diff --git a/README.md b/README.md index 989a3224..33b68334 100644 --- a/README.md +++ b/README.md @@ -4,7 +4,7 @@ The repo consists of solutions to numerous problems using different data structu contains solutions to HackerRank problems which I have solved so far and questions from Cracking the Coding Interview Book _(6th Edition)_. -You can also refer to my [Java Notes](http://java.ramswaroop.me) for a quick refresh on the Java concepts. +You can also refer to my [Java Notes](http://blog.rampatra.com/category/java) for a quick refresh on the Java concepts or if you want a break from coding then can read about my [interview experiences](https://blog.rampatra.com/category/interview/) at various companies. Lastly, feel free to connect with me on [Twitter](https://twitter.com/ram__patra) for any queries or concerns. All the best! ## Contents @@ -50,3 +50,7 @@ IntelliJ IDEA 2018.1.4 (Ultimate Edition) MacBook Pro 2.5 GHz Intel Core i7 16 GB 1600 MHz DDR3 + +--- + +_P.S. For any queries or concerns, you can reach out to me on [Twitter](https://twitter.com/rampatra_). I'll try my best to help 🙏. And, if you're keen to know what I'm currently working on then check out [Presentify](https://presentifyapp.com), [FaceScreen](https://facescreenapp.com/), [ToDoBar](https://todobarapp.com/), [SimpleFill](https://simplefillapp.com/), and [Apps.Deals](https://www.apps.deals)._ diff --git a/src/main/java/com/ctci/treesandgraphs/FirstCommonAncestor.java b/src/main/java/com/ctci/treesandgraphs/FirstCommonAncestor.java index f1fc5293..38977be3 100644 --- a/src/main/java/com/ctci/treesandgraphs/FirstCommonAncestor.java +++ b/src/main/java/com/ctci/treesandgraphs/FirstCommonAncestor.java @@ -5,6 +5,9 @@ * tree. Avoid storing additional nodes in a data structure. Also, for this question, the tree node * does NOT have access to its parent node. NOTE: This is not necessarily a binary search tree. * + * First Common Ancestor or the Least/Lowest Common Ancestor of two nodes is a node which is the + * closest to both of the nodes. + * * @author rampatra * @since 2019-02-24 */ @@ -18,7 +21,7 @@ public class FirstCommonAncestor { * - Returns null, if neither p nor q are in root's subtree. * - Else, returns the common ancestor of p and q. *

- * See {@link com.rampatra.trees.LeastCommonAncestorInBT} for a simpler version. + * See {@link com.rampatra.trees.LeastCommonAncestorInBT} for a better answer. * * @param root * @param a diff --git a/src/main/java/com/leetcode/arrays/BuySellStocks.java b/src/main/java/com/leetcode/arrays/BuySellStocks.java index d94c3296..4d4ba02b 100644 --- a/src/main/java/com/leetcode/arrays/BuySellStocks.java +++ b/src/main/java/com/leetcode/arrays/BuySellStocks.java @@ -2,7 +2,27 @@ /** * Level: Easy - * Problem: https://leetcode.com/problems/best-time-to-buy-and-sell-stock/ + * Link: https://leetcode.com/problems/best-time-to-buy-and-sell-stock/ + * Description: + * Say you have an array for which the ith element is the price of a given stock on day i. + * + * If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock), + * design an algorithm to find the maximum profit. + * + * Note that you cannot sell a stock before you buy one. + * + * Example 1: + * + * Input: [7,1,5,3,6,4] + * Output: 5 + * Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5. + * Not 7-1 = 6, as selling price needs to be larger than buying price. + * + * Example 2: + * + * Input: [7,6,4,3,1] + * Output: 0 + * Explanation: In this case, no transaction is done, i.e. max profit = 0. * * @author rampatra * @since 2019-04-23 diff --git a/src/main/java/com/leetcode/arrays/BuySellStocksII.java b/src/main/java/com/leetcode/arrays/BuySellStocksII.java index 47966b48..d215246e 100644 --- a/src/main/java/com/leetcode/arrays/BuySellStocksII.java +++ b/src/main/java/com/leetcode/arrays/BuySellStocksII.java @@ -2,8 +2,8 @@ /** * Level: Easy - * Problem Link: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/ - * Problem Description: + * Link: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/ + * Description: * Say you have an array for which the ith element is the price of a given stock on day i. *

* Design an algorithm to find the maximum profit. You may complete as many transactions as you diff --git a/src/main/java/com/leetcode/arrays/MergeSortedArray.java b/src/main/java/com/leetcode/arrays/MergeSortedArray.java index 09cd730d..975db8ec 100644 --- a/src/main/java/com/leetcode/arrays/MergeSortedArray.java +++ b/src/main/java/com/leetcode/arrays/MergeSortedArray.java @@ -4,7 +4,20 @@ /** * Level: Easy - * Problem Link: https://leetcode.com/problems/merge-sorted-array/ + * Link: https://leetcode.com/problems/merge-sorted-array/ + * Description: + * Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array. + * + * Note: + * The number of elements initialized in nums1 and nums2 are m and n respectively. + * You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold + * additional elements from nums2. + * + * Example: + * Input: + * nums1 = [1,2,3,0,0,0], m = 3 + * nums2 = [2,5,6], n = 3 + * Output: [1,2,2,3,5,6] * * @author rampatra * @since 2019-04-26 diff --git a/src/main/java/com/leetcode/arrays/RotateArray.java b/src/main/java/com/leetcode/arrays/RotateArray.java index e4f6de33..91475160 100644 --- a/src/main/java/com/leetcode/arrays/RotateArray.java +++ b/src/main/java/com/leetcode/arrays/RotateArray.java @@ -4,7 +4,28 @@ /** * Level: Easy - * Problem: https://leetcode.com/problems/rotate-array/ + * Link: https://leetcode.com/problems/rotate-array/ + * Description: + * Given an array, rotate the array to the right by k steps, where k is non-negative. + * + * Example 1: + * Input: [1,2,3,4,5,6,7] and k = 3 + * Output: [5,6,7,1,2,3,4] + * Explanation: + * rotate 1 steps to the right: [7,1,2,3,4,5,6] + * rotate 2 steps to the right: [6,7,1,2,3,4,5] + * rotate 3 steps to the right: [5,6,7,1,2,3,4] + * + * Example 2: + * Input: [-1,-100,3,99] and k = 2 + * Output: [3,99,-1,-100] + * Explanation: + * rotate 1 steps to the right: [99,-1,-100,3] + * rotate 2 steps to the right: [3,99,-1,-100] + * + * Note: + * Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem. + * Could you do it in-place with O(1) extra space? * * @author rampatra * @since 2019-04-20 diff --git a/src/main/java/com/leetcode/arrays/ShortestWordDistance.java b/src/main/java/com/leetcode/arrays/ShortestWordDistance.java index 50b198cd..5cd0c821 100644 --- a/src/main/java/com/leetcode/arrays/ShortestWordDistance.java +++ b/src/main/java/com/leetcode/arrays/ShortestWordDistance.java @@ -1,5 +1,7 @@ package com.leetcode.arrays; +import com.leetcode.hashtables.ShortestWordDistanceII; + import static org.junit.jupiter.api.Assertions.assertEquals; /** @@ -16,7 +18,7 @@ * * Note: You may assume that word1 does not equal to word2, and word1 and word2 are both in the list. * - * Lastly, for a more complex variant, see {@link com.leetcode.maps.ShortestWordDistanceII}. + * Lastly, for a more complex variant, see {@link ShortestWordDistanceII}. * * @author rampatra * @since 2019-07-31 diff --git a/src/main/java/com/leetcode/arrays/binarysearch/PowXN.java b/src/main/java/com/leetcode/arrays/binarysearch/PowXN.java new file mode 100644 index 00000000..3591c50e --- /dev/null +++ b/src/main/java/com/leetcode/arrays/binarysearch/PowXN.java @@ -0,0 +1,85 @@ +package com.leetcode.arrays.binarysearch; + +import static org.junit.jupiter.api.Assertions.assertEquals; + +/** + * Level: Medium + * Link: https://leetcode.com/problems/powx-n/ + * Description: + * Implement pow(x, n), which calculates x raised to the power n (x^n). + *

+ * Example 1: + * Input: 2.00000, 10 + * Output: 1024.00000 + *

+ * Example 2: + * Input: 2.10000, 3 + * Output: 9.26100 + *

+ * Example 3: + * Input: 2.00000, -2 + * Output: 0.25000 + * Explanation: 2^-2 = 1/22 = 1/4 = 0.25 + *

+ * Note: + * -100.0 < x < 100.0 + * n is a 32-bit signed integer, within the range [−231, 231 − 1] + * + * @author rampatra + * @since 2019-08-19 + */ +public class PowXN { + + /** + * In this approach we iterate n times and keep multiplying x with x. + * Runtime: Time limit exceeded. + * + * @param x + * @param n + * @return + */ + public static double myPowNaive(double x, int n) { + if (n == 0) { + return 1; + } + double res = x; + int absN = Math.abs(n); + for (int i = 1; i < absN; i++) { + res *= x; + } + return n < 0 ? 1 / res : res; + } + + + /** + * In this approach, we iterate log n times. We omit half of n each time. When n is odd, we store whatever product + * we have calculated so far in the final result. + *

+ * Runtime: 1 ms. + * + * @param x + * @param n + * @return + */ + public static double myPow(double x, int n) { + double res = 1; + long absN = Math.abs((long) n); // used a long so that `absN / 2` doesn't overflow + + while (absN > 0) { + if (absN % 2 == 1) res *= x; // store whatever we have calculated so far in the final result + x *= x; + absN /= 2; + } + return n < 0 ? 1 / res : res; + } + + public static void main(String[] args) { + assertEquals(1024.0, myPowNaive(2.0, 10)); + assertEquals(0.25, myPowNaive(2.0, -2)); + assertEquals(0.0, myPowNaive(0.00001, 2147483647)); + + assertEquals(1024.0, myPow(2.0, 10)); + assertEquals(0.25, myPow(2.0, -2)); + assertEquals(0.0, myPow(0.00001, 2147483647)); + } +} diff --git a/src/main/java/com/leetcode/arrays/binarysearch/SearchInsertPosition.java b/src/main/java/com/leetcode/arrays/binarysearch/SearchInsertPosition.java new file mode 100644 index 00000000..2f13214a --- /dev/null +++ b/src/main/java/com/leetcode/arrays/binarysearch/SearchInsertPosition.java @@ -0,0 +1,64 @@ +package com.leetcode.arrays.binarysearch; + +import static org.junit.jupiter.api.Assertions.assertEquals; + +/** + * Level: Easy + * Link: https://leetcode.com/problems/search-insert-position/ + * Description: + * Given a sorted array and a target value, return the index if the target is found. If not, return the index where it + * would be if it were inserted in order. + *

+ * You may assume no duplicates in the array. + *

+ * Example 1: + * Input: [1,3,5,6], 5 + * Output: 2 + *

+ * Example 2: + * Input: [1,3,5,6], 2 + * Output: 1 + *

+ * Example 3: + * Input: [1,3,5,6], 7 + * Output: 4 + *

+ * Example 4: + * Input: [1,3,5,6], 0 + * Output: 0 + *

+ * Similar question: {@link SmallestLetterGreaterThanTarget}. + * + * @author rampatra + * @since 2019-08-19 + */ +public class SearchInsertPosition { + + /** + * Runtime: 0 ms. + * + * @param nums + * @param target + * @return + */ + public static int searchInsert(int[] nums, int target) { + int low = 0; + int high = nums.length - 1; + while (low <= high) { + int mid = low + (high - low) / 2; + if (nums[mid] == target) { + return mid; + } else if (nums[mid] < target) { + low = mid + 1; + } else { + high = mid - 1; + } + } + return low; + } + + public static void main(String[] args) { + assertEquals(2, searchInsert(new int[]{1, 2}, 3)); + assertEquals(1, searchInsert(new int[]{1, 3, 5, 6}, 2)); + } +} \ No newline at end of file diff --git a/src/main/java/com/leetcode/arrays/binarysearch/SmallestLetterGreaterThanTarget.java b/src/main/java/com/leetcode/arrays/binarysearch/SmallestLetterGreaterThanTarget.java new file mode 100644 index 00000000..e44dc339 --- /dev/null +++ b/src/main/java/com/leetcode/arrays/binarysearch/SmallestLetterGreaterThanTarget.java @@ -0,0 +1,87 @@ +package com.leetcode.arrays.binarysearch; + +import static org.junit.jupiter.api.Assertions.assertEquals; + +/** + * Level: Easy + * Link: https://leetcode.com/problems/find-smallest-letter-greater-than-target/ + * Description: + * Given a list of sorted characters letters containing only lowercase letters, and given a target letter target, find + * the smallest element in the list that is larger than the given target. + * + * Letters also wrap around. For example, if the target is target = 'z' and letters = ['a', 'b'], the answer is 'a'. + * + * Examples: + * + * Input: + * letters = ["c", "f", "j"] + * target = "a" + * Output: "c" + * + * Input: + * letters = ["c", "f", "j"] + * target = "c" + * Output: "f" + * + * Input: + * letters = ["c", "f", "j"] + * target = "d" + * Output: "f" + * + * Input: + * letters = ["c", "f", "j"] + * target = "g" + * Output: "j" + * + * Input: + * letters = ["c", "f", "j"] + * target = "j" + * Output: "c" + * + * Input: + * letters = ["c", "f", "j"] + * target = "k" + * Output: "c" + * + * Note: + * - letters has a length in range [2, 10000]. + * - letters consists of lowercase letters, and contains at least 2 unique letters. + * - target is a lowercase letter. + * + * @author rampatra + * @since 2019-08-19 + */ +public class SmallestLetterGreaterThanTarget { + + /** + * Runtime: 0 ms. + * + * @param letters + * @param target + * @return + */ + public static char nextGreatestLetter(char[] letters, char target) { + int low = 0, hi = letters.length - 1; + while (low <= hi) { + int mid = low + (hi - low) / 2; + if (letters[mid] <= target) { + low = mid + 1; + } else { + hi = mid - 1; + } + } + return letters[low % letters.length]; + } + + public static void main(String[] args) { + assertEquals('a', nextGreatestLetter(new char[]{'a'}, 'z')); + assertEquals('b', nextGreatestLetter(new char[]{'a', 'b'}, 'a')); + assertEquals('b', nextGreatestLetter(new char[]{'a', 'b', 'c'}, 'a')); + assertEquals('a', nextGreatestLetter(new char[]{'a', 'b', 'c'}, 'z')); + assertEquals('c', nextGreatestLetter(new char[]{'c', 'f', 'j'}, 'a')); + assertEquals('f', nextGreatestLetter(new char[]{'c', 'f', 'j'}, 'c')); + assertEquals('f', nextGreatestLetter(new char[]{'c', 'f', 'j'}, 'd')); + assertEquals('b', nextGreatestLetter(new char[]{'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'j', 'k', 'l', + 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'}, 'a')); + } +} \ No newline at end of file diff --git a/src/main/java/com/leetcode/arrays/binarysearch/SqrtX.java b/src/main/java/com/leetcode/arrays/binarysearch/SqrtX.java new file mode 100644 index 00000000..0dde1308 --- /dev/null +++ b/src/main/java/com/leetcode/arrays/binarysearch/SqrtX.java @@ -0,0 +1,62 @@ +package com.leetcode.arrays.binarysearch; + +import static org.junit.jupiter.api.Assertions.assertEquals; + +/** + * Level: Easy + * Link: https://leetcode.com/problems/sqrtx/ + * Description: + * Implement int sqrt(int x). + * + * Compute and return the square root of x, where x is guaranteed to be a non-negative integer. + * + * Since the return type is an integer, the decimal digits are truncated and only the integer part + * of the result is returned. + * + * Example 1: + * Input: 4 + * Output: 2 + * + * Example 2: + * Input: 8 + * Output: 2 + * Explanation: The square root of 8 is 2.82842..., and since + * the decimal part is truncated, 2 is returned. + * + * @author rampatra + * @since 2019-08-19 + */ +public class SqrtX { + + /** + * Runtime: 1 ms. + * + * @param x + * @return + */ + public static int mySqrt(int x) { + if (x == 0 || x == 1) { + return x; + } + long low = 1; + long high = x / 2; + + while (low <= high) { + long mid = low + (high - low) / 2; + if (mid * mid == x) { + return (int) mid; + } else if (mid * mid < x) { + low = mid + 1; + } else { + high = mid - 1; + } + } + return (int) high; + } + + public static void main(String[] args) { + assertEquals(2, mySqrt(8)); + assertEquals(3, mySqrt(9)); + assertEquals(46339, mySqrt(2147395599)); + } +} \ No newline at end of file diff --git a/src/main/java/com/leetcode/design/AllOne.java b/src/main/java/com/leetcode/design/AllOne.java new file mode 100644 index 00000000..51f771d7 --- /dev/null +++ b/src/main/java/com/leetcode/design/AllOne.java @@ -0,0 +1,125 @@ +package com.leetcode.design; + +import java.util.HashMap; +import java.util.HashSet; +import java.util.Map; +import java.util.Set; + +import static org.junit.jupiter.api.Assertions.assertEquals; + +/** + * Level: Hard + * Link: https://leetcode.com/problems/all-oone-data-structure/ + * Description: + * Implement a data structure supporting the following operations: + * Inc(Key) - Inserts a new key with value 1. Or increments an existing key by 1. Key is guaranteed to be a non-empty + * string. + * Dec(Key) - If Key's value is 1, remove it from the data structure. Otherwise decrements an existing key by 1. If + * the key does not exist, this function does nothing. Key is guaranteed to be a non-empty string. + * GetMaxKey() - Returns one of the keys with maximal value. If no element exists, return an empty string "". + * GetMinKey() - Returns one of the keys with minimal value. If no element exists, return an empty string "". + *

+ * Challenge: Perform all these in O(1) time complexity. + * + * @author rampatra + * @since 2019-08-11 + */ +public class AllOne { + + + + Map keyToValMap; + Map> valToKeyMap; + + /** + * Initialize your data structure here. + */ + public AllOne() { + keyToValMap = new HashMap<>(); + valToKeyMap = new HashMap<>(); + } + + /** + * Inserts a new key with value 1. Or increments an existing key by 1. + */ + public void inc(String key) { + + } + + /** + * Decrements an existing key by 1. If Key's value is 1, remove it from the data structure. + */ + public void dec(String key) { + + } + + /** + * Returns one of the keys with maximal value. + */ + public String getMaxKey() { + return null; + } + + /** + * Returns one of the keys with Minimal value. + */ + public String getMinKey() { + return null; + } + + public static void main(String[] args) { + AllOne allOne = new AllOne(); + allOne.inc("r"); + allOne.inc("r"); + allOne.dec("r"); + allOne.inc("a"); + allOne.inc("b"); + allOne.inc("b"); + assertEquals("b", allOne.getMaxKey()); + assertEquals("a", allOne.getMinKey()); + + allOne = new AllOne(); + allOne.dec("hello"); + assertEquals("", allOne.getMaxKey()); + + allOne = new AllOne(); + allOne.inc("a"); + allOne.inc("b"); + allOne.inc("b"); + allOne.inc("b"); + allOne.inc("b"); + allOne.dec("b"); + allOne.dec("b"); + assertEquals("b", allOne.getMaxKey()); + assertEquals("a", allOne.getMinKey()); + + allOne = new AllOne(); + allOne.inc("hello"); + allOne.inc("hello"); + assertEquals("hello", allOne.getMaxKey()); + assertEquals("hello", allOne.getMinKey()); + allOne.inc("leet"); + assertEquals("hello", allOne.getMaxKey()); + assertEquals("leet", allOne.getMinKey()); + + allOne = new AllOne(); + allOne.inc("a"); + allOne.inc("b"); + allOne.inc("b"); + allOne.inc("c"); + allOne.inc("c"); + allOne.inc("c"); + allOne.dec("b"); + allOne.dec("b"); + assertEquals("a", allOne.getMinKey()); + allOne.dec("a"); + assertEquals("c", allOne.getMaxKey()); + //assertEquals("c", allOne.getMinKey()); + + allOne = new AllOne(); + allOne.inc("hello"); + allOne.dec("hello"); + assertEquals("", allOne.getMaxKey()); + assertEquals("", allOne.getMinKey()); + } +} \ No newline at end of file diff --git a/src/main/java/com/leetcode/design/LFUCache.java b/src/main/java/com/leetcode/design/LFUCache.java new file mode 100644 index 00000000..595b5a52 --- /dev/null +++ b/src/main/java/com/leetcode/design/LFUCache.java @@ -0,0 +1,127 @@ +package com.leetcode.design; + +import java.util.HashMap; +import java.util.LinkedHashSet; +import java.util.Map; + +import static org.junit.jupiter.api.Assertions.assertEquals; + +/** + * Level: Hard + * Link: https://leetcode.com/problems/lfu-cache/ + * Description: + * Design and implement a data structure for Least Frequently Used (LFU) cache. It should support the following + * operations: get and put. + * + * get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1. + * put(key, value) - Set or insert the value if the key is not already present. When the cache reaches its capacity, it + * should invalidate the least frequently used item before inserting a new item. For the purpose of this problem, when + * there is a tie (i.e., two or more keys that have the same frequency), the least recently used key would be evicted. + * + * Follow up: + * Could you do both operations in O(1) time complexity? + * + * Note: + * - When setting a key which is already present, reset its frequency/count to 1. + * + * @author rampatra + * @since 2019-08-20 + */ +public class LFUCache { + + private int minCount = 0; + private int capacity; + private Map keyValueMap; + private Map keyCountMap; + private Map> countKeysMap; + + + public LFUCache(int capacity) { + this.capacity = capacity; + keyValueMap = new HashMap<>(); + keyCountMap = new HashMap<>(); + countKeysMap = new HashMap<>(); + } + + + public int get(int key) { + Integer val = keyValueMap.get(key); + if (val == null) { + return -1; + } + + int prevCount = keyCountMap.get(key); + countKeysMap.getOrDefault(prevCount, new LinkedHashSet<>()).remove(key); + countKeysMap.putIfAbsent(prevCount + 1, new LinkedHashSet<>()); + countKeysMap.get(prevCount + 1).add(key); + + if (prevCount == minCount && countKeysMap.get(prevCount).size() == 0) { + minCount++; + } + + return val; + } + + + public void put(int key, int value) { + if (capacity <= 0) { + return; + } + + if (!keyValueMap.containsKey(key) && keyCountMap.size() == capacity) { + int keyToEvict = countKeysMap.get(minCount).iterator().next(); + countKeysMap.get(minCount).remove(keyToEvict); + keyValueMap.remove(keyToEvict); + keyCountMap.remove(keyToEvict); + } + + keyValueMap.put(key, value); + int prevCount = keyCountMap.getOrDefault(key, 0); + keyCountMap.put(key, 1); + + countKeysMap.getOrDefault(prevCount, new LinkedHashSet<>()).remove(key); + countKeysMap.putIfAbsent(1, new LinkedHashSet<>()); + countKeysMap.get(1).add(key); + + minCount = 1; + } + + + public static void main(String[] args) { + LFUCache lfuCache = new LFUCache(2); + lfuCache.put(2, 2); + lfuCache.put(3, 3); + lfuCache.put(4, 4); + assertEquals(-1, lfuCache.get(2)); + assertEquals(3, lfuCache.get(3)); + lfuCache.put(5, 5); + assertEquals(-1, lfuCache.get(4)); + + lfuCache = new LFUCache(2); + lfuCache.put(3, 1); + lfuCache.put(2, 1); + lfuCache.put(2, 2); + lfuCache.put(4, 4); + assertEquals(-1, lfuCache.get(3)); + assertEquals(2, lfuCache.get(2)); + + lfuCache = new LFUCache(2); + lfuCache.put(2, 1); + lfuCache.put(2, 2); + assertEquals(2, lfuCache.get(2)); + lfuCache.put(1, 1); + lfuCache.put(4, 4); + assertEquals(2, lfuCache.get(2)); + + lfuCache = new LFUCache(2); + assertEquals(-1, lfuCache.get(2)); + lfuCache.put(2, 6); + assertEquals(-1, lfuCache.get(1)); + lfuCache.put(1, 1); + lfuCache.put(1, 2); + assertEquals(2, lfuCache.get(1)); + assertEquals(6, lfuCache.get(2)); + + // todo fix some test cases https://leetcode.com/submissions/detail/253376947/ + } +} \ No newline at end of file diff --git a/src/main/java/com/leetcode/design/LRUCache.java b/src/main/java/com/leetcode/design/LRUCache.java new file mode 100644 index 00000000..162889f2 --- /dev/null +++ b/src/main/java/com/leetcode/design/LRUCache.java @@ -0,0 +1,61 @@ +package com.leetcode.design; + +import java.util.LinkedHashMap; +import java.util.Map; + +import static org.junit.jupiter.api.Assertions.assertEquals; + +/** + * Level: Medium + * Link: https://leetcode.com/problems/lru-cache/ + * Description: + * Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following + * operations: get and put. + * + * get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1. + * put(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it + * should invalidate the least recently used item before inserting a new item. + * + * The cache is initialized with a positive capacity. + * + * Follow up: + * Could you do both operations in O(1) time complexity? + * + * Runtime: 54 ms. + * + * @author rampatra + * @since 2019-08-20 + */ +public class LRUCache { + + private LinkedHashMap cache; + + public LRUCache(int capacity) { + this.cache = new LinkedHashMap(capacity, 0.75f, true) { + @Override + protected boolean removeEldestEntry(Map.Entry eldest) { + return size() > capacity; + } + }; + } + + public int get(int key) { + Integer val = cache.get(key); + return val == null ? -1 : val; + } + + public void put(int key, int value) { + cache.put(key, value); + } + + public static void main(String[] args) { + LRUCache cache = new LRUCache(2); + cache.put(1,1); + cache.put(2,2); + cache.put(1,1); + cache.put(3,3); + assertEquals(1, cache.get(1)); + assertEquals(-1, cache.get(2)); + assertEquals(3, cache.get(3)); + } +} \ No newline at end of file diff --git a/src/main/java/com/leetcode/dynamicprogramming/MaximumProductSubArray.java b/src/main/java/com/leetcode/dynamicprogramming/MaximumProductSubArray.java new file mode 100644 index 00000000..781cad13 --- /dev/null +++ b/src/main/java/com/leetcode/dynamicprogramming/MaximumProductSubArray.java @@ -0,0 +1,62 @@ +package com.leetcode.dynamicprogramming; + +import static org.junit.jupiter.api.Assertions.assertEquals; + +/** + * Level: Medium + * Link: https://leetcode.com/problems/maximum-product-subarray/ + * Description: + * Given an integer array nums, find the contiguous subarray within an array (containing at least one number) which + * has the largest product. + *

+ * Example 1: + * Input: [2,3,-2,4] + * Output: 6 + * Explanation: [2,3] has the largest product 6. + *

+ * Example 2: + * Input: [-2,0,-1] + * Output: 0 + * Explanation: The result cannot be 2, because [-2,-1] is not a subarray. + * + * @author rampatra + * @since 2019-08-18 + */ +public class MaximumProductSubArray { + + /** + * The approach is similar to {@link MaximumSubArray} where we update maxUntilIndex only if multiplying the current + * number to the product of of all numbers until index produces a larger product and if not make maxUntilIndex the + * current number. The only twist here is that we keep two such variables, one for max and one for min, and that's + * because the product of two negatives gives us a positive number. + *

+ * Runtime: 1 ms. + * + * @param nums + * @return + */ + public static int maxProduct(int[] nums) { + int maxProd = nums[0]; + int maxUntilIndex = nums[0]; + int minUntilIndex = nums[0]; + + for (int i = 1; i < nums.length; i++) { + if (nums[i] >= 0) { + maxUntilIndex = Math.max(nums[i], maxUntilIndex * nums[i]); + minUntilIndex = Math.min(nums[i], minUntilIndex * nums[i]); + } else { + int prevMaxUntilIndex = maxUntilIndex; + maxUntilIndex = Math.max(nums[i], minUntilIndex * nums[i]); // when current number is -ve then multiply with minUntilIndex to get the max as product of two negatives is a positive + minUntilIndex = Math.min(nums[i], prevMaxUntilIndex * nums[i]); + } + + maxProd = Math.max(maxProd, maxUntilIndex); + } + + return maxProd; + } + + public static void main(String[] args) { + assertEquals(24, maxProduct(new int[]{-2, 3, -4})); + } +} \ No newline at end of file diff --git a/src/main/java/com/leetcode/arrays/MaximumSubArray.java b/src/main/java/com/leetcode/dynamicprogramming/MaximumSubArray.java similarity index 98% rename from src/main/java/com/leetcode/arrays/MaximumSubArray.java rename to src/main/java/com/leetcode/dynamicprogramming/MaximumSubArray.java index f9a64a3f..90a28c93 100644 --- a/src/main/java/com/leetcode/arrays/MaximumSubArray.java +++ b/src/main/java/com/leetcode/dynamicprogramming/MaximumSubArray.java @@ -1,4 +1,4 @@ -package com.leetcode.arrays; +package com.leetcode.dynamicprogramming; /** * Level: Easy diff --git a/src/main/java/com/leetcode/graphs/WordLadder.java b/src/main/java/com/leetcode/graphs/WordLadder.java new file mode 100644 index 00000000..61e706ce --- /dev/null +++ b/src/main/java/com/leetcode/graphs/WordLadder.java @@ -0,0 +1,121 @@ +package com.leetcode.graphs; + + +import javafx.util.Pair; + +import java.util.*; + +import static org.junit.jupiter.api.Assertions.assertEquals; + +/** + * Level: Medium + * Link: https://leetcode.com/problems/word-ladder/ + * Description: + * Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation + * sequence from beginWord to endWord, such that: + *

+ * Only one letter can be changed at a time. Each transformed word must exist in the word list. Note that beginWord + * is not a transformed word. + *

+ * Note: + * - Return 0 if there is no such transformation sequence. + * - All words have the same length. + * - All words contain only lowercase alphabetic characters. + * - You may assume no duplicates in the word list. + * - You may assume beginWord and endWord are non-empty and are not the same. + *

+ * Example 1: + * Input: + * beginWord = "hit", + * endWord = "cog", + * wordList = ["hot","dot","dog","lot","log","cog"] + *

+ * Output: 5 + *

+ * Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog", + * return its length 5. + *

+ * Example 2: + * Input: + * beginWord = "hit" + * endWord = "cog" + * wordList = ["hot","dot","dog","lot","log"] + *

+ * Output: 0 + *

+ * Explanation: The endWord "cog" is not in wordList, therefore no possible transformation. + * + * @author rampatra + * @since 2019-08-15 + */ +public class WordLadder { + + /** + * Runtime: 79 ms. + * + * @param beginWord + * @param endWord + * @param wordList + * @return + */ + public static int ladderLength(String beginWord, String endWord, List wordList) { + int L = beginWord.length(); + Map> transformedToOriginalWordMap = new HashMap<>(); + Queue> queue = new LinkedList<>(); + + wordList.forEach(word -> { + String transformedWord; + for (int i = 0; i < L; i++) { + transformedWord = word.substring(0, i) + "*" + word.substring(i + 1, L); + transformedToOriginalWordMap.putIfAbsent(transformedWord, new HashSet<>()); + transformedToOriginalWordMap.get(transformedWord).add(word); + } + } + ); + + Set visited = new HashSet<>(); + queue.add(new Pair<>(beginWord, 1)); + visited.add(beginWord); + + while (!queue.isEmpty()) { + Pair currPair = queue.poll(); + String word = currPair.getKey(); + Integer level = currPair.getValue(); + + if (word.equals(endWord)) { + return level; + } + + String transformedWord; + for (int i = 0; i < L; i++) { + transformedWord = word.substring(0, i) + "*" + word.substring(i + 1, L); + + for (String originalWord : transformedToOriginalWordMap.getOrDefault(transformedWord, Collections.emptySet())) { + if (!visited.contains(originalWord)) { + queue.add(new Pair<>(originalWord, level + 1)); + visited.add(originalWord); + } + } + } + } + + return 0; + } + + /** + * TODO: Optimized both end BFS solution + * + * @param beginWord + * @param endWord + * @param wordList + * @return + */ + public static int ladderLengthOptimized(String beginWord, String endWord, List wordList) { + return -1; + } + + public static void main(String[] args) { + assertEquals(5, ladderLength("hit", "cog", Arrays.asList("hot", "dot", "dog", "lot", "log", "cog"))); + assertEquals(0, ladderLength("hit", "cog", Arrays.asList("hot", "dot", "dog", "lot", "log"))); + } +} \ No newline at end of file diff --git a/src/main/java/com/leetcode/graphs/WordLadderII.java b/src/main/java/com/leetcode/graphs/WordLadderII.java new file mode 100644 index 00000000..8265c259 --- /dev/null +++ b/src/main/java/com/leetcode/graphs/WordLadderII.java @@ -0,0 +1,152 @@ +package com.leetcode.graphs; + +import javafx.util.Pair; + +import java.util.*; + +import static org.junit.jupiter.api.Assertions.assertEquals; + +/** + * Level: Hard + * Link: https://leetcode.com/problems/word-ladder-ii/ + * Description: + * Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) + * from beginWord to endWord, such that: + *

+ * Only one letter can be changed at a time + * Each transformed word must exist in the word list. Note that beginWord is not a transformed word. + *

+ * Note: + * - Return an empty list if there is no such transformation sequence. + * - All words have the same length. + * - All words contain only lowercase alphabetic characters. + * - You may assume no duplicates in the word list. + * - You may assume beginWord and endWord are non-empty and are not the same. + *

+ * Example 1: + * Input: + * beginWord = "hit", + * endWord = "cog", + * wordList = ["hot","dot","dog","lot","log","cog"] + *

+ * Output: + * [ + * ["hit","hot","dot","dog","cog"], + * ["hit","hot","lot","log","cog"] + * ] + *

+ *

+ * Example 2: + * Input: + * beginWord = "hit" + * endWord = "cog" + * wordList = ["hot","dot","dog","lot","log"] + *

+ * Output: [] + * Explanation: The endWord "cog" is not in wordList, therefore no possible transformation. + * + * @author rampatra + * @since 2019-08-15 + */ +public class WordLadderII { + + /** + * The approach is same as {@link WordLadder}. We calculate the {@code minDistance} from start to end word and also + * keep a map of words and its adjacent words (i.e, with only character difference). After we are done calculating + * the {@code mindistance}, we perform a dfs on the map upto depth {@code minDistance} and if the last word at this + * depth is equal to the end word then we add all words to the result. + * + * @param beginWord + * @param endWord + * @param wordList + * @return + */ + public static List> findLadders(String beginWord, String endWord, List wordList) { + int L = beginWord.length(); + List> ladders = new LinkedList<>(); + Map> transformedToOriginalWordMap = new HashMap<>(); + Queue> queue = new LinkedList<>(); + + wordList.forEach(word -> { + String transformedWord; + for (int i = 0; i < L; i++) { + transformedWord = word.substring(0, i) + "*" + word.substring(i + 1, L); + transformedToOriginalWordMap.putIfAbsent(transformedWord, new HashSet<>()); + transformedToOriginalWordMap.get(transformedWord).add(word); + } + } + ); + + int minDistance = -1; + Set visited = new HashSet<>(); + queue.add(new Pair<>(beginWord, 1)); + visited.add(beginWord); + + HashMap> connectedNodes = new HashMap<>(); + + while (!queue.isEmpty()) { + Pair currPair = queue.poll(); + String word = currPair.getKey(); + Integer level = currPair.getValue(); + + if (word.equals(endWord) && minDistance == -1) { + minDistance = level - 1; + } + + String transformedWord; + for (int i = 0; i < L; i++) { + transformedWord = word.substring(0, i) + "*" + word.substring(i + 1, L); + + for (String originalWord : transformedToOriginalWordMap.getOrDefault(transformedWord, Collections.emptySet())) { + if (!visited.contains(originalWord)) { + queue.add(new Pair<>(originalWord, level + 1)); + visited.add(originalWord); + } + + if (!word.equals(originalWord)) { + connectedNodes.putIfAbsent(word, new HashSet<>()); + connectedNodes.get(word).add(originalWord); + } + } + } + } + + dfs(ladders, new LinkedHashSet<>(), connectedNodes, beginWord, endWord, 0, minDistance); + + return ladders; + } + + /** + * Perform dfs on the map which contains words and its adjacent words. + * + * @param ladders + * @param ladder + * @param connectedNodes + * @param startNode + * @param endNode + * @param distance + * @param minDistance + */ + private static void dfs(List> ladders, Set ladder, Map> connectedNodes, + String startNode, String endNode, int distance, int minDistance) { + if (distance == minDistance && startNode.equals(endNode)) { + ladder.add(endNode); + ladders.add(new ArrayList<>(ladder)); + return; + } else if (distance > minDistance) { + return; + } + + ladder.add(startNode); + for (String nextNode : connectedNodes.getOrDefault(startNode, new HashSet<>())) { + if (!ladder.contains(nextNode)) { + dfs(ladders, new LinkedHashSet<>(ladder), connectedNodes, nextNode, endNode, distance + 1, minDistance); + } + } + } + + public static void main(String[] args) { + assertEquals("[[hit, hot, lot, log, cog], [hit, hot, dot, dog, cog]]", findLadders("hit", "cog", Arrays.asList("hot", "dot", "dog", "lot", "log", "cog")).toString()); + // TODO Fix this test case System.out.println(findLadders("cet", "ism", Arrays.asList("kid", "tag", "pup", "ail", "tun", "woo", "erg", "luz", "brr", "gay", "sip", "kay", "per", "val", "mes", "ohs", "now", "boa", "cet", "pal", "bar", "die", "war", "hay", "eco", "pub", "lob", "rue", "fry", "lit", "rex", "jan", "cot", "bid", "ali", "pay", "col", "gum", "ger", "row", "won", "dan", "rum", "fad", "tut", "sag", "yip", "sui", "ark", "has", "zip", "fez", "own", "ump", "dis", "ads", "max", "jaw", "out", "btu", "ana", "gap", "cry", "led", "abe", "box", "ore", "pig", "fie", "toy", "fat", "cal", "lie", "noh", "sew", "ono", "tam", "flu", "mgm", "ply", "awe", "pry", "tit", "tie", "yet", "too", "tax", "jim", "san", "pan", "map", "ski", "ova", "wed", "non", "wac", "nut", "why", "bye", "lye", "oct", "old", "fin", "feb", "chi", "sap", "owl", "log", "tod", "dot", "bow", "fob", "for", "joe", "ivy", "fan", "age", "fax", "hip", "jib", "mel", "hus", "sob", "ifs", "tab", "ara", "dab", "jag", "jar", "arm", "lot", "tom", "sax", "tex", "yum", "pei", "wen", "wry", "ire", "irk", "far", "mew", "wit", "doe", "gas", "rte", "ian", "pot", "ask", "wag", "hag", "amy", "nag", "ron", "soy", "gin", "don", "tug", "fay", "vic", "boo", "nam", "ave", "buy", "sop", "but", "orb", "fen", "paw", "his", "sub", "bob", "yea", "oft", "inn", "rod", "yam", "pew", "web", "hod", "hun", "gyp", "wei", "wis", "rob", "gad", "pie", "mon", "dog", "bib", "rub", "ere", "dig", "era", "cat", "fox", "bee", "mod", "day", "apr", "vie", "nev", "jam", "pam", "new", "aye", "ani", "and", "ibm", "yap", "can", "pyx", "tar", "kin", "fog", "hum", "pip", "cup", "dye", "lyx", "jog", "nun", "par", "wan", "fey", "bus", "oak", "bad", "ats", "set", "qom", "vat", "eat", "pus", "rev", "axe", "ion", "six", "ila", "lao", "mom", "mas", "pro", "few", "opt", "poe", "art", "ash", "oar", "cap", "lop", "may", "shy", "rid", "bat", "sum", "rim", "fee", "bmw", "sky", "maj", "hue", "thy", "ava", "rap", "den", "fla", "auk", "cox", "ibo", "hey", "saw", "vim", "sec", "ltd", "you", "its", "tat", "dew", "eva", "tog", "ram", "let", "see", "zit", "maw", "nix", "ate", "gig", "rep", "owe", "ind", "hog", "eve", "sam", "zoo", "any", "dow", "cod", "bed", "vet", "ham", "sis", "hex", "via", "fir", "nod", "mao", "aug", "mum", "hoe", "bah", "hal", "keg", "hew", "zed", "tow", "gog", "ass", "dem", "who", "bet", "gos", "son", "ear", "spy", "kit", "boy", "due", "sen", "oaf", "mix", "hep", "fur", "ada", "bin", "nil", "mia", "ewe", "hit", "fix", "sad", "rib", "eye", "hop", "haw", "wax", "mid", "tad", "ken", "wad", "rye", "pap", "bog", "gut", "ito", "woe", "our", "ado", "sin", "mad", "ray", "hon", "roy", "dip", "hen", "iva", "lug", "asp", "hui", "yak", "bay", "poi", "yep", "bun", "try", "lad", "elm", "nat", "wyo", "gym", "dug", "toe", "dee", "wig", "sly", "rip", "geo", "cog", "pas", "zen", "odd", "nan", "lay", "pod", "fit", "hem", "joy", "bum", "rio", "yon", "dec", "leg", "put", "sue", "dim", "pet", "yaw", "nub", "bit", "bur", "sid", "sun", "oil", "red", "doc", "moe", "caw", "eel", "dix", "cub", "end", "gem", "off", "yew", "hug", "pop", "tub", "sgt", "lid", "pun", "ton", "sol", "din", "yup", "jab", "pea", "bug", "gag", "mil", "jig", "hub", "low", "did", "tin", "get", "gte", "sox", "lei", "mig", "fig", "lon", "use", "ban", "flo", "nov", "jut", "bag", "mir", "sty", "lap", "two", "ins", "con", "ant", "net", "tux", "ode", "stu", "mug", "cad", "nap", "gun", "fop", "tot", "sow", "sal", "sic", "ted", "wot", "del", "imp", "cob", "way", "ann", "tan", "mci", "job", "wet", "ism", "err", "him", "all", "pad", "hah", "hie", "aim", "ike", "jed", "ego", "mac", "baa", "min", "com", "ill", "was", "cab", "ago", "ina", "big", "ilk", "gal", "tap", "duh", "ola", "ran", "lab", "top", "gob", "hot", "ora", "tia", "kip", "han", "met", "hut", "she", "sac", "fed", "goo", "tee", "ell", "not", "act", "gil", "rut", "ala", "ape", "rig", "cid", "god", "duo", "lin", "aid", "gel", "awl", "lag", "elf", "liz", "ref", "aha", "fib", "oho", "tho", "her", "nor", "ace", "adz", "fun", "ned", "coo", "win", "tao", "coy", "van", "man", "pit", "guy", "foe", "hid", "mai", "sup", "jay", "hob", "mow", "jot", "are", "pol", "arc", "lax", "aft", "alb", "len", "air", "pug", "pox", "vow", "got", "meg", "zoe", "amp", "ale", "bud", "gee", "pin", "dun", "pat", "ten", "mob"))); + } +} diff --git a/src/main/java/com/leetcode/maps/IsomorphicStrings.java b/src/main/java/com/leetcode/hashtables/IsomorphicStrings.java similarity index 98% rename from src/main/java/com/leetcode/maps/IsomorphicStrings.java rename to src/main/java/com/leetcode/hashtables/IsomorphicStrings.java index 78bc02bd..f9a7f4eb 100644 --- a/src/main/java/com/leetcode/maps/IsomorphicStrings.java +++ b/src/main/java/com/leetcode/hashtables/IsomorphicStrings.java @@ -1,4 +1,4 @@ -package com.leetcode.maps; +package com.leetcode.hashtables; import java.util.HashMap; import java.util.Map; diff --git a/src/main/java/com/leetcode/maps/RepeatedDnaSequence.java b/src/main/java/com/leetcode/hashtables/RepeatedDnaSequence.java similarity index 98% rename from src/main/java/com/leetcode/maps/RepeatedDnaSequence.java rename to src/main/java/com/leetcode/hashtables/RepeatedDnaSequence.java index a4eddb4a..7674273d 100644 --- a/src/main/java/com/leetcode/maps/RepeatedDnaSequence.java +++ b/src/main/java/com/leetcode/hashtables/RepeatedDnaSequence.java @@ -1,4 +1,4 @@ -package com.leetcode.maps; +package com.leetcode.hashtables; import java.util.*; diff --git a/src/main/java/com/leetcode/maps/ShortestWordDistanceII.java b/src/main/java/com/leetcode/hashtables/ShortestWordDistanceII.java similarity index 99% rename from src/main/java/com/leetcode/maps/ShortestWordDistanceII.java rename to src/main/java/com/leetcode/hashtables/ShortestWordDistanceII.java index ecb6c9b9..3932cebf 100644 --- a/src/main/java/com/leetcode/maps/ShortestWordDistanceII.java +++ b/src/main/java/com/leetcode/hashtables/ShortestWordDistanceII.java @@ -1,4 +1,4 @@ -package com.leetcode.maps; +package com.leetcode.hashtables; import java.util.ArrayList; import java.util.HashMap; diff --git a/src/main/java/com/leetcode/maps/TwoSumIII.java b/src/main/java/com/leetcode/hashtables/TwoSumIII.java similarity index 98% rename from src/main/java/com/leetcode/maps/TwoSumIII.java rename to src/main/java/com/leetcode/hashtables/TwoSumIII.java index 22f0748f..88916db0 100644 --- a/src/main/java/com/leetcode/maps/TwoSumIII.java +++ b/src/main/java/com/leetcode/hashtables/TwoSumIII.java @@ -1,4 +1,4 @@ -package com.leetcode.maps; +package com.leetcode.hashtables; import java.util.HashMap; import java.util.Map; diff --git a/src/main/java/com/leetcode/maps/slidingwindow/LongestSubstringWithKDistinctCharacters.java b/src/main/java/com/leetcode/hashtables/slidingwindow/LongestSubstringWithKDistinctCharacters.java similarity index 74% rename from src/main/java/com/leetcode/maps/slidingwindow/LongestSubstringWithKDistinctCharacters.java rename to src/main/java/com/leetcode/hashtables/slidingwindow/LongestSubstringWithKDistinctCharacters.java index 21cfab23..4b209b6e 100644 --- a/src/main/java/com/leetcode/maps/slidingwindow/LongestSubstringWithKDistinctCharacters.java +++ b/src/main/java/com/leetcode/hashtables/slidingwindow/LongestSubstringWithKDistinctCharacters.java @@ -1,4 +1,4 @@ -package com.leetcode.maps.slidingwindow; +package com.leetcode.hashtables.slidingwindow; import java.util.HashMap; import java.util.Map; @@ -38,42 +38,46 @@ public static int lengthOfLongestSubstringKDistinct(String str, int k) { int length = 0; Map letterCountInWindow = new HashMap<>(); - int i = 0; // start of window - int j = i; // end of window + int left = 0; // start of window + int right = 0; // end of window - while (j < str.length()) { - char ch = str.charAt(j); + while (right < str.length()) { + char ch = str.charAt(right); letterCountInWindow.put(ch, letterCountInWindow.getOrDefault(ch, 0) + 1); // when number of distinct characters in the window exceeds k: // - update length - // - remove the first character in the window or reduce its count if the window had more than one of this character + // - remove the first character in the window or reduce its count if the window has more than one of this character // - lastly, move the window forward if (letterCountInWindow.keySet().size() > k) { - char firstChar = str.charAt(i); + char firstChar = str.charAt(left); int firstCharCount = letterCountInWindow.get(firstChar); if (firstCharCount > 1) { letterCountInWindow.put(firstChar, firstCharCount - 1); } else { letterCountInWindow.remove(firstChar); } - length = Math.max(length, j - i); - i++; + length = Math.max(length, right - left); + left++; } - j++; + right++; } - return length == 0 ? j - i : length; + return Math.max(length, right - left); } public static void main(String[] args) { assertEquals(3, lengthOfLongestSubstringKDistinct("eceba", 2)); assertEquals(7, lengthOfLongestSubstringKDistinct("eceeeeeba", 2)); + assertEquals(12, lengthOfLongestSubstringKDistinct("bbbeeeeebaaaaaaaaaaa", 2)); assertEquals(2, lengthOfLongestSubstringKDistinct("abcdef", 2)); assertEquals(1, lengthOfLongestSubstringKDistinct("a", 1)); + assertEquals(0, lengthOfLongestSubstringKDistinct("aa", 0)); assertEquals(2, lengthOfLongestSubstringKDistinct("aa", 1)); assertEquals(3, lengthOfLongestSubstringKDistinct("aaa", 1)); - assertEquals(0, lengthOfLongestSubstringKDistinct("aa", 0)); + assertEquals(3, lengthOfLongestSubstringKDistinct("aab", 2)); + assertEquals(8, lengthOfLongestSubstringKDistinct("abcabcbb", 3)); + assertEquals(5, lengthOfLongestSubstringKDistinct("pwwkew", 3)); } } diff --git a/src/main/java/com/leetcode/hashtables/slidingwindow/LongestSubstringWithoutRepeatingCharacters.java b/src/main/java/com/leetcode/hashtables/slidingwindow/LongestSubstringWithoutRepeatingCharacters.java new file mode 100644 index 00000000..93a940e4 --- /dev/null +++ b/src/main/java/com/leetcode/hashtables/slidingwindow/LongestSubstringWithoutRepeatingCharacters.java @@ -0,0 +1,128 @@ +package com.leetcode.hashtables.slidingwindow; + +import java.util.HashSet; +import java.util.Set; + +import static org.junit.jupiter.api.Assertions.assertEquals; + +/** + * Level: Medium + * Link: https://leetcode.com/problems/longest-substring-without-repeating-characters/ + * Description: + * Given a string, find the length of the longest substring without repeating characters. + *

+ * Example 1: + * Input: "abcabcbb" + * Output: 3 + * Explanation: The answer is "abc", with the length of 3. + *

+ * Example 2: + * Input: "bbbbb" + * Output: 1 + * Explanation: The answer is "b", with the length of 1. + *

+ * Example 3: + * Input: "pwwkew" + * Output: 3 + * Explanation: The answer is "wke", with the length of 3. + * Note that the answer must be a substring, "pwke" is a subsequence and not a substring. + * + * @author rampatra + * @since 2019-08-15 + */ +public class LongestSubstringWithoutRepeatingCharacters { + + /** + * Sliding Window Approach (using map). + *

+ * Note: + * If we know that the charset is rather small, we can replace the Map with an integer array as direct access table. + *

+ * Commonly used tables are: + *

+ * int[26] for Letters 'a' - 'z' or 'A' - 'Z' + * int[128] for ASCII + * int[256] for Extended ASCII + *

+ * Runtime: 8 ms. + * + * @param s + * @return + */ + public static int lengthOfLongestSubstring(String s) { + int left = 0; + int right = 0; + int longestSubstringLen = 0; + Set charsInWindow = new HashSet<>(); + + + while (right < s.length()) { + + if (charsInWindow.contains(s.charAt(right))) { + while (s.charAt(left) != s.charAt(right)) { + longestSubstringLen = Math.max(longestSubstringLen, right - left); + charsInWindow.remove(s.charAt(left)); + left++; + } + left++; + } + + charsInWindow.add(s.charAt(right)); + right++; + } + + return Math.max(longestSubstringLen, right - left); + } + + /** + * Sliding Window Approach using int array. + * + * Runtime: 2 ms. + * + * @param s + * @return + */ + public static int lengthOfLongestSubstringNoMap(String s) { + int left = 0; + int right = 0; + int longestSubstringLen = 0; + int[] charsInWindow = new int[128]; + + // keep moving forward the right pointer and adding characters to the window + while (right < s.length()) { + + // once we encounter repeated characters, move the left pointer until the repeated character is removed + if (charsInWindow[s.charAt(right)] == 1) { + while (s.charAt(left) != s.charAt(right)) { + longestSubstringLen = Math.max(longestSubstringLen, right - left); + charsInWindow[s.charAt(left)] = 0; + left++; + } + left++; + } + + charsInWindow[s.charAt(right)] = 1; + right++; + } + + return Math.max(longestSubstringLen, right - left); + } + + public static void main(String[] args) { + assertEquals(0, lengthOfLongestSubstring("")); + assertEquals(1, lengthOfLongestSubstring(" ")); + assertEquals(1, lengthOfLongestSubstring("a")); + assertEquals(2, lengthOfLongestSubstring("aab")); + assertEquals(3, lengthOfLongestSubstring("abcabcbb")); + assertEquals(1, lengthOfLongestSubstring("bbbbb")); + assertEquals(3, lengthOfLongestSubstring("pwwkew")); + + assertEquals(0, lengthOfLongestSubstringNoMap("")); + assertEquals(1, lengthOfLongestSubstringNoMap(" ")); + assertEquals(1, lengthOfLongestSubstringNoMap("a")); + assertEquals(2, lengthOfLongestSubstringNoMap("aab")); + assertEquals(3, lengthOfLongestSubstringNoMap("abcabcbb")); + assertEquals(1, lengthOfLongestSubstringNoMap("bbbbb")); + assertEquals(3, lengthOfLongestSubstringNoMap("pwwkew")); + } +} \ No newline at end of file diff --git a/src/main/java/com/leetcode/hashtables/slidingwindow/MinimumWindowSubstring.java b/src/main/java/com/leetcode/hashtables/slidingwindow/MinimumWindowSubstring.java new file mode 100644 index 00000000..5414cdc1 --- /dev/null +++ b/src/main/java/com/leetcode/hashtables/slidingwindow/MinimumWindowSubstring.java @@ -0,0 +1,104 @@ +package com.leetcode.hashtables.slidingwindow; + +import java.util.HashMap; +import java.util.Map; + +import static org.junit.jupiter.api.Assertions.assertEquals; + +/** + * Level: Hard + * Link: https://leetcode.com/problems/minimum-window-substring/ + * Description: + * Given a string S and a string T, find the minimum window in S which will contain all the characters in T in + * complexity O(n). + *

+ * Example: + *

+ * Input: S = "ADOBECODEBANC", T = "ABC" + * Output: "BANC" + *

+ * Note: + * - If there is no such window in S that covers all characters in T, return the empty string "". + * - If there is such window, you are guaranteed that there will always be only one unique minimum window in S. + * + * @author rampatra + * @since 2019-08-13 + */ +public class MinimumWindowSubstring { + + /** + * Sliding Window Approach (using map). + * + * Note: + * If we know that the charset is rather small, we can replace the Map with an integer array as direct access table. + * + * Commonly used tables are: + * + * int[26] for Letters 'a' - 'z' or 'A' - 'Z' + * int[128] for ASCII + * int[256] for Extended ASCII + * + * Runtime: 22 ms. + * + * @param s + * @param t + * @return + */ + public static String minWindow(String s, String t) { + + int left = 0; // start of window + int right = 0; // end of window + int begin = 0; + int windowSize = Integer.MAX_VALUE; + int charsInWindow = 0; // to check whether the window has all the characters in t with the required frequency + + // characters and their counts in t + Map dictT = new HashMap<>(); + for (int i = 0; i < t.length(); i++) { + char ch = t.charAt(i); + dictT.put(ch, dictT.getOrDefault(ch, 0) + 1); + } + + // characters and their counts in the window + Map dictWindow = new HashMap<>(); + + while (right < s.length()) { + char rightChar = s.charAt(right); + int rightCharCount; + dictWindow.put(rightChar, (rightCharCount = dictWindow.getOrDefault(rightChar, 0) + 1)); + + // once the window has a character in t with the required frequency, increment charsInWindow + if (dictT.get(rightChar) != null && dictT.get(rightChar).equals(rightCharCount)) { + charsInWindow++; + } + + // once the window has all characters in t with required frequency then shorten the window by moving the + // left window forward until the window no longer has all characters of t + while (left <= right && charsInWindow == dictT.size()) { + if (right - left < windowSize) { + windowSize = right - left + 1; + begin = left; + } + + char leftChar = s.charAt(left); + Integer leftCharCount = dictWindow.get(leftChar); + dictWindow.put(leftChar, leftCharCount - 1); + + if (dictT.get(leftChar) != null && leftCharCount - 1 < dictT.get(leftChar)) { + charsInWindow--; + } + left++; + } + right++; + } + + return windowSize == Integer.MAX_VALUE ? "" : s.substring(begin, begin + windowSize); + } + + public static void main(String[] args) { + assertEquals("BANC", minWindow("ADOBECODEBANC", "ABC")); + assertEquals("BAC", minWindow("ADOBECODEBAC", "ABC")); + assertEquals("ba", minWindow("bba", "ab")); + assertEquals("baca", minWindow("acbbaca", "aba")); + } +} \ No newline at end of file diff --git a/src/main/java/com/leetcode/heaps/KthLargestElementInArray.java b/src/main/java/com/leetcode/heaps/KthLargestElementInArray.java new file mode 100644 index 00000000..2422de08 --- /dev/null +++ b/src/main/java/com/leetcode/heaps/KthLargestElementInArray.java @@ -0,0 +1,143 @@ +package com.leetcode.heaps; + +import com.rampatra.base.MaxHeap; + +import java.util.PriorityQueue; + +import static org.junit.jupiter.api.Assertions.assertEquals; + +/** + * Level: Medium + * Link: https://leetcode.com/problems/kth-largest-element-in-an-array/ + * Description: + * Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not + * the kth distinct element. + *

+ * Example 1: + * Input: [3,2,1,5,6,4] and k = 2 + * Output: 5 + *

+ * Example 2: + * Input: [3,2,3,1,2,4,5,5,6] and k = 4 + * Output: 4 + *

+ * Note: + * You may assume k is always valid, 1 ≤ k ≤ array's length. + * + * @author rampatra + * @since 2019-08-19 + */ +public class KthLargestElementInArray { + + /** + * Runtime: 1 ms. + * + * @param nums + * @param k + * @return + */ + public static int findKthLargest(int[] nums, int k) { + return heapSortUntilK(nums, k); + } + + /** + * In heapsort, after each iteration we have the max element at the end of the array. Ergo, if we run the algorithm + * k times then we would have our kth largest element. + * + * @param a + * @param k + * @return + */ + public static int heapSortUntilK(int[] a, int k) { + buildMaxHeap(a); + int count = 0; + + for (int i = a.length - 1; i > 0; i--) { + if (count++ == k) { + break; + } + swap(a, 0, i); + maxHeapify(a, 0, i); + } + + return a[a.length - k]; + } + + /** + * Makes the array {@param a} satisfy the max heap property starting from + * {@param index} till {@param end} position in array. + *

+ * See this {@link MaxHeap#maxHeapify} for a basic version of maxHeapify. + *

+ * Time complexity: O(log n). + * + * @param a + * @param index + * @param end + */ + public static void maxHeapify(int[] a, int index, int end) { + int largest = index; + int leftIndex = 2 * index + 1; + int rightIndex = 2 * index + 2; + + if (leftIndex < end && a[index] < a[leftIndex]) { + largest = leftIndex; + } + if (rightIndex < end && a[largest] < a[rightIndex]) { + largest = rightIndex; + } + + if (largest != index) { + swap(a, index, largest); + maxHeapify(a, largest, end); + } + } + + /** + * Converts array {@param a} in to a max heap. + *

+ * Time complexity: O(n) and is not O(n log n). + */ + private static void buildMaxHeap(int[] a) { + for (int i = a.length / 2 - 1; i >= 0; i--) { + maxHeapify(a, i, a.length); + } + } + + + /** + * When you poll() on a PriorityQueue the smallest number in the queue is removed. Based on this property, we can + * iterate over the entire array and in the end we would be left with the k largest element in the queue. + * + * @param nums + * @param k + * @return + */ + public static int findKthLargestUsingPriorityQueue(int[] nums, int k) { + PriorityQueue priorityQueue = new PriorityQueue<>(); + + for (int num : nums) { + priorityQueue.add(num); + + if (priorityQueue.size() > k) { + priorityQueue.poll(); + } + } + + return priorityQueue.isEmpty() ? -1 : priorityQueue.peek(); + } + + private static void swap(int[] a, int firstIndex, int secondIndex) { + a[firstIndex] = a[firstIndex] + a[secondIndex]; + a[secondIndex] = a[firstIndex] - a[secondIndex]; + a[firstIndex] = a[firstIndex] - a[secondIndex]; + } + + public static void main(String[] args) { + assertEquals(5, findKthLargest(new int[]{3, 2, 1, 5, 6, 4}, 2)); + assertEquals(3, findKthLargest(new int[]{3, 2, 1, 5, 6, 4}, 4)); + + assertEquals(5, findKthLargestUsingPriorityQueue(new int[]{3, 2, 1, 5, 6, 4}, 2)); + assertEquals(3, findKthLargestUsingPriorityQueue(new int[]{3, 2, 1, 5, 6, 4}, 4)); + } +} diff --git a/src/main/java/com/leetcode/heaps/TopKFrequentElements.java b/src/main/java/com/leetcode/heaps/TopKFrequentElements.java new file mode 100644 index 00000000..5a684325 --- /dev/null +++ b/src/main/java/com/leetcode/heaps/TopKFrequentElements.java @@ -0,0 +1,68 @@ +package com.leetcode.heaps; + +import javafx.util.Pair; + +import java.util.*; +import java.util.stream.Collectors; + +import static org.junit.jupiter.api.Assertions.assertEquals; + +/** + * Level: Medium + * Link: https://leetcode.com/problems/top-k-frequent-elements/ + * Description: + * Given a non-empty array of integers, return the k most frequent elements. + *

+ * Example 1: + * Input: nums = [1,1,1,2,2,3], k = 2 + * Output: [1,2] + *

+ * Example 2: + * Input: nums = [1], k = 1 + * Output: [1] + *

+ * Note: + * - You may assume k is always valid, 1 ≤ k ≤ number of unique elements. + * - Your algorithm's time complexity must be better than O(n log n), where n is the array's size. + * + * @author rampatra + * @since 2019-08-19 + */ +public class TopKFrequentElements { + + /** + * TODO: A faster approach without using Pair. + *

+ * Runtime: 51 ms. + * + * @param nums + * @param k + * @return + */ + public static List topKFrequent(int[] nums, int k) { + Map numCount = new HashMap<>(); + PriorityQueue> pq = new PriorityQueue<>(Comparator.comparing(Pair::getValue)); + + for (int num : nums) { + numCount.put(num, numCount.getOrDefault(num, 0) + 1); + } + + for (Map.Entry entry : numCount.entrySet()) { + pq.add(new Pair<>(entry.getKey(), entry.getValue())); + + if (pq.size() > k) { + pq.poll(); + } + } + + return pq.stream().map(Pair::getKey).collect(Collectors.toList()); + } + + public static void main(String[] args) { + assertEquals("[2, 1]", topKFrequent(new int[]{1, 1, 1, 2, 2, 3}, 2).toString()); + assertEquals("[0]", topKFrequent(new int[]{3, 0, 1, 0}, 1).toString()); + assertEquals("[1]", topKFrequent(new int[]{1}, 1).toString()); + assertEquals("[1, 2]", topKFrequent(new int[]{1, 2}, 2).toString()); + assertEquals("[2, -1]", topKFrequent(new int[]{4, 1, -1, 2, -1, 2, 3}, 2).toString()); + } +} diff --git a/src/main/java/com/leetcode/linkedlists/LinkedListCycleII.java b/src/main/java/com/leetcode/linkedlists/LinkedListCycleII.java new file mode 100644 index 00000000..d5fe0e50 --- /dev/null +++ b/src/main/java/com/leetcode/linkedlists/LinkedListCycleII.java @@ -0,0 +1,73 @@ +package com.leetcode.linkedlists; + +/** + * Level: Medium + * Link: https://leetcode.com/problems/linked-list-cycle-ii/ + * Description: + * Given a linked list, return the node where the cycle begins. If there is no cycle, return null. + * + * To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in + * the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list. + * + * Note: Do not modify the linked list. + * + * Example 1: + * + * Input: head = [3,2,0,-4], pos = 1 + * Output: tail connects to node index 1 + * Explanation: There is a cycle in the linked list, where tail connects to the second node. + * + * + * Example 2: + * + * Input: head = [1,2], pos = 0 + * Output: tail connects to node index 0 + * Explanation: There is a cycle in the linked list, where tail connects to the first node. + * + * + * Example 3: + * + * Input: head = [1], pos = -1 + * Output: no cycle + * Explanation: There is no cycle in the linked list. + * + * Follow-up: + * Can you solve it without using extra space? + * + * @author rampatra + * @since 2019-08-18 + */ +public class LinkedListCycleII { + + /** + * Runtime: 0 ms. + * + * @param head + * @return + */ + public Node detectCycle(Node head) { + Node slow = head; + Node fast = head; + + while (fast != null && fast.next != null) { + slow = slow.next; + fast = fast.next.next; + if (slow == fast) { + break; + } + } + + if (fast == null || fast.next == null) { + return null; + } else { + slow = head; + + while (slow != fast) { + slow = slow.next; + fast = fast.next; + } + + return slow; + } + } +} \ No newline at end of file diff --git a/src/main/java/com/leetcode/stacks/ExclusiveTimeOfFunctions.java b/src/main/java/com/leetcode/stacks/ExclusiveTimeOfFunctions.java new file mode 100644 index 00000000..63c61dc2 --- /dev/null +++ b/src/main/java/com/leetcode/stacks/ExclusiveTimeOfFunctions.java @@ -0,0 +1,89 @@ +package com.leetcode.stacks; + +import javafx.util.Pair; + +import java.util.Arrays; +import java.util.List; +import java.util.Stack; + +import static org.junit.jupiter.api.Assertions.assertEquals; + +/** + * Level: Medium + * Link: https://leetcode.com/problems/exclusive-time-of-functions/ + * Description: + * On a single threaded CPU, we execute some functions. Each function has a unique id between 0 and N-1. + * + * We store logs in timestamp order that describe when a function is entered or exited. + * + * Each log is a string with this format: "{function_id}:{"start" | "end"}:{timestamp}". For example, "0:start:3" + * means the function with id 0 started at the beginning of timestamp 3. "1:end:2" means the function with id 1 ended + * at the end of timestamp 2. + * + * A function's exclusive time is the number of units of time spent in this function. Note that this does not include + * any recursive calls to child functions. + * + * The CPU is single threaded which means that only one function is being executed at a given time unit. + * + * Return the exclusive time of each function, sorted by their function id. + * + * Input: + * n = 2 + * logs = ["0:start:0","1:start:2","1:end:5","0:end:6"] + * Output: [3, 4] + * Explanation: + * Function 0 starts at the beginning of time 0, then it executes 2 units of time and reaches the end of time 1. + * Now function 1 starts at the beginning of time 2, executes 4 units of time and ends at time 5. + * Function 0 is running again at the beginning of time 6, and also ends at the end of time 6, thus executing for 1 unit of time. + * So function 0 spends 2 + 1 = 3 units of total time executing, and function 1 spends 4 units of total time executing. + * + * + * Note: + * -> 1 <= n <= 100 + * -> Two functions won't start or end at the same time. + * -> Functions will always log when they exit. + * + * @author rampatra + * @since 2019-08-17 + */ +public class ExclusiveTimeOfFunctions { + + /** + * Runtime: 18 ms. + * + * @param n + * @param logs + * @return + */ + public static int[] exclusiveTime(int n, List logs) { + int[] times = new int[n]; + Stack> stack = new Stack<>(); + + for (String log : logs) { + String[] l = log.split(":"); + int id = Integer.parseInt(l[0]); + String operation = l[1]; + int timestamp = Integer.parseInt(l[2]); + + if (operation.equals("start")) { + if (!stack.empty()) { // if there are other processes started before, calculate their time until now + times[stack.peek().getKey()] += (timestamp - stack.peek().getValue() - 1); + } + stack.push(new Pair<>(id, timestamp)); + } else { + times[id] += timestamp - stack.pop().getValue() + 1; + if (!stack.isEmpty()) { // if there are other processes, make their start time to now + stack.push(new Pair<>(stack.pop().getKey(), timestamp)); + } + } + } + + return times; + } + + public static void main(String[] args) { + assertEquals("[4]", Arrays.toString(exclusiveTime(1, Arrays.asList("0:start:0", "0:start:1", "0:end:2", "0:end:3")))); + assertEquals("[6]", Arrays.toString(exclusiveTime(1, Arrays.asList("0:start:0", "0:start:1", "0:start:2", "0:end:3", "0:end:4", "0:end:5")))); + assertEquals("[3, 4]", Arrays.toString(exclusiveTime(2, Arrays.asList("0:start:0", "1:start:2", "1:end:5", "0:end:6")))); + } +} \ No newline at end of file diff --git a/src/main/java/com/leetcode/strings/AnagramsInString.java b/src/main/java/com/leetcode/strings/AnagramsInString.java index afdb3f9f..12e7d766 100644 --- a/src/main/java/com/leetcode/strings/AnagramsInString.java +++ b/src/main/java/com/leetcode/strings/AnagramsInString.java @@ -5,8 +5,40 @@ import java.util.List; /** - * Level: Easy + * Level: Medium * Problem: https://leetcode.com/problems/find-all-anagrams-in-a-string/ + * Description: + * Given a string s and a non-empty string p, find all the start indices of p's anagrams in s. + * + * Strings consists of lowercase English letters only and the length of both strings s and p will not be larger + * than 20,100. + * + * The order of output does not matter. + * + * Example 1: + * + * Input: + * s: "cbaebabacd" p: "abc" + * + * Output: + * [0, 6] + * + * Explanation: + * The substring with start index = 0 is "cba", which is an anagram of "abc". + * The substring with start index = 6 is "bac", which is an anagram of "abc". + * + * Example 2: + * + * Input: + * s: "abab" p: "ab" + * + * Output: + * [0, 1, 2] + * + * Explanation: + * The substring with start index = 0 is "ab", which is an anagram of "ab". + * The substring with start index = 1 is "ba", which is an anagram of "ab". + * The substring with start index = 2 is "ab", which is an anagram of "ab". * * @author rampatra * @since 2019-04-13 @@ -55,7 +87,7 @@ private static List findAllAnagramsInTextNaive(String text, String patt * where, * n = length of text or number of characters in text *

- * Runtime: 7 ms on leetcode. + * Runtime: 6 ms. * * @param text * @param pattern diff --git a/src/main/java/com/leetcode/trees/SerializeDeserializeBinaryTree.java b/src/main/java/com/leetcode/trees/SerializeDeserializeBinaryTree.java new file mode 100644 index 00000000..690de39d --- /dev/null +++ b/src/main/java/com/leetcode/trees/SerializeDeserializeBinaryTree.java @@ -0,0 +1,179 @@ +package com.leetcode.trees; + + +import java.util.LinkedList; +import java.util.Queue; + +/** + * Level: Hard + * Link: https://leetcode.com/problems/serialize-and-deserialize-binary-tree/ + * Description: + * Serialization is the process of converting a data structure or object into a sequence of bits so that it can be + * stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in + * the same or another computer environment. + * + * Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your + * serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized + * to a string and this string can be deserialized to the original tree structure. + * + * Example: + * + * You may serialize the following tree: + * + * 1 + * / \ + * 2 3 + * / \ + * 4 5 + * + * as "[1,2,3,null,null,4,5]" + * + * Clarification: The above format is the same as how LeetCode serializes a binary tree. You do not necessarily need + * to follow this format, so please be creative and come up with different approaches yourself. + * + * Note: Do not use class member/global/static variables to store states. Your serialize and deserialize algorithms + * should be stateless. + * + * @author rampatra + * @since 2019-08-17 + */ +public class SerializeDeserializeBinaryTree { + + /** + * Runtime: 31 ms. + * + * @param root + * @return + */ + public static String serialize(TreeNode root) { + if (root == null) { + return "[]"; + } + + StringBuilder sb = new StringBuilder(); + sb.append("["); + + Queue queue = new LinkedList<>(); + queue.add(root); + + while (!queue.isEmpty()) { + TreeNode node = queue.poll(); + + if (sb.length() > 1) { + sb.append(", "); + } + if (node == null) { + sb.append("null"); + continue; + } + + sb.append(node.val); + + queue.add(node.left); + queue.add(node.right); + } + + sb.append("]"); + return removeExtraNulls(sb.toString()); + } + + private static String removeExtraNulls(String data) { + int i = data.length() - 1; + while (!(data.charAt(i) >= 48 && data.charAt(i) <= 57)) { + i--; + } + return data.substring(0, i + 1) + "]"; + } + + /** + * + * @param data + * @return + */ + public static TreeNode deserialize(String data) { + data = data.substring(1, data.length() - 1); + + if (data.length() == 0) { + return null; + } + + String[] values = data.split(", "); + + TreeNode root = new TreeNode(Integer.parseInt(values[0])); + + Queue queue = new LinkedList<>(); + queue.add(root); + + for (int i = 0; i < values.length && !queue.isEmpty(); i += 2) { + TreeNode currNode = queue.poll(); + + if (i + 1 < values.length && !values[i + 1].equals("null")) { + TreeNode leftNode = new TreeNode(Integer.parseInt(values[i + 1])); + currNode.left = leftNode; + queue.add(leftNode); + } + + if (i + 2 < values.length && !values[i + 2].equals("null")) { + TreeNode rightNode = new TreeNode(Integer.parseInt(values[i + 2])); + currNode.right = rightNode; + queue.add(rightNode); + } + } + + return root; + } + + public static void main(String[] args) { + // TODO Convert the print statements to asserts + + System.out.println(serialize(new TreeNode(1))); + + /* + Binary Tree + + 1 + / \ + 2 3 + / \ + 4 5 + */ + TreeNode tree = new TreeNode(1); + tree.left = new TreeNode(2); + tree.right = new TreeNode(3); + tree.left.left = new TreeNode(4); + tree.left.right = new TreeNode(5); + + System.out.println(serialize(tree)); + + System.out.println(serialize(deserialize(serialize(tree)))); + + System.out.println(serialize(deserialize(serialize(null)))); + + TreeNode tree2 = new TreeNode(1); + tree2.right = new TreeNode(2); + tree2.right.right = new TreeNode(3); + tree2.right.right.right = new TreeNode(4); + tree2.right.right.right.right = new TreeNode(5); + tree2.right.right.right.right.right = new TreeNode(6); + tree2.right.right.right.right.right.right = new TreeNode(7); + tree2.right.right.right.right.right.right.right = new TreeNode(8); + + System.out.println(serialize(tree2)); + System.out.println(serialize(deserialize(serialize(tree2)))); + + System.out.println("---"); + + System.out.println(serialize(deserialize("[1, 2]"))); + System.out.println(serialize(deserialize("[1, 2, 3]"))); + System.out.println(serialize(deserialize("[3, 2, 4, 1]"))); + System.out.println(serialize(deserialize("[3, 2, 4, 1, 5, 6]"))); + System.out.println(serialize(deserialize("[1, 2, 3, null, null, 4, 5]"))); + System.out.println(serialize(deserialize("[5, 2, 3, null, null, 2, 4, 3, 1]"))); + + System.out.println(serialize(deserialize("[1, null, 2, null, 3, null, 4, null, 5]"))); + System.out.println(serialize(deserialize("[1, null, 2, null, 3, null, 4, null, 5, null, 6]"))); + System.out.println(serialize(deserialize("[1, null, 2, null, 3, null, 4, null, 5, null, 6, null, 7]"))); + System.out.println(serialize(deserialize("[1, null, 2, null, 3, null, 4, null, 5, null, 6, null, 7, null, 8]"))); + System.out.println(serialize(deserialize("[1, null, 2, null, 3, null, 4, null, 5, null, 6, null, 7, null, 8, null, 9]"))); + } +} \ No newline at end of file diff --git a/src/main/java/com/rampatra/linkedlists/DetectAndRemoveLoop.java b/src/main/java/com/rampatra/linkedlists/DetectAndRemoveLoop.java index ade19e90..5ad549d4 100644 --- a/src/main/java/com/rampatra/linkedlists/DetectAndRemoveLoop.java +++ b/src/main/java/com/rampatra/linkedlists/DetectAndRemoveLoop.java @@ -34,6 +34,11 @@ public class DetectAndRemoveLoop { * ii. Now, move both slow and fast pointer at same pace and where they meet is the starting point of the loop. * iii. Lastly, to remove the loop make the next of the node (before the starting point of loop) to null. * + * Proof for Floyd's Cycle Detection: Consider a cyclic list and imagine the slow and fast pointers are two runners + * racing around a circle track. The fast runner will eventually meet the slow runner. Why? Consider this case - + * The fast runner is just one step behind the slow runner. In the next iteration, they both increment one and two + * steps respectively and meet each other. + * * @param list * @param * @return {@code true} if loop exists {@code false} otherwise. @@ -58,7 +63,7 @@ public static > boolean detectAndRemoveLoop(SingleLinked while (true) { slow = slow.next; fast = fast.next; - if (slow.next == fast.next) { + if (slow == fast) { fast.next = null; break; } @@ -78,5 +83,23 @@ public static void main(String[] args) { linkedList.getNode(4).next = linkedList.getNode(2); System.out.println(detectAndRemoveLoop(linkedList)); linkedList.printList(); + + linkedList = new SingleLinkedList<>(); + linkedList.add(0); + linkedList.add(1); + linkedList.getNode(1).next = linkedList.getNode(0); + System.out.println(detectAndRemoveLoop(linkedList)); + linkedList.printList(); + + linkedList = new SingleLinkedList<>(); + linkedList.add(0); + System.out.println(detectAndRemoveLoop(linkedList)); + linkedList.printList(); + + linkedList = new SingleLinkedList<>(); + linkedList.add(0); + linkedList.getNode(0).next = linkedList.getNode(0); + System.out.println(detectAndRemoveLoop(linkedList)); + linkedList.printList(); } } \ No newline at end of file diff --git a/src/main/java/com/rampatra/linkedlists/LRUCache.java b/src/main/java/com/rampatra/linkedlists/LRUCache.java index 5a8d12f1..ed046035 100644 --- a/src/main/java/com/rampatra/linkedlists/LRUCache.java +++ b/src/main/java/com/rampatra/linkedlists/LRUCache.java @@ -2,6 +2,7 @@ import java.util.Iterator; import java.util.LinkedHashMap; +import java.util.LinkedHashSet; import java.util.Map; /** @@ -20,7 +21,7 @@ */ public class LRUCache { - LinkedHashMap linkedHashMap; + private LinkedHashMap linkedHashMap; // initialize cache LRUCache(final int size) { @@ -37,7 +38,7 @@ V add(E key, V value) { } V get(E key) { - return linkedHashMap.get(key); + return linkedHashMap.get(key); } private void print() { @@ -52,12 +53,13 @@ public static void main(String[] args) { cache.add(1, 1); cache.add(2, 2); cache.add(3, 3); + cache.print(); // initial cache contents + + cache.add(4, 4); // should remove 1 as it was accessed last cache.print(); - if (cache.get(4) == null) { - cache.add(4, 4); - } - cache.print(); - cache.add(5, 5); + + cache.get(2); + cache.add(5, 5); // should remove 3 as 2 was recently accessed cache.print(); } }