xor.js
@@ -0,0 +1,28 @@
+/**
+ * 2433. Find The Original Array of Prefix Xor
+ * https://leetcode.com/problems/find-the-original-array-of-prefix-xor/
+ * Difficulty: Medium
+ *
+ * You are given an integer array pref of size n. Find and return the array arr of size
+ * n that satisfies:
+ * - pref[i] = arr[0] ^ arr[1] ^ ... ^ arr[i].
+ *
+ * Note that ^ denotes the bitwise-xor operation.
+ *
+ * It can be proven that the answer is unique.
+ */
+
+/**
+ * @param {number[]} pref
+ * @return {number[]}
+ */
+var findArray = function(pref) {
+  const result = new Array(pref.length);
+  result[0] = pref[0];
+
+  for (let i = 1; i < pref.length; i++) {
+    result[i] = pref[i] ^ pref[i - 1];
+  }
+
+  return result;
+};
diff --git a/solutions/2434-using-a-robot-to-print-the-lexicographically-smallest-string.js b/solutions/2434-using-a-robot-to-print-the-lexicographically-smallest-string.js
new file mode 100644
index 00000000..430e0fa6
--- /dev/null
+++ b/solutions/2434-using-a-robot-to-print-the-lexicographically-smallest-string.js
@@ -0,0 +1,49 @@
+/**
+ * 2434. Using a Robot to Print the Lexicographically Smallest String
+ * https://leetcode.com/problems/using-a-robot-to-print-the-lexicographically-smallest-string/
+ * Difficulty: Medium
+ *
+ * You are given a string s and a robot that currently holds an empty string t. Apply one of the
+ * following operations until s and t are both empty:
+ * - Remove the first character of a string s and give it to the robot. The robot will append
+ *   this character to the string t.
+ * - Remove the last character of a string t and give it to the robot. The robot will write
+ *   this character on paper.
+ *
+ * Return the lexicographically smallest string that can be written on the paper.
+ */
+
+/**
+ * @param {string} s
+ * @return {string}
+ */
+var robotWithString = function(s) {
+  const charCount = Array(26).fill(0);
+  for (const char of s) {
+    charCount[char.charCodeAt(0) - 97]++;
+  }
+
+  const stack = [];
+  let minCharIndex = 0;
+  let result = '';
+
+  for (const char of s) {
+    stack.push(char);
+    charCount[char.charCodeAt(0) - 97]--;
+
+    while (minCharIndex < 26 && charCount[minCharIndex] === 0) {
+      minCharIndex++;
+    }
+
+    while (stack.length
+      && (minCharIndex === 26 || stack[stack.length - 1].charCodeAt(0) - 97 <= minCharIndex)) {
+      result += stack.pop();
+    }
+  }
+
+  while (stack.length) {
+    result += stack.pop();
+  }
+
+  return result;
+};
diff --git a/solutions/2435-paths-in-matrix-whose-sum-is-divisible-by-k.js b/solutions/2435-paths-in-matrix-whose-sum-is-divisible-by-k.js
new file mode 100644
index 00000000..741526c7
--- /dev/null
+++ b/solutions/2435-paths-in-matrix-whose-sum-is-divisible-by-k.js
@@ -0,0 +1,45 @@
+/**
+ * 2435. Paths in Matrix Whose Sum Is Divisible by K
+ * https://leetcode.com/problems/paths-in-matrix-whose-sum-is-divisible-by-k/
+ * Difficulty: Hard
+ *
+ * You are given a 0-indexed m x n integer matrix grid and an integer k. You are currently at
+ * position (0, 0) and you want to reach position (m - 1, n - 1) moving only down or right.
+ *
+ * Return the number of paths where the sum of the elements on the path is divisible by k.
+ * Since the answer may be very large, return it modulo 109 + 7.
+ */
+
+/**
+ * @param {number[][]} grid
+ * @param {number} k
+ * @return {number}
+ */
+var numberOfPaths = function(grid, k) {
+  const rows = grid.length;
+  const cols = grid[0].length;
+  const modulo = 1e9 + 7;
+  const dp = Array.from({ length: rows }, () => {
+    return Array.from({ length: cols }, () => Array(k).fill(0));
+  });
+
+  dp[0][0][grid[0][0] % k] = 1;
+
+  for (let row = 0; row < rows; row++) {
+    for (let col = 0; col < cols; col++) {
+      const currentValue = grid[row][col] % k;
+      for (let sum = 0; sum < k; sum++) {
+        if (row > 0) {
+          const prevSum = (sum - currentValue + k) % k;
+          dp[row][col][sum] = (dp[row][col][sum] + dp[row - 1][col][prevSum]) % modulo;
+        }
+        if (col > 0) {
+          const prevSum = (sum - currentValue + k) % k;
+          dp[row][col][sum] = (dp[row][col][sum] + dp[row][col - 1][prevSum]) % modulo;
+        }
+      }
+    }
+  }
+
+  return dp[rows - 1][cols - 1][0];
+};
diff --git a/solutions/2437-number-of-valid-clock-times.js b/solutions/2437-number-of-valid-clock-times.js
new file mode 100644
index 00000000..8b4d26bb
--- /dev/null
+++ b/solutions/2437-number-of-valid-clock-times.js
@@ -0,0 +1,42 @@
+/**
+ * 2437. Number of Valid Clock Times
+ * https://leetcode.com/problems/number-of-valid-clock-times/
+ * Difficulty: Easy
+ *
+ * You are given a string of length 5 called time, representing the current time on a digital
+ * clock in the format "hh:mm". The earliest possible time is "00:00" and the latest possible
+ * time is "23:59".
+ *
+ * In the string time, the digits represented by the ? symbol are unknown, and must be replaced
+ * with a digit from 0 to 9.
+ *
+ * Return an integer answer, the number of valid clock times that can be created by replacing
+ * every ? with a digit from 0 to 9.
+ */
+
+/**
+ * @param {string} time
+ * @return {number}
+ */
+var countTime = function(time) {
+  let hourChoices = 1;
+  let minuteChoices = 1;
+
+  if (time[0] === '?' && time[1] === '?') {
+    hourChoices = 24;
+  } else if (time[0] === '?') {
+    hourChoices = time[1] <= '3' ? 3 : 2;
+  } else if (time[1] === '?') {
+    hourChoices = time[0] === '2' ? 4 : 10;
+  }
+
+  if (time[3] === '?' && time[4] === '?') {
+    minuteChoices = 60;
+  } else if (time[3] === '?') {
+    minuteChoices = 6;
+  } else if (time[4] === '?') {
+    minuteChoices = 10;
+  }
+
+  return hourChoices * minuteChoices;
+};
diff --git a/solutions/2438-range-product-queries-of-powers.js b/solutions/2438-range-product-queries-of-powers.js
new file mode 100644
index 00000000..ebebead8
--- /dev/null
+++ b/solutions/2438-range-product-queries-of-powers.js
@@ -0,0 +1,45 @@
+/**
+ * 2438. Range Product Queries of Powers
+ * https://leetcode.com/problems/range-product-queries-of-powers/
+ * Difficulty: Medium
+ *
+ * Given a positive integer n, there exists a 0-indexed array called powers, composed of the
+ * minimum number of powers of 2 that sum to n. The array is sorted in non-decreasing order,
+ * and there is only one way to form the array.
+ *
+ * You are also given a 0-indexed 2D integer array queries, where queries[i] = [lefti, righti].
+ * Each queries[i] represents a query where you have to find the product of all powers[j] with
+ * lefti <= j <= righti.
+ *
+ * Return an array answers, equal in length to queries, where answers[i] is the answer to the
+ * ith query. Since the answer to the ith query may be too large, each answers[i] should be
+ * returned modulo 109 + 7.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} queries
+ * @return {number[]}
+ */
+var productQueries = function(n, queries) {
+  const modulo = 1e9 + 7;
+  const powers = [];
+  while (n > 0) {
+    const power = Math.floor(Math.log2(n));
+    powers.push(1 << power);
+    n -= 1 << power;
+  }
+  powers.reverse();
+
+  const result = new Array(queries.length);
+  for (let i = 0; i < queries.length; i++) {
+    const [start, end] = queries[i];
+    let product = 1;
+    for (let j = start; j <= end; j++) {
+      product = (product * powers[j]) % modulo;
+    }
+    result[i] = product;
+  }
+
+  return result;
+};
diff --git a/solutions/2439-minimize-maximum-of-array.js b/solutions/2439-minimize-maximum-of-array.js
new file mode 100644
index 00000000..5d147f37
--- /dev/null
+++ b/solutions/2439-minimize-maximum-of-array.js
@@ -0,0 +1,53 @@
+/**
+ * 2439. Minimize Maximum of Array
+ * https://leetcode.com/problems/minimize-maximum-of-array/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed array nums comprising of n non-negative integers.
+ *
+ * In one operation, you must:
+ * - Choose an integer i such that 1 <= i < n and nums[i] > 0.
+ * - Decrease nums[i] by 1.
+ * - Increase nums[i - 1] by 1.
+ *
+ * Return the minimum possible value of the maximum integer of nums after performing
+ * any number of operations.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var minimizeArrayValue = function(nums) {
+  let left = 0;
+  let right = Math.max(...nums);
+  let result = right;
+
+  while (left <= right) {
+    const mid = Math.floor((left + right) / 2);
+    let carry = 0;
+    let valid = true;
+
+    for (let i = nums.length - 1; i >= 0; i--) {
+      const total = nums[i] + carry;
+      if (total > mid) {
+        if (i === 0) {
+          valid = false;
+          break;
+        }
+        carry = total - mid;
+      } else {
+        carry = 0;
+      }
+    }
+
+    if (valid) {
+      result = mid;
+      right = mid - 1;
+    } else {
+      left = mid + 1;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2440-create-components-with-same-value.js b/solutions/2440-create-components-with-same-value.js
new file mode 100644
index 00000000..d0a6391f
--- /dev/null
+++ b/solutions/2440-create-components-with-same-value.js
@@ -0,0 +1,64 @@
+/**
+ * 2440. Create Components With Same Value
+ * https://leetcode.com/problems/create-components-with-same-value/
+ * Difficulty: Hard
+ *
+ * There is an undirected tree with n nodes labeled from 0 to n - 1.
+ *
+ * You are given a 0-indexed integer array nums of length n where nums[i] represents the value
+ * of the ith node. You are also given a 2D integer array edges of length n - 1 where
+ * edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree.
+ *
+ * You are allowed to delete some edges, splitting the tree into multiple connected components.
+ * Let the value of a component be the sum of all nums[i] for which node i is in the component.
+ *
+ * Return the maximum number of edges you can delete, such that every connected component in
+ * the tree has the same value.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number[][]} edges
+ * @return {number}
+ */
+var componentValue = function(nums, edges) {
+  const n = nums.length;
+  const graph = Array.from({ length: n }, () => []);
+  const totalSum = nums.reduce((sum, val) => sum + val, 0);
+
+  for (const [u, v] of edges) {
+    graph[u].push(v);
+    graph[v].push(u);
+  }
+
+  for (let i = n; i >= 1; i--) {
+    if (totalSum % i === 0) {
+      const target = totalSum / i;
+      const [, components] = countNodes(0, -1, target);
+      if (components === i) {
+        return i - 1;
+      }
+    }
+  }
+
+  return 0;
+
+  function countNodes(node, parent, target) {
+    let sum = nums[node];
+    let components = 0;
+
+    for (const neighbor of graph[node]) {
+      if (neighbor !== parent) {
+        const [childSum, childComponents] = countNodes(neighbor, node, target);
+        sum += childSum;
+        components += childComponents;
+      }
+    }
+
+    if (sum === target) {
+      return [0, components + 1];
+    }
+
+    return [sum, components];
+  }
+};
diff --git a/solutions/2441-largest-positive-integer-that-exists-with-its-negative.js b/solutions/2441-largest-positive-integer-that-exists-with-its-negative.js
new file mode 100644
index 00000000..e80f0ac9
--- /dev/null
+++ b/solutions/2441-largest-positive-integer-that-exists-with-its-negative.js
@@ -0,0 +1,28 @@
+/**
+ * 2441. Largest Positive Integer That Exists With Its Negative
+ * https://leetcode.com/problems/largest-positive-integer-that-exists-with-its-negative/
+ * Difficulty: Easy
+ *
+ * Given an integer array nums that does not contain any zeros, find the largest positive
+ * integer k such that -k also exists in the array.
+ *
+ * Return the positive integer k. If there is no such integer, return -1.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var findMaxK = function(nums) {
+  const set = new Set();
+  let result = -1;
+
+  for (const num of nums) {
+    if (set.has(-num)) {
+      result = Math.max(result, Math.abs(num));
+    }
+    set.add(num);
+  }
+
+  return result;
+};
diff --git a/solutions/2442-count-number-of-distinct-integers-after-reverse-operations.js b/solutions/2442-count-number-of-distinct-integers-after-reverse-operations.js
new file mode 100644
index 00000000..d1cc0abd
--- /dev/null
+++ b/solutions/2442-count-number-of-distinct-integers-after-reverse-operations.js
@@ -0,0 +1,32 @@
+/**
+ * 2442. Count Number of Distinct Integers After Reverse Operations
+ * https://leetcode.com/problems/count-number-of-distinct-integers-after-reverse-operations/
+ * Difficulty: Medium
+ *
+ * You are given an array nums consisting of positive integers.
+ *
+ * You have to take each integer in the array, reverse its digits, and add it to the end of the
+ * array. You should apply this operation to the original integers in nums.
+ *
+ * Return the number of distinct integers in the final array.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var countDistinctIntegers = function(nums) {
+  const set = new Set(nums);
+
+  for (const num of nums) {
+    let reversed = 0;
+    let temp = num;
+    while (temp > 0) {
+      reversed = reversed * 10 + temp % 10;
+      temp = Math.floor(temp / 10);
+    }
+    set.add(reversed);
+  }
+
+  return set.size;
+};
diff --git a/solutions/2443-sum-of-number-and-its-reverse.js b/solutions/2443-sum-of-number-and-its-reverse.js
new file mode 100644
index 00000000..7e17944a
--- /dev/null
+++ b/solutions/2443-sum-of-number-and-its-reverse.js
@@ -0,0 +1,28 @@
+/**
+ * 2443. Sum of Number and Its Reverse
+ * https://leetcode.com/problems/sum-of-number-and-its-reverse/
+ * Difficulty: Medium
+ *
+ * Given a non-negative integer num, return true if num can be expressed as the sum of any
+ * non-negative integer and its reverse, or false otherwise.
+ */
+
+/**
+ * @param {number} num
+ * @return {boolean}
+ */
+var sumOfNumberAndReverse = function(num) {
+  for (let i = 0; i <= num; i++) {
+    let reversed = 0;
+    let temp = i;
+    while (temp > 0) {
+      reversed = reversed * 10 + temp % 10;
+      temp = Math.floor(temp / 10);
+    }
+    if (i + reversed === num) {
+      return true;
+    }
+  }
+
+  return false;
+};
diff --git a/solutions/2446-determine-if-two-events-have-conflict.js b/solutions/2446-determine-if-two-events-have-conflict.js
new file mode 100644
index 00000000..3a70a834
--- /dev/null
+++ b/solutions/2446-determine-if-two-events-have-conflict.js
@@ -0,0 +1,28 @@
+/**
+ * 2446. Determine if Two Events Have Conflict
+ * https://leetcode.com/problems/determine-if-two-events-have-conflict/
+ * Difficulty: Easy
+ *
+ * You are given two arrays of strings that represent two inclusive events that happened on the
+ * same day, event1 and event2, where:
+ * - event1 = [startTime1, endTime1] and
+ * - event2 = [startTime2, endTime2].
+ *
+ * Event times are valid 24 hours format in the form of HH:MM.
+ *
+ * A conflict happens when two events have some non-empty intersection (i.e., some moment is common
+ * to both events).
+ *
+ * Return true if there is a conflict between two events. Otherwise, return false.
+ */
+
+/**
+ * @param {string[]} event1
+ * @param {string[]} event2
+ * @return {boolean}
+ */
+var haveConflict = function(event1, event2) {
+  const [start1, end1] = event1;
+  const [start2, end2] = event2;
+  return start1 <= end2 && start2 <= end1;
+};
diff --git a/solutions/2447-number-of-subarrays-with-gcd-equal-to-k.js b/solutions/2447-number-of-subarrays-with-gcd-equal-to-k.js
new file mode 100644
index 00000000..efc4339c
--- /dev/null
+++ b/solutions/2447-number-of-subarrays-with-gcd-equal-to-k.js
@@ -0,0 +1,41 @@
+/**
+ * 2447. Number of Subarrays With GCD Equal to K
+ * https://leetcode.com/problems/number-of-subarrays-with-gcd-equal-to-k/
+ * Difficulty: Medium
+ *
+ * Given an integer array nums and an integer k, return the number of subarrays of nums where
+ * the greatest common divisor of the subarray's elements is k.
+ *
+ * A subarray is a contiguous non-empty sequence of elements within an array.
+ *
+ * The greatest common divisor of an array is the largest integer that evenly divides all the
+ * array elements.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+var subarrayGCD = function(nums, k) {
+  let result = 0;
+  for (let start = 0; start < nums.length; start++) {
+    let currentGCD = nums[start];
+    for (let end = start; end < nums.length; end++) {
+      currentGCD = gcd(currentGCD, nums[end]);
+      if (currentGCD === k) {
+        result++;
+      }
+    }
+  }
+
+  return result;
+
+  function gcd(a, b) {
+    while (b) {
+      a %= b;
+      [a, b] = [b, a];
+    }
+    return a;
+  }
+};
diff --git a/solutions/2448-minimum-cost-to-make-array-equal.js b/solutions/2448-minimum-cost-to-make-array-equal.js
new file mode 100644
index 00000000..1bf9d6fd
--- /dev/null
+++ b/solutions/2448-minimum-cost-to-make-array-equal.js
@@ -0,0 +1,45 @@
+/**
+ * 2448. Minimum Cost to Make Array Equal
+ * https://leetcode.com/problems/minimum-cost-to-make-array-equal/
+ * Difficulty: Hard
+ *
+ * You are given two 0-indexed arrays nums and cost consisting each of n positive integers.
+ *
+ * You can do the following operation any number of times:
+ * - Increase or decrease any element of the array nums by 1.
+ *
+ * The cost of doing one operation on the ith element is cost[i].
+ *
+ * Return the minimum total cost such that all the elements of the array nums become equal.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number[]} cost
+ * @return {number}
+ */
+var minCost = function(nums, cost) {
+  const pairs = nums.map((num, i) => [num, cost[i]]).sort((a, b) => a[0] - b[0]);
+  const n = nums.length;
+  const prefixSum = Array(n).fill(0);
+  const prefixCost = Array(n).fill(0);
+
+  prefixSum[0] = pairs[0][0] * pairs[0][1];
+  prefixCost[0] = pairs[0][1];
+
+  for (let i = 1; i < n; i++) {
+    prefixSum[i] = prefixSum[i - 1] + pairs[i][0] * pairs[i][1];
+    prefixCost[i] = prefixCost[i - 1] + pairs[i][1];
+  }
+
+  let result = Infinity;
+  for (let i = 0; i < n; i++) {
+    const target = pairs[i][0];
+    const leftCost = i > 0 ? target * prefixCost[i - 1] - prefixSum[i - 1] : 0;
+    const rightCost = i < n - 1 ? prefixSum[n - 1] - prefixSum[i]
+      - target * (prefixCost[n - 1] - prefixCost[i]) : 0;
+    result = Math.min(result, leftCost + rightCost);
+  }
+
+  return result;
+};
diff --git a/solutions/2449-minimum-number-of-operations-to-make-arrays-similar.js b/solutions/2449-minimum-number-of-operations-to-make-arrays-similar.js
new file mode 100644
index 00000000..29c2cacf
--- /dev/null
+++ b/solutions/2449-minimum-number-of-operations-to-make-arrays-similar.js
@@ -0,0 +1,44 @@
+/**
+ * 2449. Minimum Number of Operations to Make Arrays Similar
+ * https://leetcode.com/problems/minimum-number-of-operations-to-make-arrays-similar/
+ * Difficulty: Hard
+ *
+ * You are given two positive integer arrays nums and target, of the same length.
+ *
+ * In one operation, you can choose any two distinct indices i and j where 0 <= i,
+ * j < nums.length and:
+ * - set nums[i] = nums[i] + 2 and
+ * - set nums[j] = nums[j] - 2.
+ *
+ * Two arrays are considered to be similar if the frequency of each element is the same.
+ *
+ * Return the minimum number of operations required to make nums similar to target. The
+ * test cases are generated such that nums can always be similar to target.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number[]} target
+ * @return {number}
+ */
+var makeSimilar = function(nums, target) {
+  const evenNums = nums.filter(num => num % 2 === 0).sort((a, b) => a - b);
+  const oddNums = nums.filter(num => num % 2 === 1).sort((a, b) => a - b);
+  const evenTarget = target.filter(num => num % 2 === 0).sort((a, b) => a - b);
+  const oddTarget = target.filter(num => num % 2 === 1).sort((a, b) => a - b);
+
+  let result = 0;
+  for (let i = 0; i < evenNums.length; i++) {
+    if (evenNums[i] > evenTarget[i]) {
+      result += (evenNums[i] - evenTarget[i]) / 2;
+    }
+  }
+
+  for (let i = 0; i < oddNums.length; i++) {
+    if (oddNums[i] > oddTarget[i]) {
+      result += (oddNums[i] - oddTarget[i]) / 2;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2451-odd-string-difference.js b/solutions/2451-odd-string-difference.js
new file mode 100644
index 00000000..8556415c
--- /dev/null
+++ b/solutions/2451-odd-string-difference.js
@@ -0,0 +1,53 @@
+/**
+ * 2451. Odd String Difference
+ * https://leetcode.com/problems/odd-string-difference/
+ * Difficulty: Easy
+ *
+ * You are given an array of equal-length strings words. Assume that the length of each
+ * string is n.
+ *
+ * Each string words[i] can be converted into a difference integer array difference[i] of
+ * length n - 1 where difference[i][j] = words[i][j+1] - words[i][j] where 0 <= j <= n - 2.
+ * Note that the difference between two letters is the difference between their positions in
+ * the alphabet i.e. the position of 'a' is 0, 'b' is 1, and 'z' is 25.
+ * - For example, for the string "acb", the difference integer array is [2 - 0, 1 - 2] = [2, -1].
+ *
+ * All the strings in words have the same difference integer array, except one. You should find
+ * that string.
+ *
+ * Return the string in words that has different difference integer array.
+ */
+
+/**
+ * @param {string[]} words
+ * @return {string}
+ */
+var oddString = function(words) {
+  const map = new Map();
+  for (const word of words) {
+    const diff = getDiff(word);
+    map.set(diff, (map.get(diff) || 0) + 1);
+  }
+
+  let oddDiff;
+  for (const [diff, count] of map) {
+    if (count === 1) {
+      oddDiff = diff;
+      break;
+    }
+  }
+
+  for (const word of words) {
+    if (getDiff(word) === oddDiff) {
+      return word;
+    }
+  }
+
+  function getDiff(word) {
+    const diff = [];
+    for (let i = 1; i < word.length; i++) {
+      diff.push(word.charCodeAt(i) - word.charCodeAt(i - 1));
+    }
+    return diff.join(',');
+  }
+};
diff --git a/solutions/2452-words-within-two-edits-of-dictionary.js b/solutions/2452-words-within-two-edits-of-dictionary.js
new file mode 100644
index 00000000..237dac21
--- /dev/null
+++ b/solutions/2452-words-within-two-edits-of-dictionary.js
@@ -0,0 +1,41 @@
+/**
+ * 2452. Words Within Two Edits of Dictionary
+ * https://leetcode.com/problems/words-within-two-edits-of-dictionary/
+ * Difficulty: Medium
+ *
+ * You are given two string arrays, queries and dictionary. All words in each array comprise of
+ * lowercase English letters and have the same length.
+ *
+ * In one edit you can take a word from queries, and change any letter in it to any other letter.
+ * Find all words from queries that, after a maximum of two edits, equal some word from dictionary.
+ *
+ * Return a list of all words from queries, that match with some word from dictionary after a
+ * maximum of two edits. Return the words in the same order they appear in queries.
+ */
+
+/**
+ * @param {string[]} queries
+ * @param {string[]} dictionary
+ * @return {string[]}
+ */
+var twoEditWords = function(queries, dictionary) {
+  const result = [];
+
+  for (const query of queries) {
+    for (const word of dictionary) {
+      let edits = 0;
+      for (let i = 0; i < query.length; i++) {
+        if (query[i] !== word[i]) {
+          edits++;
+          if (edits > 2) break;
+        }
+      }
+      if (edits <= 2) {
+        result.push(query);
+        break;
+      }
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2453-destroy-sequential-targets.js b/solutions/2453-destroy-sequential-targets.js
new file mode 100644
index 00000000..d1c9f856
--- /dev/null
+++ b/solutions/2453-destroy-sequential-targets.js
@@ -0,0 +1,40 @@
+/**
+ * 2453. Destroy Sequential Targets
+ * https://leetcode.com/problems/destroy-sequential-targets/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed array nums consisting of positive integers, representing targets
+ * on a number line. You are also given an integer space.
+ *
+ * You have a machine which can destroy targets. Seeding the machine with some nums[i] allows
+ * it to destroy all targets with values that can be represented as nums[i] + c * space, where
+ * c is any non-negative integer. You want to destroy the maximum number of targets in nums.
+ *
+ * Return the minimum value of nums[i] you can seed the machine with to destroy the maximum
+ * number of targets.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} space
+ * @return {number}
+ */
+var destroyTargets = function(nums, space) {
+  const map = new Map();
+  let maxCount = 0;
+
+  for (const num of nums) {
+    const remainder = num % space;
+    map.set(remainder, (map.get(remainder) || 0) + 1);
+    maxCount = Math.max(maxCount, map.get(remainder));
+  }
+
+  let result = Infinity;
+  for (const num of nums) {
+    if (map.get(num % space) === maxCount) {
+      result = Math.min(result, num);
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2455-average-value-of-even-numbers-that-are-divisible-by-three.js b/solutions/2455-average-value-of-even-numbers-that-are-divisible-by-three.js
new file mode 100644
index 00000000..e793e8fe
--- /dev/null
+++ b/solutions/2455-average-value-of-even-numbers-that-are-divisible-by-three.js
@@ -0,0 +1,29 @@
+/**
+ * 2455. Average Value of Even Numbers That Are Divisible by Three
+ * https://leetcode.com/problems/average-value-of-even-numbers-that-are-divisible-by-three/
+ * Difficulty: Easy
+ *
+ * Given an integer array nums of positive integers, return the average value of all even integers
+ * that are divisible by 3.
+ *
+ * Note that the average of n elements is the sum of the n elements divided by n and rounded down
+ * to the nearest integer.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var averageValue = function(nums) {
+  let sum = 0;
+  let count = 0;
+
+  for (const num of nums) {
+    if (num % 2 === 0 && num % 3 === 0) {
+      sum += num;
+      count++;
+    }
+  }
+
+  return count > 0 ? Math.floor(sum / count) : 0;
+};
diff --git a/solutions/2458-height-of-binary-tree-after-subtree-removal-queries.js b/solutions/2458-height-of-binary-tree-after-subtree-removal-queries.js
new file mode 100644
index 00000000..418ee0ff
--- /dev/null
+++ b/solutions/2458-height-of-binary-tree-after-subtree-removal-queries.js
@@ -0,0 +1,68 @@
+/**
+ * 2458. Height of Binary Tree After Subtree Removal Queries
+ * https://leetcode.com/problems/height-of-binary-tree-after-subtree-removal-queries/
+ * Difficulty: Hard
+ *
+ * You are given the root of a binary tree with n nodes. Each node is assigned a unique value
+ * from 1 to n. You are also given an array queries of size m.
+ *
+ * You have to perform m independent queries on the tree where in the ith query you do the
+ * following:
+ * - Remove the subtree rooted at the node with the value queries[i] from the tree. It is
+ *   guaranteed that queries[i] will not be equal to the value of the root.
+ *
+ * Return an array answer of size m where answer[i] is the height of the tree after performing
+ * the ith query.
+ *
+ * Note:
+ * - The queries are independent, so the tree returns to its initial state after each query.
+ * - The height of a tree is the number of edges in the longest simple path from the root to
+ *   some node in the tree.
+ */
+
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @param {number[]} queries
+ * @return {number[]}
+ */
+var treeQueries = function(root, queries) {
+  const heights = new Map();
+  const maxHeightWithout = new Map();
+
+  computeHeight(root);
+  computeMaxHeight(root, 0, -1);
+
+  const result = new Array(queries.length);
+  for (let i = 0; i < queries.length; i++) {
+    result[i] = maxHeightWithout.get(queries[i]);
+  }
+
+  return result;
+
+  function computeHeight(node) {
+    if (!node) return -1;
+    const leftHeight = computeHeight(node.left);
+    const rightHeight = computeHeight(node.right);
+    heights.set(node.val, Math.max(leftHeight, rightHeight) + 1);
+    return heights.get(node.val);
+  }
+
+  function computeMaxHeight(node, depth, cousinHeight) {
+    if (!node) return;
+    const leftHeight = node.left ? heights.get(node.left.val) : -1;
+    const rightHeight = node.right ? heights.get(node.right.val) : -1;
+    const maxAlternative = Math.max(cousinHeight, depth - 1);
+    maxHeightWithout.set(node.val, maxAlternative);
+
+    computeMaxHeight(node.left, depth + 1, Math.max(cousinHeight, rightHeight + depth + 1));
+    computeMaxHeight(node.right, depth + 1, Math.max(cousinHeight, leftHeight + depth + 1));
+  }
+};
diff --git a/solutions/2461-maximum-sum-of-distinct-subarrays-with-length-k.js b/solutions/2461-maximum-sum-of-distinct-subarrays-with-length-k.js
new file mode 100644
index 00000000..a1212952
--- /dev/null
+++ b/solutions/2461-maximum-sum-of-distinct-subarrays-with-length-k.js
@@ -0,0 +1,45 @@
+/**
+ * 2461. Maximum Sum of Distinct Subarrays With Length K
+ * https://leetcode.com/problems/maximum-sum-of-distinct-subarrays-with-length-k/
+ * Difficulty: Medium
+ *
+ * You are given an integer array nums and an integer k. Find the maximum subarray sum of all the
+ * subarrays of nums that meet the following conditions:
+ * - The length of the subarray is k, and
+ * - All the elements of the subarray are distinct.
+ *
+ * Return the maximum subarray sum of all the subarrays that meet the conditions. If no subarray
+ * meets the conditions, return 0.
+ *
+ * A subarray is a contiguous non-empty sequence of elements within an array.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+var maximumSubarraySum = function(nums, k) {
+  let result = 0;
+  const seen = new Map();
+  let windowSum = 0;
+
+  for (let i = 0; i < nums.length; i++) {
+    seen.set(nums[i], (seen.get(nums[i]) || 0) + 1);
+    windowSum += nums[i];
+
+    if (i >= k - 1) {
+      if (seen.size === k) {
+        result = Math.max(result, windowSum);
+      }
+      const leftNum = nums[i - k + 1];
+      windowSum -= leftNum;
+      seen.set(leftNum, seen.get(leftNum) - 1);
+      if (seen.get(leftNum) === 0) {
+        seen.delete(leftNum);
+      }
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2463-minimum-total-distance-traveled.js b/solutions/2463-minimum-total-distance-traveled.js
new file mode 100644
index 00000000..cb71bbc0
--- /dev/null
+++ b/solutions/2463-minimum-total-distance-traveled.js
@@ -0,0 +1,71 @@
+/**
+ * 2463. Minimum Total Distance Traveled
+ * https://leetcode.com/problems/minimum-total-distance-traveled/
+ * Difficulty: Hard
+ *
+ * There are some robots and factories on the X-axis. You are given an integer array robot where
+ * robot[i] is the position of the ith robot. You are also given a 2D integer array factory where
+ * factory[j] = [positionj, limitj] indicates that positionj is the position of the jth factory
+ * and that the jth factory can repair at most limitj robots.
+ *
+ * The positions of each robot are unique. The positions of each factory are also unique. Note
+ * that a robot can be in the same position as a factory initially.
+ *
+ * All the robots are initially broken; they keep moving in one direction. The direction could be
+ * the negative or the positive direction of the X-axis. When a robot reaches a factory that did
+ * not reach its limit, the factory repairs the robot, and it stops moving.
+ *
+ * At any moment, you can set the initial direction of moving for some robot. Your target is to
+ * minimize the total distance traveled by all the robots.
+ *
+ * Return the minimum total distance traveled by all the robots. The test cases are generated
+ * such that all the robots can be repaired.
+ *
+ * Note that:
+ * - All robots move at the same speed.
+ * - If two robots move in the same direction, they will never collide.
+ * - If two robots move in opposite directions and they meet at some point, they do not collide.
+ *   They cross each other.
+ * - If a robot passes by a factory that reached its limits, it crosses it as if it does not exist.
+ * - If the robot moved from a position x to a position y, the distance it moved is |y - x|.
+ */
+
+/**
+ * @param {number[]} robot
+ * @param {number[][]} factory
+ * @return {number}
+ */
+var minimumTotalDistance = function(robot, factory) {
+  robot.sort((a, b) => a - b);
+  factory.sort((a, b) => a[0] - b[0]);
+
+  const robotCount = robot.length;
+  const factoryCount = factory.length;
+  const memo = new Array(robotCount + 1).fill(null).map(() => new Array(factoryCount + 1).fill(-1));
+
+  return calculateMinDistance(0, 0);
+
+  function calculateMinDistance(robotIndex, factoryIndex) {
+    if (robotIndex === robotCount) return 0;
+    if (factoryIndex === factoryCount) return 1e18;
+
+    if (memo[robotIndex][factoryIndex] !== -1) {
+      return memo[robotIndex][factoryIndex];
+    }
+
+    let result = calculateMinDistance(robotIndex, factoryIndex + 1);
+
+    let totalDistance = 0;
+    for (let robotsTaken = 0; robotsTaken < factory[factoryIndex][1]
+         && robotIndex + robotsTaken < robotCount; robotsTaken++) {
+      totalDistance += Math.abs(robot[robotIndex + robotsTaken] - factory[factoryIndex][0]);
+      result = Math.min(
+        result,
+        totalDistance + calculateMinDistance(robotIndex + robotsTaken + 1, factoryIndex + 1)
+      );
+    }
+
+    memo[robotIndex][factoryIndex] = result;
+    return result;
+  }
+};
diff --git a/solutions/2465-number-of-distinct-averages.js b/solutions/2465-number-of-distinct-averages.js
new file mode 100644
index 00000000..768feffd
--- /dev/null
+++ b/solutions/2465-number-of-distinct-averages.js
@@ -0,0 +1,34 @@
+/**
+ * 2465. Number of Distinct Averages
+ * https://leetcode.com/problems/number-of-distinct-averages/
+ * Difficulty: Easy
+ *
+ * You are given a 0-indexed integer array nums of even length.
+ *
+ * As long as nums is not empty, you must repetitively:
+ * - Find the minimum number in nums and remove it.
+ * - Find the maximum number in nums and remove it.
+ * - Calculate the average of the two removed numbers.
+ *
+ * The average of two numbers a and b is (a + b) / 2.
+ * - For example, the average of 2 and 3 is (2 + 3) / 2 = 2.5.
+ *
+ * Return the number of distinct averages calculated using the above process.
+ *
+ * Note that when there is a tie for a minimum or maximum number, any can be removed.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var distinctAverages = function(nums) {
+  nums.sort((a, b) => a - b);
+  const averages = new Set();
+
+  for (let i = 0, j = nums.length - 1; i < j; i++, j--) {
+    averages.add((nums[i] + nums[j]) / 2);
+  }
+
+  return averages.size;
+};
diff --git a/solutions/2466-count-ways-to-build-good-strings.js b/solutions/2466-count-ways-to-build-good-strings.js
new file mode 100644
index 00000000..5bea8af2
--- /dev/null
+++ b/solutions/2466-count-ways-to-build-good-strings.js
@@ -0,0 +1,47 @@
+/**
+ * 2466. Count Ways To Build Good Strings
+ * https://leetcode.com/problems/count-ways-to-build-good-strings/
+ * Difficulty: Medium
+ *
+ * Given the integers zero, one, low, and high, we can construct a string by starting with an
+ * empty string, and then at each step perform either of the following:
+ * - Append the character '0' zero times.
+ * - Append the character '1' one times.
+ *
+ * This can be performed any number of times.
+ *
+ * A good string is a string constructed by the above process having a length between low and
+ * high (inclusive).
+ *
+ * Return the number of different good strings that can be constructed satisfying these properties.
+ * Since the answer can be large, return it modulo 109 + 7.
+ */
+
+/**
+ * @param {number} low
+ * @param {number} high
+ * @param {number} zero
+ * @param {number} one
+ * @return {number}
+ */
+var countGoodStrings = function(low, high, zero, one) {
+  const modulo = 1e9 + 7;
+  const dp = new Array(high + 1).fill(0);
+  dp[0] = 1;
+
+  for (let length = 1; length <= high; length++) {
+    if (length >= zero) {
+      dp[length] = (dp[length] + dp[length - zero]) % modulo;
+    }
+    if (length >= one) {
+      dp[length] = (dp[length] + dp[length - one]) % modulo;
+    }
+  }
+
+  let result = 0;
+  for (let length = low; length <= high; length++) {
+    result = (result + dp[length]) % modulo;
+  }
+
+  return result;
+};
diff --git a/solutions/2468-split-message-based-on-limit.js b/solutions/2468-split-message-based-on-limit.js
new file mode 100644
index 00000000..71fc5d00
--- /dev/null
+++ b/solutions/2468-split-message-based-on-limit.js
@@ -0,0 +1,52 @@
+/**
+ * 2468. Split Message Based on Limit
+ * https://leetcode.com/problems/split-message-based-on-limit/
+ * Difficulty: Hard
+ *
+ * You are given a string, message, and a positive integer, limit.
+ *
+ * You must split message into one or more parts based on limit. Each resulting part should have
+ * the suffix "", where "b" is to be replaced with the total number of parts and "a" is to
+ * be replaced with the index of the part, starting from 1 and going up to b. Additionally, the
+ * length of each resulting part (including its suffix) should be equal to limit, except for the
+ * last part whose length can be at most limit.
+ *
+ * The resulting parts should be formed such that when their suffixes are removed and they are
+ * all concatenated in order, they should be equal to message. Also, the result should contain
+ * as few parts as possible.
+ *
+ * Return the parts message would be split into as an array of strings. If it is impossible to
+ * split message as required, return an empty array.
+ */
+
+/**
+ * @param {string} message
+ * @param {number} limit
+ * @return {string[]}
+ */
+var splitMessage = function(message, limit) {
+  const n = message.length;
+  let digitSum = 0;
+
+  for (let parts = 1; parts <= n; parts++) {
+    digitSum += String(parts).length;
+    const indexLength = String(parts).length * parts;
+    const formatLength = 3 * parts;
+
+    if (limit * parts - (digitSum + indexLength + formatLength) >= n) {
+      const result = [];
+      let index = 0;
+
+      for (let i = 1; i <= parts; i++) {
+        const suffix = `<${i}/${parts}>`;
+        const chars = limit - suffix.length;
+        result.push(message.slice(index, index + chars) + suffix);
+        index += chars;
+      }
+
+      return result;
+    }
+  }
+
+  return [];
+};
diff --git a/solutions/2471-minimum-number-of-operations-to-sort-a-binary-tree-by-level.js b/solutions/2471-minimum-number-of-operations-to-sort-a-binary-tree-by-level.js
new file mode 100644
index 00000000..7d26a4c1
--- /dev/null
+++ b/solutions/2471-minimum-number-of-operations-to-sort-a-binary-tree-by-level.js
@@ -0,0 +1,66 @@
+/**
+ * 2471. Minimum Number of Operations to Sort a Binary Tree by Level
+ * https://leetcode.com/problems/minimum-number-of-operations-to-sort-a-binary-tree-by-level/
+ * Difficulty: Medium
+ *
+ * You are given the root of a binary tree with unique values.
+ *
+ * In one operation, you can choose any two nodes at the same level and swap their values.
+ *
+ * Return the minimum number of operations needed to make the values at each level sorted in a
+ * strictly increasing order.
+ *
+ * The level of a node is the number of edges along the path between it and the root node.
+ */
+
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @return {number}
+ */
+var minimumOperations = function(root) {
+  let result = 0;
+  const queue = [root];
+  const levelValues = [];
+
+  while (queue.length) {
+    const levelSize = queue.length;
+    const values = [];
+
+    for (let i = 0; i < levelSize; i++) {
+      const node = queue.shift();
+      values.push(node.val);
+      if (node.left) queue.push(node.left);
+      if (node.right) queue.push(node.right);
+    }
+
+    levelValues.push(values);
+  }
+
+  for (const values of levelValues) {
+    const sorted = [...values].sort((a, b) => a - b);
+    const indexMap = new Map(values.map((val, i) => [val, i]));
+    let swaps = 0;
+
+    for (let i = 0; i < values.length; i++) {
+      if (values[i] !== sorted[i]) {
+        const targetIndex = indexMap.get(sorted[i]);
+        [values[i], values[targetIndex]] = [values[targetIndex], values[i]];
+        indexMap.set(values[targetIndex], targetIndex);
+        indexMap.set(values[i], i);
+        swaps++;
+      }
+    }
+
+    result += swaps;
+  }
+
+  return result;
+};
diff --git a/solutions/2472-maximum-number-of-non-overlapping-palindrome-substrings.js b/solutions/2472-maximum-number-of-non-overlapping-palindrome-substrings.js
new file mode 100644
index 00000000..6244a364
--- /dev/null
+++ b/solutions/2472-maximum-number-of-non-overlapping-palindrome-substrings.js
@@ -0,0 +1,51 @@
+/**
+ * 2472. Maximum Number of Non-overlapping Palindrome Substrings
+ * https://leetcode.com/problems/maximum-number-of-non-overlapping-palindrome-substrings/
+ * Difficulty: Hard
+ *
+ * You are given a string s and a positive integer k.
+ *
+ * Select a set of non-overlapping substrings from the string s that satisfy the following
+ * conditions:
+ * - The length of each substring is at least k.
+ * - Each substring is a palindrome.
+ *
+ * Return the maximum number of substrings in an optimal selection.
+ *
+ * A substring is a contiguous sequence of characters within a string.
+ */
+
+/**
+ * @param {string} s
+ * @param {number} k
+ * @return {number}
+ */
+var maxPalindromes = function(s, k) {
+  const n = s.length;
+  const isPal = new Array(n).fill().map(() => new Array(n).fill(false));
+  const dp = new Array(n + 1).fill(0);
+
+  for (let len = 1; len <= n; len++) {
+    for (let start = 0; start + len <= n; start++) {
+      const end = start + len - 1;
+      if (len === 1) {
+        isPal[start][end] = true;
+      } else if (len === 2) {
+        isPal[start][end] = s[start] === s[end];
+      } else {
+        isPal[start][end] = s[start] === s[end] && isPal[start + 1][end - 1];
+      }
+    }
+  }
+
+  for (let end = k; end <= n; end++) {
+    dp[end] = dp[end - 1];
+    for (let start = end - k; start >= 0; start--) {
+      if (isPal[start][end - 1]) {
+        dp[end] = Math.max(dp[end], dp[start] + 1);
+      }
+    }
+  }
+
+  return dp[n];
+};
diff --git a/solutions/2475-number-of-unequal-triplets-in-array.js b/solutions/2475-number-of-unequal-triplets-in-array.js
new file mode 100644
index 00000000..8717363b
--- /dev/null
+++ b/solutions/2475-number-of-unequal-triplets-in-array.js
@@ -0,0 +1,36 @@
+/**
+ * 2475. Number of Unequal Triplets in Array
+ * https://leetcode.com/problems/number-of-unequal-triplets-in-array/
+ * Difficulty: Easy
+ *
+ * You are given a 0-indexed array of positive integers nums. Find the number of
+ * triplets (i, j, k) that meet the following conditions:
+ * - 0 <= i < j < k < nums.length
+ * - nums[i], nums[j], and nums[k] are pairwise distinct.
+ *   - In other words, nums[i] != nums[j], nums[i] != nums[k], and nums[j] != nums[k].
+ *
+ * Return the number of triplets that meet the conditions.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var unequalTriplets = function(nums) {
+  const map = new Map();
+  for (const num of nums) {
+    map.set(num, (map.get(num) || 0) + 1);
+  }
+
+  let result = 0;
+  let left = 0;
+  const total = nums.length;
+
+  for (const count of map.values()) {
+    const right = total - count - left;
+    result += left * count * right;
+    left += count;
+  }
+
+  return result;
+};
diff --git a/solutions/2476-closest-nodes-queries-in-a-binary-search-tree.js b/solutions/2476-closest-nodes-queries-in-a-binary-search-tree.js
new file mode 100644
index 00000000..8f9a487a
--- /dev/null
+++ b/solutions/2476-closest-nodes-queries-in-a-binary-search-tree.js
@@ -0,0 +1,67 @@
+/**
+ * 2476. Closest Nodes Queries in a Binary Search Tree
+ * https://leetcode.com/problems/closest-nodes-queries-in-a-binary-search-tree/
+ * Difficulty: Medium
+ *
+ * You are given the root of a binary search tree and an array queries of size n consisting of
+ * positive integers.
+ *
+ * Find a 2D array answer of size n where answer[i] = [mini, maxi]:
+ * - mini is the largest value in the tree that is smaller than or equal to queries[i]. If a such
+ *   value does not exist, add -1 instead.
+ * - maxi is the smallest value in the tree that is greater than or equal to queries[i]. If a such
+ *   value does not exist, add -1 instead.
+ *
+ * Return the array answer.
+ */
+
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @param {number[]} queries
+ * @return {number[][]}
+ */
+var closestNodes = function(root, queries) {
+  const values = [];
+
+  inorder(root);
+
+  const result = queries.map(query => {
+    let minVal = -1;
+    let maxVal = -1;
+
+    let left = 0;
+    let right = values.length - 1;
+
+    while (left <= right) {
+      const mid = Math.floor((left + right) / 2);
+      if (values[mid] === query) {
+        return [query, query];
+      } else if (values[mid] < query) {
+        minVal = values[mid];
+        left = mid + 1;
+      } else {
+        maxVal = values[mid];
+        right = mid - 1;
+      }
+    }
+
+    return [minVal, maxVal];
+  });
+
+  return result;
+
+  function inorder(node) {
+    if (!node) return;
+    inorder(node.left);
+    values.push(node.val);
+    inorder(node.right);
+  }
+};
diff --git a/solutions/2477-minimum-fuel-cost-to-report-to-the-capital.js b/solutions/2477-minimum-fuel-cost-to-report-to-the-capital.js
new file mode 100644
index 00000000..fe195062
--- /dev/null
+++ b/solutions/2477-minimum-fuel-cost-to-report-to-the-capital.js
@@ -0,0 +1,51 @@
+/**
+ * 2477. Minimum Fuel Cost to Report to the Capital
+ * https://leetcode.com/problems/minimum-fuel-cost-to-report-to-the-capital/
+ * Difficulty: Medium
+ *
+ * There is a tree (i.e., a connected, undirected graph with no cycles) structure country network
+ * consisting of n cities numbered from 0 to n - 1 and exactly n - 1 roads. The capital city is
+ * city 0. You are given a 2D integer array roads where roads[i] = [ai, bi] denotes that there
+ * exists a bidirectional road connecting cities ai and bi.
+ *
+ * There is a meeting for the representatives of each city. The meeting is in the capital city.
+ *
+ * There is a car in each city. You are given an integer seats that indicates the number of seats
+ * in each car.
+ *
+ * A representative can use the car in their city to travel or change the car and ride with another
+ * representative. The cost of traveling between two cities is one liter of fuel.
+ *
+ * Return the minimum number of liters of fuel to reach the capital city.
+ */
+
+/**
+ * @param {number[][]} roads
+ * @param {number} seats
+ * @return {number}
+ */
+var minimumFuelCost = function(roads, seats) {
+  const n = roads.length + 1;
+  const graph = Array.from({ length: n }, () => []);
+  for (const [a, b] of roads) {
+    graph[a].push(b);
+    graph[b].push(a);
+  }
+
+  let fuel = 0;
+  dfs(0, -1);
+  return fuel;
+
+  function dfs(node, parent) {
+    let representatives = 1;
+    for (const neighbor of graph[node]) {
+      if (neighbor !== parent) {
+        representatives += dfs(neighbor, node);
+      }
+    }
+    if (node !== 0) {
+      fuel += Math.ceil(representatives / seats);
+    }
+    return representatives;
+  }
+};
diff --git a/solutions/2481-minimum-cuts-to-divide-a-circle.js b/solutions/2481-minimum-cuts-to-divide-a-circle.js
new file mode 100644
index 00000000..805ed6ac
--- /dev/null
+++ b/solutions/2481-minimum-cuts-to-divide-a-circle.js
@@ -0,0 +1,24 @@
+/**
+ * 2481. Minimum Cuts to Divide a Circle
+ * https://leetcode.com/problems/minimum-cuts-to-divide-a-circle/
+ * Difficulty: Easy
+ *
+ * A valid cut in a circle can be:
+ * - A cut that is represented by a straight line that touches two points on the edge of the
+ *   circle and passes through its center, or
+ * - A cut that is represented by a straight line that touches one point on the edge of the
+ *   circle and its center.
+ *
+ * Some valid and invalid cuts are shown in the figures below.
+ *
+ * Given the integer n, return the minimum number of cuts needed to divide a circle into n
+ * equal slices.
+ */
+
+/**
+ * @param {number} n
+ * @return {number}
+ */
+var numberOfCuts = function(n) {
+  return n === 1 ? 0 : n % 2 === 0 ? n / 2 : n;
+};
diff --git a/solutions/2483-minimum-penalty-for-a-shop.js b/solutions/2483-minimum-penalty-for-a-shop.js
new file mode 100644
index 00000000..0819d961
--- /dev/null
+++ b/solutions/2483-minimum-penalty-for-a-shop.js
@@ -0,0 +1,38 @@
+/**
+ * 2483. Minimum Penalty for a Shop
+ * https://leetcode.com/problems/minimum-penalty-for-a-shop/
+ * Difficulty: Medium
+ *
+ * You are given the customer visit log of a shop represented by a 0-indexed string customers
+ * consisting only of characters 'N' and 'Y':
+ * - if the ith character is 'Y', it means that customers come at the ith hour
+ * - whereas 'N' indicates that no customers come at the ith hour.
+ *
+ * If the shop closes at the jth hour (0 <= j <= n), the penalty is calculated as follows:
+ * - For every hour when the shop is open and no customers come, the penalty increases by 1.
+ * - For every hour when the shop is closed and customers come, the penalty increases by 1.
+ *
+ * Return the earliest hour at which the shop must be closed to incur a minimum penalty.
+ *
+ * Note that if a shop closes at the jth hour, it means the shop is closed at the hour j.
+ */
+
+/**
+ * @param {string} customers
+ * @return {number}
+ */
+var bestClosingTime = function(customers) {
+  let minPenalty = 0;
+  let currentPenalty = 0;
+  let result = 0;
+
+  for (let i = 0; i < customers.length; i++) {
+    currentPenalty += customers[i] === 'Y' ? -1 : 1;
+    if (currentPenalty < minPenalty) {
+      minPenalty = currentPenalty;
+      result = i + 1;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2485-find-the-pivot-integer.js b/solutions/2485-find-the-pivot-integer.js
new file mode 100644
index 00000000..73a95a4a
--- /dev/null
+++ b/solutions/2485-find-the-pivot-integer.js
@@ -0,0 +1,22 @@
+/**
+ * 2485. Find the Pivot Integer
+ * https://leetcode.com/problems/find-the-pivot-integer/
+ * Difficulty: Easy
+ *
+ * Given a positive integer n, find the pivot integer x such that:
+ * - The sum of all elements between 1 and x inclusively equals the sum of all elements between
+ *   x and n inclusively.
+ *
+ * Return the pivot integer x. If no such integer exists, return -1. It is guaranteed that there
+ * will be at most one pivot index for the given input.
+ */
+
+/**
+ * @param {number} n
+ * @return {number}
+ */
+var pivotInteger = function(n) {
+  const totalSum = n * (n + 1) / 2;
+  const pivot = Math.sqrt(totalSum);
+  return Number.isInteger(pivot) && pivot <= n ? pivot : -1;
+};
diff --git a/solutions/2486-append-characters-to-string-to-make-subsequence.js b/solutions/2486-append-characters-to-string-to-make-subsequence.js
new file mode 100644
index 00000000..09bfb362
--- /dev/null
+++ b/solutions/2486-append-characters-to-string-to-make-subsequence.js
@@ -0,0 +1,32 @@
+/**
+ * 2486. Append Characters to String to Make Subsequence
+ * https://leetcode.com/problems/append-characters-to-string-to-make-subsequence/
+ * Difficulty: Medium
+ *
+ * You are given two strings s and t consisting of only lowercase English letters.
+ *
+ * Return the minimum number of characters that need to be appended to the end of s so that
+ * t becomes a subsequence of s.
+ *
+ * A subsequence is a string that can be derived from another string by deleting some or no
+ * characters without changing the order of the remaining characters.
+ */
+
+/**
+ * @param {string} s
+ * @param {string} t
+ * @return {number}
+ */
+var appendCharacters = function(s, t) {
+  let sIndex = 0;
+  let tIndex = 0;
+
+  while (sIndex < s.length && tIndex < t.length) {
+    if (s[sIndex] === t[tIndex]) {
+      tIndex++;
+    }
+    sIndex++;
+  }
+
+  return t.length - tIndex;
+};
diff --git a/solutions/2487-remove-nodes-from-linked-list.js b/solutions/2487-remove-nodes-from-linked-list.js
new file mode 100644
index 00000000..b6d6cb2c
--- /dev/null
+++ b/solutions/2487-remove-nodes-from-linked-list.js
@@ -0,0 +1,44 @@
+/**
+ * 2487. Remove Nodes From Linked List
+ * https://leetcode.com/problems/remove-nodes-from-linked-list/
+ * Difficulty: Medium
+ *
+ * You are given the head of a linked list.
+ *
+ * Remove every node which has a node with a greater value anywhere to the right side of it.
+ *
+ * Return the head of the modified linked list.
+ */
+
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @return {ListNode}
+ */
+var removeNodes = function(head) {
+  const stack = [];
+  let current = head;
+
+  while (current) {
+    while (stack.length && stack[stack.length - 1].val < current.val) {
+      stack.pop();
+    }
+    stack.push(current);
+    current = current.next;
+  }
+
+  let result = null;
+  while (stack.length) {
+    current = stack.pop();
+    current.next = result;
+    result = current;
+  }
+
+  return result;
+};
diff --git a/solutions/2488-count-subarrays-with-median-k.js b/solutions/2488-count-subarrays-with-median-k.js
new file mode 100644
index 00000000..694d9ea9
--- /dev/null
+++ b/solutions/2488-count-subarrays-with-median-k.js
@@ -0,0 +1,43 @@
+/**
+ * 2488. Count Subarrays With Median K
+ * https://leetcode.com/problems/count-subarrays-with-median-k/
+ * Difficulty: Hard
+ *
+ * You are given an array nums of size n consisting of distinct integers from 1 to n and
+ * a positive integer k.
+ *
+ * Return the number of non-empty subarrays in nums that have a median equal to k.
+ *
+ * Note:
+ * - The median of an array is the middle element after sorting the array in ascending order.
+ *   If the array is of even length, the median is the left middle element.
+ *   - For example, the median of [2,3,1,4] is 2, and the median of [8,4,3,5,1] is 4.
+ * - A subarray is a contiguous part of an array.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+var countSubarrays = function(nums, k) {
+  const n = nums.length;
+  const kIndex = nums.indexOf(k);
+  const balanceMap = new Map([[0, 1]]);
+  let balance = 0;
+  let result = 0;
+
+  for (let i = kIndex + 1; i < n; i++) {
+    balance += nums[i] > k ? 1 : -1;
+    balanceMap.set(balance, (balanceMap.get(balance) || 0) + 1);
+  }
+
+  balance = 0;
+  for (let i = kIndex; i >= 0; i--) {
+    balance += nums[i] > k ? -1 : nums[i] < k ? 1 : 0;
+    result += balanceMap.get(balance) || 0;
+    result += balanceMap.get(balance + 1) || 0;
+  }
+
+  return result;
+};
diff --git a/solutions/2491-divide-players-into-teams-of-equal-skill.js b/solutions/2491-divide-players-into-teams-of-equal-skill.js
new file mode 100644
index 00000000..5f481cf0
--- /dev/null
+++ b/solutions/2491-divide-players-into-teams-of-equal-skill.js
@@ -0,0 +1,32 @@
+/**
+ * 2491. Divide Players Into Teams of Equal Skill
+ * https://leetcode.com/problems/divide-players-into-teams-of-equal-skill/
+ * Difficulty: Medium
+ *
+ * You are given a positive integer array skill of even length n where skill[i] denotes the
+ * skill of the ith player. Divide the players into n / 2 teams of size 2 such that the total
+ * skill of each team is equal.
+ *
+ * The chemistry of a team is equal to the product of the skills of the players on that team.
+ *
+ * Return the sum of the chemistry of all the teams, or return -1 if there is no way to divide
+ * the players into teams such that the total skill of each team is equal.
+ */
+
+/**
+ * @param {number[]} skill
+ * @return {number}
+ */
+var dividePlayers = function(skill) {
+  skill.sort((a, b) => a - b);
+  const n = skill.length;
+  const targetSum = skill[0] + skill[n - 1];
+  let result = 0;
+
+  for (let i = 0, j = n - 1; i < j; i++, j--) {
+    if (skill[i] + skill[j] !== targetSum) return -1;
+    result += skill[i] * skill[j];
+  }
+
+  return result;
+};
diff --git a/solutions/2615-sum-of-distances.js b/solutions/2615-sum-of-distances.js
new file mode 100644
index 00000000..470c10cf
--- /dev/null
+++ b/solutions/2615-sum-of-distances.js
@@ -0,0 +1,44 @@
+/**
+ * 2615. Sum of Distances
+ * https://leetcode.com/problems/sum-of-distances/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed integer array nums. There exists an array arr of length nums.length,
+ * where arr[i] is the sum of |i - j| over all j such that nums[j] == nums[i] and j != i. If there
+ * is no such j, set arr[i] to be 0.
+ *
+ * Return the array arr.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var distance = function(nums) {
+  const valueIndices = new Map();
+  const result = new Array(nums.length).fill(0);
+
+  for (let i = 0; i < nums.length; i++) {
+    if (!valueIndices.has(nums[i])) {
+      valueIndices.set(nums[i], []);
+    }
+    valueIndices.get(nums[i]).push(i);
+  }
+
+  for (const indices of valueIndices.values()) {
+    let prefixSum = 0;
+    for (let i = 1; i < indices.length; i++) {
+      prefixSum += indices[i] - indices[0];
+    }
+
+    result[indices[0]] = prefixSum;
+
+    for (let i = 1; i < indices.length; i++) {
+      const diff = indices[i] - indices[i - 1];
+      prefixSum += diff * (i - (indices.length - i));
+      result[indices[i]] = prefixSum;
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2900-longest-unequal-adjacent-groups-subsequence-i.js b/solutions/2900-longest-unequal-adjacent-groups-subsequence-i.js
new file mode 100644
index 00000000..b4df7c52
--- /dev/null
+++ b/solutions/2900-longest-unequal-adjacent-groups-subsequence-i.js
@@ -0,0 +1,40 @@
+/**
+ * 2900. Longest Unequal Adjacent Groups Subsequence I
+ * https://leetcode.com/problems/longest-unequal-adjacent-groups-subsequence-i/
+ * Difficulty: Easy
+ *
+ * You are given a string array words and a binary array groups both of length n, where words[i]
+ * is associated with groups[i].
+ *
+ * Your task is to select the longest alternating subsequence from words. A subsequence of words
+ * is alternating if for any two consecutive strings in the sequence, their corresponding elements
+ * in the binary array groups differ. Essentially, you are to choose strings such that adjacent
+ * elements have non-matching corresponding bits in the groups array.
+ *
+ * Formally, you need to find the longest subsequence of an array of indices [0, 1, ..., n - 1]
+ * denoted as [i0, i1, ..., ik-1], such that groups[ij] != groups[ij+1] for each 0 <= j < k - 1
+ * and then find the words corresponding to these indices.
+ *
+ * Return the selected subsequence. If there are multiple answers, return any of them.
+ *
+ * Note: The elements in words are distinct.
+ */
+
+/**
+ * @param {string[]} words
+ * @param {number[]} groups
+ * @return {string[]}
+ */
+var getLongestSubsequence = function(words, groups) {
+  const result = [words[0]];
+  let lastGroup = groups[0];
+
+  for (let i = 1; i < words.length; i++) {
+    if (groups[i] !== lastGroup) {
+      result.push(words[i]);
+      lastGroup = groups[i];
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/2901-longest-unequal-adjacent-groups-subsequence-ii.js b/solutions/2901-longest-unequal-adjacent-groups-subsequence-ii.js
new file mode 100644
index 00000000..1a290423
--- /dev/null
+++ b/solutions/2901-longest-unequal-adjacent-groups-subsequence-ii.js
@@ -0,0 +1,68 @@
+/**
+ * 2901. Longest Unequal Adjacent Groups Subsequence II
+ * https://leetcode.com/problems/longest-unequal-adjacent-groups-subsequence-ii/
+ * Difficulty: Medium
+ *
+ * You are given a string array words, and an array groups, both arrays having length n.
+ *
+ * The hamming distance between two strings of equal length is the number of positions at which the
+ * corresponding characters are different.
+ *
+ * You need to select the longest subsequence from an array of indices [0, 1, ..., n - 1], such
+ * that for the subsequence denoted as [i0, i1, ..., ik-1] having length k, the following holds:
+ * - For adjacent indices in the subsequence, their corresponding groups are unequal, i.e.,
+ *   groups[ij] != groups[ij+1], for each j where 0 < j + 1 < k.
+ * - words[ij] and words[ij+1] are equal in length, and the hamming distance between them is 1,
+ *   where 0 < j + 1 < k, for all indices in the subsequence.
+ *
+ * Return a string array containing the words corresponding to the indices (in order) in the
+ * selected subsequence. If there are multiple answers, return any of them.
+ *
+ * Note: strings in words may be unequal in length.
+ */
+
+/**
+ * @param {string[]} words
+ * @param {number[]} groups
+ * @return {string[]}
+ */
+var getWordsInLongestSubsequence = function(words, groups) {
+  const n = words.length;
+  const dp = new Array(n).fill(1);
+  const prev = new Array(n).fill(-1);
+  let maxLen = 1;
+  let lastIndex = 0;
+
+  for (let i = 1; i < n; i++) {
+    for (let j = 0; j < i; j++) {
+      if (groups[i] !== groups[j] && helper(words[i], words[j])) {
+        if (dp[j] + 1 > dp[i]) {
+          dp[i] = dp[j] + 1;
+          prev[i] = j;
+        }
+      }
+    }
+    if (dp[i] > maxLen) {
+      maxLen = dp[i];
+      lastIndex = i;
+    }
+  }
+
+  const result = [];
+  while (lastIndex !== -1) {
+    result.push(words[lastIndex]);
+    lastIndex = prev[lastIndex];
+  }
+
+  return result.reverse();
+
+  function helper(s1, s2) {
+    if (s1.length !== s2.length) return false;
+    let diff = 0;
+    for (let i = 0; i < s1.length; i++) {
+      if (s1[i] !== s2[i]) diff++;
+      if (diff > 1) return false;
+    }
+    return diff === 1;
+  }
+};
diff --git a/solutions/2918-minimum-equal-sum-of-two-arrays-after-replacing-zeros.js b/solutions/2918-minimum-equal-sum-of-two-arrays-after-replacing-zeros.js
new file mode 100644
index 00000000..15ea4af7
--- /dev/null
+++ b/solutions/2918-minimum-equal-sum-of-two-arrays-after-replacing-zeros.js
@@ -0,0 +1,40 @@
+/**
+ * 2918. Minimum Equal Sum of Two Arrays After Replacing Zeros
+ * https://leetcode.com/problems/minimum-equal-sum-of-two-arrays-after-replacing-zeros/
+ * Difficulty: Medium
+ *
+ * You are given two arrays nums1 and nums2 consisting of positive integers.
+ *
+ * You have to replace all the 0's in both arrays with strictly positive integers such that the
+ * sum of elements of both arrays becomes equal.
+ *
+ * Return the minimum equal sum you can obtain, or -1 if it is impossible.
+ */
+
+/**
+ * @param {number[]} nums1
+ * @param {number[]} nums2
+ * @return {number}
+ */
+var minSum = function(nums1, nums2) {
+  let sum1 = 0;
+  let zeros1 = 0;
+  for (const num of nums1) {
+    sum1 += num;
+    if (num === 0) zeros1++;
+  }
+
+  let sum2 = 0;
+  let zeros2 = 0;
+  for (const num of nums2) {
+    sum2 += num;
+    if (num === 0) zeros2++;
+  }
+
+  const minSum1 = sum1 + zeros1;
+  const minSum2 = sum2 + zeros2;
+
+  if (minSum1 > sum2 && zeros2 === 0 || minSum2 > sum1 && zeros1 === 0) return -1;
+
+  return Math.max(minSum1, minSum2);
+};
diff --git a/solutions/2942-find-words-containing-character.js b/solutions/2942-find-words-containing-character.js
new file mode 100644
index 00000000..9fadea6b
--- /dev/null
+++ b/solutions/2942-find-words-containing-character.js
@@ -0,0 +1,28 @@
+/**
+ * 2942. Find Words Containing Character
+ * https://leetcode.com/problems/find-words-containing-character/
+ * Difficulty: Easy
+ *
+ * You are given a 0-indexed array of strings words and a character x.
+ *
+ * Return an array of indices representing the words that contain the character x.
+ *
+ * Note that the returned array may be in any order.
+ */
+
+/**
+ * @param {string[]} words
+ * @param {character} x
+ * @return {number[]}
+ */
+var findWordsContaining = function(words, x) {
+  const result = [];
+
+  for (let i = 0; i < words.length; i++) {
+    if (words[i].includes(x)) {
+      result.push(i);
+    }
+  }
+
+  return result;
+};
diff --git a/solutions/3024-type-of-triangle.js b/solutions/3024-type-of-triangle.js
new file mode 100644
index 00000000..0a62411d
--- /dev/null
+++ b/solutions/3024-type-of-triangle.js
@@ -0,0 +1,27 @@
+/**
+ * 3024. Type of Triangle
+ * https://leetcode.com/problems/type-of-triangle/
+ * Difficulty: Easy
+ *
+ * You are given a 0-indexed integer array nums of size 3 which can form the sides of a triangle.
+ * - A triangle is called equilateral if it has all sides of equal length.
+ * - A triangle is called isosceles if it has exactly two sides of equal length.
+ * - A triangle is called scalene if all its sides are of different lengths.
+ *
+ * Return a string representing the type of triangle that can be formed or "none" if it cannot
+ * form a triangle.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {string}
+ */
+var triangleType = function(nums) {
+  const [sideA, sideB, sideC] = nums.sort((a, b) => a - b);
+
+  if (sideA + sideB <= sideC) return 'none';
+  if (sideA === sideB && sideB === sideC) return 'equilateral';
+  if (sideA === sideB || sideB === sideC) return 'isosceles';
+
+  return 'scalene';
+};
diff --git a/solutions/3068-find-the-maximum-sum-of-node-values.js b/solutions/3068-find-the-maximum-sum-of-node-values.js
new file mode 100644
index 00000000..27666217
--- /dev/null
+++ b/solutions/3068-find-the-maximum-sum-of-node-values.js
@@ -0,0 +1,52 @@
+/**
+ * 3068. Find the Maximum Sum of Node Values
+ * https://leetcode.com/problems/find-the-maximum-sum-of-node-values/
+ * Difficulty: Hard
+ *
+ * There exists an undirected tree with n nodes numbered 0 to n - 1. You are given a 0-indexed
+ * 2D integer array edges of length n - 1, where edges[i] = [ui, vi] indicates that there is an
+ * edge between nodes ui and vi in the tree. You are also given a positive integer k, and a
+ * 0-indexed array of non-negative integers nums of length n, where nums[i] represents the
+ * value of the node numbered i.
+ *
+ * Alice wants the sum of values of tree nodes to be maximum, for which Alice can perform the
+ * following operation any number of times (including zero) on the tree:
+ * - Choose any edge [u, v] connecting the nodes u and v, and update their values as follows:
+ *   - nums[u] = nums[u] XOR k
+ *   - nums[v] = nums[v] XOR k
+ *
+ * Return the maximum possible sum of the values Alice can achieve by performing the operation
+ * any number of times.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @param {number[][]} edges
+ * @return {number}
+ */
+var maximumValueSum = function(nums, k, edges) {
+  let totalSum = 0;
+  let minChange = Infinity;
+  let changeCount = 0;
+
+  for (const num of nums) {
+    const xored = num ^ k;
+    const change = xored - num;
+
+    if (change > 0) {
+      totalSum += xored;
+      changeCount++;
+    } else {
+      totalSum += num;
+    }
+
+    minChange = Math.min(minChange, Math.abs(change));
+  }
+
+  if (changeCount % 2 === 0) {
+    return totalSum;
+  }
+
+  return totalSum - minChange;
+};
diff --git a/solutions/3335-total-characters-in-string-after-transformations-i.js b/solutions/3335-total-characters-in-string-after-transformations-i.js
new file mode 100644
index 00000000..e62e76ea
--- /dev/null
+++ b/solutions/3335-total-characters-in-string-after-transformations-i.js
@@ -0,0 +1,47 @@
+/**
+ * 3335. Total Characters in String After Transformations I
+ * https://leetcode.com/problems/total-characters-in-string-after-transformations-i/
+ * Difficulty: Medium
+ *
+ * You are given a string s and an integer t, representing the number of transformations to
+ * perform. In one transformation, every character in s is replaced according to the following
+ * rules:
+ * - If the character is 'z', replace it with the string "ab".
+ * - Otherwise, replace it with the next character in the alphabet. For example, 'a' is replaced
+ *   with 'b', 'b' is replaced with 'c', and so on.
+ *
+ * Return the length of the resulting string after exactly t transformations.
+ *
+ * Since the answer may be very large, return it modulo 109 + 7.
+ */
+
+/**
+ * @param {string} s
+ * @param {number} t
+ * @return {number}
+ */
+var lengthAfterTransformations = function(s, t) {
+  const MOD = 1e9 + 7;
+  const freq = new Array(26).fill(0);
+
+  for (const char of s) {
+    freq[char.charCodeAt(0) - 97]++;
+  }
+
+  for (let i = 0; i < t; i++) {
+    const newFreq = new Array(26).fill(0);
+    for (let j = 0; j < 25; j++) {
+      newFreq[j + 1] = freq[j];
+    }
+    newFreq[0] = (newFreq[0] + freq[25]) % MOD;
+    newFreq[1] = (newFreq[1] + freq[25]) % MOD;
+    freq.splice(0, 26, ...newFreq);
+  }
+
+  let result = 0;
+  for (const count of freq) {
+    result = (result + count) % MOD;
+  }
+
+  return result;
+};
diff --git a/solutions/3337-total-characters-in-string-after-transformations-ii.js b/solutions/3337-total-characters-in-string-after-transformations-ii.js
new file mode 100644
index 00000000..91d4d6aa
--- /dev/null
+++ b/solutions/3337-total-characters-in-string-after-transformations-ii.js
@@ -0,0 +1,78 @@
+/**
+ * 3337. Total Characters in String After Transformations II
+ * https://leetcode.com/problems/total-characters-in-string-after-transformations-ii/
+ * Difficulty: Hard
+ *
+ * You are given a string s consisting of lowercase English letters, an integer t representing
+ * the number of transformations to perform, and an array nums of size 26. In one transformation,
+ * every character in s is replaced according to the following rules:
+ * - Replace s[i] with the next nums[s[i] - 'a'] consecutive characters in the alphabet. For
+ *   example, if s[i] = 'a' and nums[0] = 3, the character 'a' transforms into the next 3
+ *   consecutive characters ahead of it, which results in "bcd".
+ * - The transformation wraps around the alphabet if it exceeds 'z'. For example, if s[i] = 'y'
+ *   and nums[24] = 3, the character 'y' transforms into the next 3 consecutive characters ahead
+ *   of it, which results in "zab".
+ *
+ * Return the length of the resulting string after exactly t transformations.
+ *
+ * Since the answer may be very large, return it modulo 109 + 7.
+ */
+
+/**
+ * @param {string} s
+ * @param {number} t
+ * @param {number[]} nums
+ * @return {number}
+ */
+var lengthAfterTransformations = function(s, t, nums) {
+  const MOD = 1000000007n;
+  const freq = new Array(26).fill(0n);
+
+  for (const char of s) {
+    freq[char.charCodeAt(0) - 97]++;
+  }
+
+  const matrix = new Array(26).fill().map(() => new Array(26).fill(0n));
+  for (let i = 0; i < 26; i++) {
+    for (let j = 0; j < nums[i]; j++) {
+      matrix[(i + j + 1) % 26][i] = 1n;
+    }
+  }
+
+  const finalMatrix = matrixPower(matrix, t);
+  let total = 0n;
+
+  for (let i = 0; i < 26; i++) {
+    let sum = 0n;
+    for (let j = 0; j < 26; j++) {
+      sum = (sum + finalMatrix[i][j] * freq[j]) % MOD;
+    }
+    total = (total + sum) % MOD;
+  }
+
+  return Number(total);
+
+  function matrixMultiply(a, b) {
+    const result = new Array(26).fill().map(() => new Array(26).fill(0n));
+    for (let i = 0; i < 26; i++) {
+      for (let j = 0; j < 26; j++) {
+        for (let k = 0; k < 26; k++) {
+          result[i][j] = (result[i][j] + a[i][k] * b[k][j]) % MOD;
+        }
+      }
+    }
+    return result;
+  }
+
+  function matrixPower(mat, power) {
+    let result = new Array(26).fill().map(() => new Array(26).fill(0n));
+    for (let i = 0; i < 26; i++) result[i][i] = 1n;
+
+    while (power > 0) {
+      if (power & 1) result = matrixMultiply(result, mat);
+      mat = matrixMultiply(mat, mat);
+      power >>= 1;
+    }
+    return result;
+  }
+};
diff --git a/solutions/3341-find-minimum-time-to-reach-last-room-i.js b/solutions/3341-find-minimum-time-to-reach-last-room-i.js
new file mode 100644
index 00000000..fae5957b
--- /dev/null
+++ b/solutions/3341-find-minimum-time-to-reach-last-room-i.js
@@ -0,0 +1,55 @@
+/**
+ * 3341. Find Minimum Time to Reach Last Room I
+ * https://leetcode.com/problems/find-minimum-time-to-reach-last-room-i/
+ * Difficulty: Medium
+ *
+ * There is a dungeon with n x m rooms arranged as a grid.
+ *
+ * You are given a 2D array moveTime of size n x m, where moveTime[i][j] represents the minimum
+ * time in seconds when you can start moving to that room. You start from the room (0, 0) at
+ * time t = 0 and can move to an adjacent room. Moving between adjacent rooms takes exactly
+ * one second.
+ *
+ * Return the minimum time to reach the room (n - 1, m - 1).
+ *
+ * Two rooms are adjacent if they share a common wall, either horizontally or vertically.
+ */
+
+/**
+ * @param {number[][]} moveTime
+ * @return {number}
+ */
+var minTimeToReach = function(moveTime) {
+  const rows = moveTime.length;
+  const cols = moveTime[0].length;
+  const distances = Array.from({ length: rows }, () => new Array(cols).fill(Infinity));
+  const pq = new PriorityQueue((a, b) => a[0] - b[0]);
+  pq.enqueue([0, 0, 0]);
+  const directions = [[0, 1], [1, 0], [0, -1], [-1, 0]];
+
+  distances[0][0] = 0;
+
+  while (!pq.isEmpty()) {
+    const [time, row, col] = pq.dequeue();
+
+    if (row === rows - 1 && col === cols - 1) return time;
+
+    if (time > distances[row][col]) continue;
+
+    for (const [dr, dc] of directions) {
+      const newRow = row + dr;
+      const newCol = col + dc;
+
+      if (newRow < 0 || newRow >= rows || newCol < 0 || newCol >= cols) continue;
+
+      const newTime = Math.max(time, moveTime[newRow][newCol]) + 1;
+
+      if (newTime < distances[newRow][newCol]) {
+        distances[newRow][newCol] = newTime;
+        pq.enqueue([newTime, newRow, newCol]);
+      }
+    }
+  }
+
+  return distances[rows - 1][cols - 1];
+};
diff --git a/solutions/3343-count-number-of-balanced-permutations.js b/solutions/3343-count-number-of-balanced-permutations.js
new file mode 100644
index 00000000..0e27eae1
--- /dev/null
+++ b/solutions/3343-count-number-of-balanced-permutations.js
@@ -0,0 +1,70 @@
+/**
+ * 3343. Count Number of Balanced Permutations
+ * https://leetcode.com/problems/count-number-of-balanced-permutations/
+ * Difficulty: Hard
+ *
+ * You are given a string num. A string of digits is called balanced if the sum of the digits
+ * at even indices is equal to the sum of the digits at odd indices.
+ *
+ * Return the number of distinct permutations of num that are balanced.
+ *
+ * Since the answer may be very large, return it modulo 109 + 7.
+ *
+ * A permutation is a rearrangement of all the characters of a string.
+ */
+
+/**
+ * @param {string} num
+ * @return {number}
+ */
+var countBalancedPermutations = function(num) {
+  const MOD = 1e9 + 7;
+  const n = num.length;
+  const digitCounts = new Array(10).fill(0);
+  let totalSum = 0;
+
+  for (const char of num) {
+    const digit = parseInt(char);
+    digitCounts[digit]++;
+    totalSum += digit;
+  }
+
+  if (totalSum % 2) return 0;
+
+  const memo = new Map();
+
+  function combination(n, r) {
+    if (r > n) return 0;
+    if (r === 0 || r === n) return 1;
+    if (r > n - r) r = n - r;
+
+    let result = 1n;
+    for (let i = 0; i < r; i++) {
+      result = result * BigInt(n - i) / BigInt(i + 1);
+    }
+    return Number(result % BigInt(MOD));
+  }
+
+  function exploreDigits(digit, oddSlots, evenSlots, targetBalance) {
+    if (oddSlots === 0 && evenSlots === 0 && targetBalance === 0) return 1;
+    if (digit < 0 || oddSlots < 0 || evenSlots < 0 || targetBalance < 0) return 0;
+
+    const key = `${digit},${oddSlots},${evenSlots},${targetBalance}`;
+    if (memo.has(key)) return memo.get(key);
+
+    let result = 0;
+    for (let oddCount = 0; oddCount <= digitCounts[digit]; oddCount++) {
+      const evenCount = digitCounts[digit] - oddCount;
+      const ways = (BigInt(combination(oddSlots, oddCount))
+        * BigInt(combination(evenSlots, evenCount))) % BigInt(MOD);
+      const next = BigInt(exploreDigits(digit - 1, oddSlots - oddCount,
+        evenSlots - evenCount, targetBalance - digit * oddCount));
+      result = (result + Number((ways * next) % BigInt(MOD))) % MOD;
+    }
+
+    memo.set(key, result);
+    return result;
+  }
+
+  return exploreDigits(9, n - Math.floor(n / 2), Math.floor(n / 2), totalSum / 2);
+};
diff --git a/solutions/3355-zero-array-transformation-i.js b/solutions/3355-zero-array-transformation-i.js
new file mode 100644
index 00000000..21b9bc13
--- /dev/null
+++ b/solutions/3355-zero-array-transformation-i.js
@@ -0,0 +1,43 @@
+/**
+ * 3355. Zero Array Transformation I
+ * https://leetcode.com/problems/zero-array-transformation-i/
+ * Difficulty: Medium
+ *
+ * You are given an integer array nums of length n and a 2D array queries, where
+ * queries[i] = [li, ri].
+ *
+ * For each queries[i]:
+ * - Select a subset of indices within the range [li, ri] in nums.
+ * - Decrement the values at the selected indices by 1.
+ *
+ * A Zero Array is an array where all elements are equal to 0.
+ *
+ * Return true if it is possible to transform nums into a Zero Array after processing all the
+ * queries sequentially, otherwise return false.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number[][]} queries
+ * @return {boolean}
+ */
+var isZeroArray = function(nums, queries) {
+  const maxDecrements = new Array(nums.length).fill(0);
+
+  for (const [left, right] of queries) {
+    maxDecrements[left]++;
+    if (right + 1 < nums.length) {
+      maxDecrements[right + 1]--;
+    }
+  }
+
+  let currentDecrements = 0;
+  for (let i = 0; i < nums.length; i++) {
+    currentDecrements += maxDecrements[i];
+    if (nums[i] > currentDecrements) {
+      return false;
+    }
+  }
+
+  return true;
+};
diff --git a/solutions/3362-zero-array-transformation-iii.js b/solutions/3362-zero-array-transformation-iii.js
new file mode 100644
index 00000000..f6d979bb
--- /dev/null
+++ b/solutions/3362-zero-array-transformation-iii.js
@@ -0,0 +1,53 @@
+/**
+ * 3362. Zero Array Transformation III
+ * https://leetcode.com/problems/zero-array-transformation-iii/
+ * Difficulty: Medium
+ *
+ * You are given an integer array nums of length n and a 2D array queries where
+ * queries[i] = [li, ri].
+ *
+ * Each queries[i] represents the following action on nums:
+ * - Decrement the value at each index in the range [li, ri] in nums by at most 1.
+ * - The amount by which the value is decremented can be chosen independently for each index.
+ *
+ * A Zero Array is an array with all its elements equal to 0.
+ *
+ * Return the maximum number of elements that can be removed from queries, such that nums can
+ * still be converted to a zero array using the remaining queries. If it is not possible to
+ * convert nums to a zero array, return -1.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number[][]} queries
+ * @return {number}
+ */
+var maxRemoval = function(nums, queries) {
+  queries.sort((a, b) => a[0] - b[0]);
+  const endIndexHeap = new MaxPriorityQueue();
+  const expiredQueries = new Array(nums.length + 1).fill(0);
+  let operationCount = 0;
+  let queryIndex = 0;
+
+  for (let numIndex = 0; numIndex < nums.length; numIndex++) {
+    operationCount -= expiredQueries[numIndex];
+
+    while (queryIndex < queries.length && queries[queryIndex][0] === numIndex) {
+      endIndexHeap.push(queries[queryIndex][1]);
+      queryIndex++;
+    }
+
+    while (
+      operationCount < nums[numIndex] && !endIndexHeap.isEmpty() && endIndexHeap.front() >= numIndex
+    ) {
+      operationCount++;
+      expiredQueries[endIndexHeap.pop() + 1]++;
+    }
+
+    if (operationCount < nums[numIndex]) {
+      return -1;
+    }
+  }
+
+  return endIndexHeap.size();
+};