> entry : parentToChildMap.entrySet()) {
+
+ TreeNode parent = (TreeNode) nodes.get(entry.getKey());
+
+ for (Integer childIndex : entry.getValue()) {
+ parent.addChild(nodes.get(childIndex));
+ }
+ }
+
+ return nodes.get(0);
+ }
+
+
+ public static void main(String[] args) {
+ Tree root = solve();
+ SumInLeavesVisitor vis1 = new SumInLeavesVisitor();
+ ProductOfRedNodesVisitor vis2 = new ProductOfRedNodesVisitor();
+ FancyVisitor vis3 = new FancyVisitor();
+
+ root.accept(vis1);
+ root.accept(vis2);
+ root.accept(vis3);
+
+ int res1 = vis1.getResult();
+ int res2 = vis2.getResult();
+ int res3 = vis3.getResult();
+
+ System.out.println(res1);
+ System.out.println(res2);
+ System.out.println(res3);
+ }
+}
\ No newline at end of file
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/CanPlaceFlowers.java b/src/main/java/com/leetcode/arrays/CanPlaceFlowers.java
new file mode 100644
index 00000000..916499ac
--- /dev/null
+++ b/src/main/java/com/leetcode/arrays/CanPlaceFlowers.java
@@ -0,0 +1,68 @@
+package com.leetcode.arrays;
+
+
+import static org.junit.jupiter.api.Assertions.assertFalse;
+import static org.junit.jupiter.api.Assertions.assertTrue;
+
+/**
+ * Level: Easy
+ * Problem Link: https://leetcode.com/problems/can-place-flowers/
+ * Problem Description:
+ * Suppose you have a long flowerBed in which some of the plots are planted and some are not. However, flowers cannot
+ * be planted in adjacent plots - they would compete for water and both would die.
+ *
+ * Given a flowerBed (represented as an array containing 0 and 1, where 0 means empty and 1 means not empty), and a
+ * number n, return if n new flowers can be planted in it without violating the no-adjacent-flowers rule.
+ *
+ * Example 1:
+ * Input: flowerBed = [1,0,0,0,1], n = 1
+ * Output: True
+ *
+ * Example 2:
+ * Input: flowerBed = [1,0,0,0,1], n = 2
+ * Output: False
+ *
+ * Note:
+ * The input array won't violate no-adjacent-flowers rule.
+ * The input array size is in the range of [1, 20000].
+ * n is a non-negative integer which won't exceed the input array size.
+ *
+ * @author rampatra
+ * @since 2019-07-24
+ */
+public class CanPlaceFlowers {
+
+ /**
+ * Time Complexity: O(n)
+ * Space Complexity: O(1)
+ * Runtime: 1 ms.
+ *
+ * @param flowerBed
+ * @param n
+ * @return
+ */
+ public static boolean canPlaceFlowers(int[] flowerBed, int n) {
+ int i = 0, count = 0;
+ while (i < flowerBed.length) {
+ if (flowerBed[i] == 0 && (i == 0 || flowerBed[i - 1] == 0) && (i == flowerBed.length - 1 || flowerBed[i + 1] == 0)) {
+ flowerBed[i++] = 1;
+ count++;
+ }
+ if (count >= n)
+ return true;
+ i++;
+ }
+ return false;
+ }
+
+ public static void main(String[] args) {
+ assertTrue(canPlaceFlowers(new int[]{0}, 0));
+ assertTrue(canPlaceFlowers(new int[]{0}, 1));
+ assertTrue(canPlaceFlowers(new int[]{1}, 0));
+ assertFalse(canPlaceFlowers(new int[]{1}, 1));
+ assertTrue(canPlaceFlowers(new int[]{1, 0, 0, 0, 1}, 1));
+ assertFalse(canPlaceFlowers(new int[]{1, 0, 0, 0, 1}, 2));
+ assertFalse(canPlaceFlowers(new int[]{1, 0, 0, 0, 0, 1}, 2));
+ assertTrue(canPlaceFlowers(new int[]{1, 0, 0, 0, 1, 0, 0}, 2));
+ }
+}
\ No newline at end of file
diff --git a/src/main/java/com/leetcode/arrays/FindTheCelebrity.java b/src/main/java/com/leetcode/arrays/FindTheCelebrity.java
new file mode 100644
index 00000000..1b15d996
--- /dev/null
+++ b/src/main/java/com/leetcode/arrays/FindTheCelebrity.java
@@ -0,0 +1,105 @@
+package com.leetcode.arrays;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/**
+ * Level: Medium
+ * Problem Link: https://leetcode.com/problems/find-the-celebrity/
+ * Problem Description:
+ * Suppose you are at a party with n people (labeled from 0 to n - 1) and among them, there may exist one celebrity.
+ * The definition of a celebrity is that all the other n - 1 people know him/her but he/she does not know any of them.
+ *
+ * Now you want to find out who the celebrity is or verify that there is not one. The only thing you are allowed to do
+ * is to ask questions like: "Hi, A. Do you know B?" to get information of whether A knows B. You need to find out the
+ * celebrity (or verify there is not one) by asking as few questions as possible (in the asymptotic sense).
+ *
+ * You are given a helper function bool knows(a, b) which tells you whether A knows B. Implement a
+ * function int findCelebrity(n). There will be exactly one celebrity if he/she is in the party. Return the celebrity's
+ * label if there is a celebrity in the party. If there is no celebrity, return -1.
+ *
+ * Example 1:
+ *
+ * Input: graph = [
+ * [1,1,0],
+ * [0,1,0],
+ * [1,1,1]
+ * ]
+ * Output: 1
+ * Explanation: There are three persons labeled with 0, 1 and 2. graph[i][j] = 1 means person i knows person j, otherwise
+ * graph[i][j] = 0 means person i does not know person j. The celebrity is the person labeled as 1 because both 0 and 2
+ * know him but 1 does not know anybody.
+ *
+ *
+ * Example 2:
+ *
+ * Input: graph = [
+ * [1,0,1],
+ * [1,1,0],
+ * [0,1,1]
+ * ]
+ * Output: -1
+ * Explanation: There is no celebrity.
+ *
+ *
+ * Note: The directed graph is represented as an adjacency matrix, which is an n x n matrix where a[i][j] = 1 means
+ * person i knows person j while a[i][j] = 0 means the contrary. Remember that you won't have direct access to the
+ * adjacency matrix.
+ *
+ * @author rampatra
+ * @since 2019-08-04
+ */
+public class FindTheCelebrity {
+
+ private int[][] knowsMatrix;
+
+ FindTheCelebrity(int[][] knowsMatrix) {
+ this.knowsMatrix = knowsMatrix;
+ }
+
+ public boolean knows(int a, int b) {
+ return knowsMatrix[a][b] == 1;
+ }
+
+ /**
+ * Time Complexity: O(n)
+ * Space Complexity: O(1)
+ * Runtime: 6 ms.
+ *
+ * @param n
+ * @return
+ */
+ public int findCelebrity(int n) {
+ int celebrityIndex = 0;
+
+ for (int i = 1; i < n; i++) {
+ // if a person doesn't know another person then he maybe a celebrity
+ if (!knows(i, celebrityIndex)) {
+ celebrityIndex = i;
+ }
+ }
+
+ for (int i = 0; i < n; i++) {
+ // verify whether the celebrity only knows himself and all other people know the celebrity
+ if ((knows(celebrityIndex, i) && i != celebrityIndex) || !knows(i, celebrityIndex)) {
+ return -1;
+ }
+ }
+
+ return celebrityIndex;
+ }
+
+ public static void main(String[] args) {
+ FindTheCelebrity findTheCelebrity = new FindTheCelebrity(new int[][]{
+ {1, 1, 0},
+ {0, 1, 0},
+ {1, 1, 1}});
+
+ assertEquals(1, findTheCelebrity.findCelebrity(3));
+
+ findTheCelebrity = new FindTheCelebrity(new int[][]{
+ {1, 0},
+ {0, 1}});
+
+ assertEquals(-1, findTheCelebrity.findCelebrity(2));
+ }
+}
diff --git a/src/main/java/com/leetcode/arrays/InsertInterval.java b/src/main/java/com/leetcode/arrays/InsertInterval.java
new file mode 100644
index 00000000..52155f19
--- /dev/null
+++ b/src/main/java/com/leetcode/arrays/InsertInterval.java
@@ -0,0 +1,78 @@
+package com.leetcode.arrays;
+
+import java.util.Arrays;
+
+/**
+ * Level: Hard
+ * Problem Link: https://leetcode.com/problems/insert-interval/
+ * Problem Description:
+ * Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
+ *
+ * You may assume that the intervals were initially sorted according to their start times.
+ *
+ * Example 1:
+ * Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
+ * Output: [[1,5],[6,9]]
+ *
+ * Example 2:
+ * Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
+ * Output: [[1,2],[3,10],[12,16]]
+ * Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].
+ *
+ * Companies: LinkedIn.
+ * Related: {@link MergeIntervals}.
+ *
+ * @author rampatra
+ * @since 2019-07-23
+ */
+public class InsertInterval {
+
+ /**
+ * Time Complexity: O(n)
+ * Space Complexity: O(n)
+ * Runtime: 2 ms.
+ *
+ * @param intervals
+ * @param newInterval
+ * @return
+ */
+ public static int[][] insert(int[][] intervals, int[] newInterval) {
+ if (intervals.length == 0 && newInterval.length == 0) {
+ return new int[][]{};
+ } else if (intervals.length == 0) {
+ return new int[][]{newInterval};
+ }
+
+ int[][] mergedIntervals = new int[intervals.length + 1][2];
+ int j = 0;
+
+ for (int i = 0; i < intervals.length; i++) {
+ if (newInterval == null || newInterval[0] > intervals[i][1]) { // newInterval is after the ith interval
+ mergedIntervals[j++] = intervals[i];
+ } else if (newInterval[1] < intervals[i][0]) { // newInterval is before the ith interval
+ mergedIntervals[j++] = newInterval;
+ mergedIntervals[j++] = intervals[i];
+ newInterval = null;
+ } else { // if overlapping
+ newInterval[0] = Math.min(newInterval[0], intervals[i][0]);
+ newInterval[1] = Math.max(newInterval[1], intervals[i][1]);
+ }
+ }
+
+ if (newInterval != null) {
+ mergedIntervals[j++] = newInterval;
+ }
+
+ return Arrays.copyOfRange(mergedIntervals, 0, j);
+ }
+
+ public static void main(String[] args) {
+ System.out.println(Arrays.deepToString(insert(new int[][]{}, new int[]{})));
+ System.out.println(Arrays.deepToString(insert(new int[][]{}, new int[]{5, 7})));
+ System.out.println(Arrays.deepToString(insert(new int[][]{{1, 5}}, new int[]{0, 0})));
+ System.out.println(Arrays.deepToString(insert(new int[][]{{1, 5}}, new int[]{2, 3})));
+ System.out.println(Arrays.deepToString(insert(new int[][]{{2, 5}, {6, 7}, {8, 9}}, new int[]{0, 1})));
+ System.out.println(Arrays.deepToString(insert(new int[][]{{1, 3}, {6, 9}}, new int[]{2, 5})));
+ System.out.println(Arrays.deepToString(insert(new int[][]{{1, 2}, {3, 5}, {6, 7}, {8, 10}, {12, 16}}, new int[]{4, 8})));
+ }
+}
\ No newline at end of file
diff --git a/src/main/java/com/leetcode/arrays/MergeIntervals.java b/src/main/java/com/leetcode/arrays/MergeIntervals.java
new file mode 100644
index 00000000..58f56ebc
--- /dev/null
+++ b/src/main/java/com/leetcode/arrays/MergeIntervals.java
@@ -0,0 +1,78 @@
+package com.leetcode.arrays;
+
+import java.util.Arrays;
+import java.util.Comparator;
+
+/**
+ * Level: Medium
+ * Problem Link: https://leetcode.com/problems/merge-intervals/
+ * Problem Description:
+ *
+ * Given a collection of intervals, merge all overlapping intervals.
+ *
+ * Example 1:
+ * Input: [[1,3],[2,6],[8,10],[15,18]]
+ * Output: [[1,6],[8,10],[15,18]]
+ * Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
+ *
+ * Example 2:
+ * Input: [[1,4],[4,5]]
+ * Output: [[1,5]]
+ * Explanation: Intervals [1,4] and [4,5] are considered overlapping.
+ *
+ * Companies: LinkedIn
+ * Related: {@link InsertInterval}.
+ *
+ * @author rampatra
+ * @since 2019-07-22
+ */
+public class MergeIntervals {
+
+ /**
+ * Time complexity: O(n log n)
+ * Space complexity: O(n)
+ * Runtime: 6 ms
+ *
+ * @param intervals a list of intervals, may not be sorted
+ * @return a list of intervals, with overlapping intervals merged
+ */
+ public static int[][] merge(int[][] intervals) {
+ // some validations
+ if (intervals.length == 0) return new int[0][2];
+
+ // we first sort the intervals based on their start times
+ Arrays.sort(intervals, new IntervalComparator());
+
+ int[][] mergedIntervals = new int[intervals.length][2];
+ int lastMergedIndex = 0;
+ mergedIntervals[lastMergedIndex] = intervals[0];
+
+ for (int i = 1; i < intervals.length; i++) {
+ if (isOverlap(mergedIntervals[lastMergedIndex], intervals[i])) {
+ // if two intervals overlap, then merge the two
+ mergedIntervals[lastMergedIndex] = new int[]{Math.min(mergedIntervals[lastMergedIndex][0], intervals[i][0]),
+ Math.max(mergedIntervals[lastMergedIndex][1], intervals[i][1])};
+ } else {
+ mergedIntervals[++lastMergedIndex] = intervals[i];
+ }
+ }
+
+ return Arrays.copyOfRange(mergedIntervals, 0, lastMergedIndex + 1);
+ }
+
+ private static boolean isOverlap(int[] interval1, int[] interval2) {
+ return interval1[0] <= interval2[0] && interval1[1] >= interval2[0];
+ }
+
+ private static class IntervalComparator implements Comparator {
+ @Override
+ public int compare(int[] interval1, int[] interval2) {
+ return interval1[0] - interval2[0];
+ }
+ }
+
+ public static void main(String[] args) {
+ System.out.println(Arrays.deepToString(merge(new int[][]{{1, 3}, {2, 6}, {8, 10}, {15, 18}})));
+ System.out.println(Arrays.deepToString(merge(new int[][]{{1, 4}, {4, 5}})));
+ }
+}
\ No newline at end of file
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/NumberOfIslands.java b/src/main/java/com/leetcode/arrays/NumberOfIslands.java
new file mode 100644
index 00000000..04ac6831
--- /dev/null
+++ b/src/main/java/com/leetcode/arrays/NumberOfIslands.java
@@ -0,0 +1,107 @@
+package com.leetcode.arrays;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/**
+ * Level: Medium
+ * Link: https://leetcode.com/problems/number-of-islands/
+ * Description:
+ * Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water
+ * and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid
+ * are all surrounded by water.
+ *
+ * Example 1:
+ * Input:
+ * 11110
+ * 11010
+ * 11000
+ * 00000
+ * Output: 1
+ *
+ * Example 2:
+ * Input:
+ * 11000
+ * 11000
+ * 00100
+ * 00011
+ * Output: 3
+ *
+ * @author rampatra
+ * @since 2019-08-07
+ */
+public class NumberOfIslands {
+
+ /**
+ * The idea is simple and straightforward. Once we encounter land ('1' in grid) we drown the island or change the
+ * neighboring '1's to '0's. Therefore, the number of '1's we encounter, we can say that we have that many islands.
+ *
+ * Time Complexity: O(n)
+ * Space Complexity: O(n)
+ * Runtime: 1 ms.
+ *
+ * @param grid
+ * @return
+ */
+ public static int numIslands(char[][] grid) {
+ int count = 0;
+
+ for (int i = 0; i < grid.length; i++) {
+ for (int j = 0; j < grid[0].length; j++) {
+ if (grid[i][j] == '1') {
+ drownTheIsland(grid, i, j);
+ count++;
+ }
+ }
+ }
+
+ return count;
+ }
+
+ private static void drownTheIsland(char[][] grid, int i, int j) {
+ if (i < 0 || j < 0 || i >= grid.length || j >= grid[0].length || grid[i][j] == '0') {
+ return;
+ }
+
+ grid[i][j] = '0';
+
+ drownTheIsland(grid, i, j + 1);
+ drownTheIsland(grid, i, j - 1);
+ drownTheIsland(grid, i + 1, j);
+ drownTheIsland(grid, i - 1, j);
+ }
+
+ public static void main(String[] args) {
+ assertEquals(1, numIslands(new char[][]{
+ {'1', '1', '1', '1', '0'},
+ {'1', '1', '0', '1', '0'},
+ {'1', '1', '0', '0', '0'},
+ {'0', '0', '0', '0', '0'}
+ }));
+
+ assertEquals(2, numIslands(new char[][]{
+ {'1', '1', '1', '1', '0'},
+ {'1', '1', '0', '1', '0'},
+ {'1', '1', '0', '0', '0'},
+ {'0', '0', '0', '1', '0'}
+ }));
+
+ assertEquals(1, numIslands(new char[][]{
+ {'1', '1', '1', '1', '1'},
+ {'1', '1', '1', '1', '1'},
+ {'1', '1', '1', '1', '1'},
+ {'1', '1', '1', '1', '1'}
+ }));
+
+ assertEquals(1, numIslands(new char[][]{
+ {'1'}
+ }));
+
+ assertEquals(0, numIslands(new char[][]{
+ {'0'}
+ }));
+
+ assertEquals(0, numIslands(new char[][]{
+ {}
+ }));
+ }
+}
\ No newline at end of file
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
new file mode 100644
index 00000000..5cd0c821
--- /dev/null
+++ b/src/main/java/com/leetcode/arrays/ShortestWordDistance.java
@@ -0,0 +1,71 @@
+package com.leetcode.arrays;
+
+import com.leetcode.hashtables.ShortestWordDistanceII;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/**
+ * Level: Easy
+ * Problem Link: https://leetcode.com/problems/shortest-word-distance/
+ * Problem Description:
+ * Given a list of words and two words word1 and word2, return the shortest distance between these two words in the
+ * list of words.
+ *
+ * Example 1:
+ * Assume that words = ["practice", "makes", "perfect", "coding", "makes"].
+ * Given word1 = "coding", word2 = "practice", return 3.
+ * Given word1 = "makes", word2 = "coding", return 1.
+ *
+ * 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 ShortestWordDistanceII}.
+ *
+ * @author rampatra
+ * @since 2019-07-31
+ */
+public class ShortestWordDistance {
+
+ /**
+ * Time Complexity:
+ * Space Complexity:
+ * Runtime: 1 ms.
+ *
+ * @param words
+ * @param word1
+ * @param word2
+ * @return
+ */
+ public static int findShortestDistance(String[] words, String word1, String word2) {
+ int indexWord1 = -1;
+ int indexWord2 = -1;
+ int minDistance = Integer.MAX_VALUE;
+
+ for (int i = 0; i < words.length; i++) {
+ if (words[i].equals(word1)) {
+ indexWord1 = i;
+ } else if (words[i].equals(word2)) {
+ indexWord2 = i;
+ }
+ if (indexWord1 != -1 && indexWord2 != -1) {
+ minDistance = Math.min(minDistance, Math.abs(indexWord1 - indexWord2));
+ }
+ }
+
+ return minDistance;
+ }
+
+ public static void main(String[] args) {
+ assertEquals(3, findShortestDistance(new String[]{"practice", "makes", "perfect", "coding", "makes"},
+ "practice", "coding"));
+ assertEquals(3, findShortestDistance(new String[]{"practice", "makes", "perfect", "coding", "makes"},
+ "coding", "practice"));
+ assertEquals(1, findShortestDistance(new String[]{"practice", "makes", "perfect", "coding", "makes"},
+ "makes", "coding"));
+ assertEquals(1, findShortestDistance(new String[]{"practice", "makes", "perfect", "coding", "makes"},
+ "makes", "perfect"));
+ assertEquals(0, findShortestDistance(new String[]{"practice", "makes", "perfect", "coding", "makes"},
+ "perfect", "perfect"));
+ assertEquals(0, findShortestDistance(new String[]{"practice", "makes", "perfect", "coding", "makes"},
+ "makes", "makes"));
+ }
+}
diff --git a/src/main/java/com/leetcode/arrays/ShortestWordDistanceIII.java b/src/main/java/com/leetcode/arrays/ShortestWordDistanceIII.java
new file mode 100644
index 00000000..0d404633
--- /dev/null
+++ b/src/main/java/com/leetcode/arrays/ShortestWordDistanceIII.java
@@ -0,0 +1,80 @@
+package com.leetcode.arrays;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/**
+ * Level: Easy
+ * Problem Link: https://leetcode.com/problems/shortest-word-distance-iii/
+ * Problem Description:
+ * This is a follow-up problem of {@link ShortestWordDistance}. The only difference is that now word1 could be the
+ * same as word2.
+ *
+ * Given a list of words and two words word1 and word2, return the shortest distance between these two words in the list.
+ * word1 and word2 may be the same and they represent two individual words in the list.
+ *
+ * For example,
+ * Assume that words = ["practice", "makes", "perfect", "coding", "makes"].
+ * Given word1 = “makes”, word2 = “coding”, return 1.
+ * Given word1 = "makes", word2 = "makes", return 3.
+ *
+ * Note: You may assume word1 and word2 are both in the list. If they are same then it's guaranteed that there are
+ * two occurrences of the same.
+ *
+ * @author rampatra
+ * @since 2019-07-31
+ */
+public class ShortestWordDistanceIII {
+
+ /**
+ * Time Complexity:
+ * Space Complexity:
+ * TODO
+ *
+ * @param words
+ * @param word1
+ * @param word2
+ * @return
+ */
+ public static int findShortestDistance(String[] words, String word1, String word2) {
+ int indexWord1 = -1;
+ int indexWord2 = -1;
+ int minDistance = Integer.MAX_VALUE;
+
+ for (int i = 0; i < words.length; i++) {
+ if (words[i].equals(word1)) {
+ // if both words are same and the first index is already set then do nothing
+ if (word1.equals(word2) && indexWord1 != -1) {
+
+ } else {
+ indexWord1 = i;
+ }
+ }
+ if (words[i].equals(word2)) {
+ // if both words are same and i is same as first index then it implies its the
+ // first occurrence, skip and continue look for the second occurrence
+ if (word1.equals(word2) && i == indexWord1) {
+ continue;
+ }
+ indexWord2 = i;
+ }
+ if (indexWord1 != -1 && indexWord2 != -1) {
+ minDistance = Math.min(minDistance, Math.abs(indexWord1 - indexWord2));
+ }
+ }
+
+ return minDistance;
+ }
+
+ public static void main(String[] args) {
+ assertEquals(3, findShortestDistance(new String[]{"practice", "makes", "perfect", "coding", "makes"},
+ "makes", "makes"));
+ assertEquals(3, findShortestDistance(new String[]{"practice", "makes", "perfect", "coding", "makes"},
+ "coding", "practice"));
+ assertEquals(3, findShortestDistance(new String[]{"practice", "makes", "perfect", "coding", "makes"},
+ "practice", "coding"));
+ assertEquals(1, findShortestDistance(new String[]{"practice", "makes", "perfect", "coding", "makes"},
+ "makes", "coding"));
+ assertEquals(1, findShortestDistance(new String[]{"practice", "makes", "perfect", "coding", "makes"},
+ "makes", "perfect"));
+ }
+}
diff --git a/src/main/java/com/leetcode/arrays/SparseMatrixMultiplication.java b/src/main/java/com/leetcode/arrays/SparseMatrixMultiplication.java
new file mode 100644
index 00000000..78c884e4
--- /dev/null
+++ b/src/main/java/com/leetcode/arrays/SparseMatrixMultiplication.java
@@ -0,0 +1,87 @@
+package com.leetcode.arrays;
+
+import java.util.Arrays;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/**
+ * Level: Medium
+ * Link: https://leetcode.com/problems/sparse-matrix-multiplication/
+ * Description:
+ * Given two sparse matrices A and B, return the result of AB.
+ *
+ * You may assume that A's column number is equal to B's row number.
+ *
+ * Example:
+ *
+ * Input:
+ *
+ * A = [
+ * [ 1, 0, 0],
+ * [-1, 0, 3]
+ * ]
+ *
+ * B = [
+ * [ 7, 0, 0 ],
+ * [ 0, 0, 0 ],
+ * [ 0, 0, 1 ]
+ * ]
+ *
+ * Output:
+ *
+ * | 1 0 0 | | 7 0 0 | | 7 0 0 |
+ * AB = | -1 0 3 | x | 0 0 0 | = | -7 0 3 |
+ * | 0 0 1 |
+ *
+ * @author rampatra
+ * @since 2019-08-09
+ */
+public class SparseMatrixMultiplication {
+
+ /**
+ * Time Complexity: O(Arow * Acol * Bcol)
+ * Space Complexity: O(Arow * Bcol)
+ *
+ * @param A
+ * @param B
+ * @return
+ */
+ public static int[][] multiply(int[][] A, int[][] B) {
+ int[][] AB = new int[A.length][B[0].length];
+
+ for (int Bcol = 0; Bcol < B[0].length; Bcol++) {
+ for (int Arow = 0; Arow < A.length; Arow++) {
+ int sum = 0;
+ for (int Acol = 0; Acol < A[0].length; Acol++) {
+ sum += A[Arow][Acol] * B[Acol][Bcol];
+ }
+ AB[Arow][Bcol] = sum;
+ }
+ }
+
+ return AB;
+ }
+
+ public static void main(String[] args) {
+ assertEquals(Arrays.deepToString(new int[][]{
+ {7, 0, 0},
+ {-7, 0, 3}
+ }), Arrays.deepToString(multiply(new int[][]{
+ {1, 0, 0},
+ {-1, 0, 3}
+ }, new int[][]{
+ {7, 0, 0},
+ {0, 0, 0},
+ {0, 0, 1}
+ })));
+
+ assertEquals(Arrays.deepToString(new int[][]{
+ {0}
+ }), Arrays.deepToString(multiply(new int[][]{
+ {0, 1}
+ }, new int[][]{
+ {1},
+ {0}
+ })));
+ }
+}
\ No newline at end of file
diff --git a/src/main/java/com/leetcode/arrays/ValidTriangleNumber.java b/src/main/java/com/leetcode/arrays/ValidTriangleNumber.java
new file mode 100644
index 00000000..cdbfb0c5
--- /dev/null
+++ b/src/main/java/com/leetcode/arrays/ValidTriangleNumber.java
@@ -0,0 +1,117 @@
+package com.leetcode.arrays;
+
+import java.util.Arrays;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/**
+ * Level: Medium
+ * Problem Link: https://leetcode.com/problems/valid-triangle-number/
+ * Problem Description:
+ * Given an array consists of non-negative integers, your task is to count the number of triplets chosen from the array
+ * that can make triangles if we take them as side lengths of a triangle.
+ *
+ * Example 1:
+ * Input: [2,2,3,4]
+ * Output: 3
+ * Explanation:
+ * Valid combinations are:
+ * 2,3,4 (using the first 2)
+ * 2,3,4 (using the second 2)
+ * 2,2,3
+ *
+ * Note:
+ * - The length of the given array won't exceed 1000.
+ * - The integers in the given array are in the range of [0, 1000].
+ * - Triangle Property: Sum of any 2 sides must be greater than the 3rd side.
+ *
+ * @author rampatra
+ * @since 2019-08-07
+ */
+public class ValidTriangleNumber {
+
+ /**
+ * Time complexity : O(n^2 log n). In worst case, the inner loop will take n log n (binary search applied n times).
+ * Space complexity : O(log n). Sorting takes O(log n) space.
+ * Runtime: 13 ms.
+ *
+ * @param nums
+ * @return
+ */
+ public static int triangleNumberUsingBinarySearch(int[] nums) {
+ int noOfTriangles = 0;
+
+ Arrays.sort(nums);
+
+ for (int i = 0; i < nums.length - 2; i++) {
+ int k = i + 2;
+ for (int j = i + 1; j < nums.length - 1; j++) {
+ k = binarySearch(nums, k, nums.length - 1, nums[i] + nums[j]);
+ if (k - j - 1 > 0) {
+ noOfTriangles += k - j - 1;
+ }
+ }
+ }
+
+ return noOfTriangles;
+ }
+
+ private static int binarySearch(int[] nums, int low, int high, int num) {
+ while (low <= high) {
+ int mid = (low + high) / 2;
+ if (nums[mid] < num) {
+ low = mid + 1;
+ } else {
+ high = mid - 1;
+ }
+ }
+
+ return low;
+ }
+
+ /**
+ * The concept is simple. For each pair (i,j), find the value of k such that nums[i] + nums[j] > nums[k] (as per
+ * triangle property). Once we find k then we can form k- j - 1 triangles.
+ *
+ * Time Complexity: O(n^2) Loop of k and j will be executed O(n^2) times in total, because, we do
+ * not reinitialize the value of k for a new value of j chosen(for the same i). Thus, the complexity
+ * will be O(n^2 + n^2) = O(n^2).
+ * Space Complexity: O(log n). Sorting takes O(log n) space.
+ * Runtime: 5 ms.
+ *
+ * @param nums
+ * @return
+ */
+ public static int triangleNumber(int[] nums) {
+ int noOfTriangles = 0;
+ Arrays.sort(nums);
+
+ for (int i = 0; i < nums.length - 2; i++) {
+ int k = i + 2;
+ for (int j = i + 1; j < nums.length - 1; j++) {
+ while (k < nums.length && nums[i] + nums[j] > nums[k]) {
+ k++;
+ }
+ if (k - j - 1 > 0) {
+ noOfTriangles += k - j - 1;
+ }
+ }
+ }
+
+ return noOfTriangles;
+ }
+
+ public static void main(String[] args) {
+ assertEquals(0, triangleNumberUsingBinarySearch(new int[]{}));
+ assertEquals(0, triangleNumberUsingBinarySearch(new int[]{1}));
+ assertEquals(3, triangleNumberUsingBinarySearch(new int[]{2, 2, 3, 4}));
+ assertEquals(0, triangleNumberUsingBinarySearch(new int[]{0, 1, 0, 1}));
+ assertEquals(7, triangleNumberUsingBinarySearch(new int[]{1, 2, 3, 4, 5, 6}));
+
+ assertEquals(0, triangleNumber(new int[]{}));
+ assertEquals(0, triangleNumber(new int[]{1}));
+ assertEquals(3, triangleNumber(new int[]{2, 2, 3, 4}));
+ assertEquals(0, triangleNumber(new int[]{0, 1, 0, 1}));
+ assertEquals(7, triangleNumber(new int[]{1, 2, 3, 4, 5, 6}));
+ }
+}
\ No newline at end of file
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/DesignHitCounter.java b/src/main/java/com/leetcode/design/DesignHitCounter.java
new file mode 100644
index 00000000..62133196
--- /dev/null
+++ b/src/main/java/com/leetcode/design/DesignHitCounter.java
@@ -0,0 +1,117 @@
+package com.leetcode.design;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/**
+ * Level: Medium
+ * Problem Link: https://leetcode.com/problems/design-hit-counter/
+ * Problem Description:
+ * Design a hit counter which counts the number of hits received in the past 5 minutes.
+ *
+ * Each function accepts a timestamp parameter (in seconds granularity) and you may assume that calls are being made
+ * to the system in chronological order (ie, the timestamp is monotonically increasing). You may assume that the
+ * earliest timestamp starts at 1.
+ *
+ * It is possible that several hits arrive roughly at the same time.
+ *
+ * Example:
+ *
+ * HitCounter counter = new HitCounter();
+ *
+ * // hit at timestamp 1.
+ * counter.hit(1);
+ *
+ * // hit at timestamp 2.
+ * counter.hit(2);
+ *
+ * // hit at timestamp 3.
+ * counter.hit(3);
+ *
+ * // get hits at timestamp 4, should return 3.
+ * counter.getHits(4);
+ *
+ * // hit at timestamp 300.
+ * counter.hit(300);
+ *
+ * // get hits at timestamp 300, should return 4.
+ * counter.getHits(300);
+ *
+ * // get hits at timestamp 301, should return 3.
+ * counter.getHits(301);
+ *
+ * Follow up:
+ * What if the number of hits per second could be very large? Does your design scale?
+ *
+ * Runtime: 42 ms (better than ~98%).
+ *
+ * @author rampatra
+ * @since 2019-08-04
+ */
+public class DesignHitCounter {
+
+ private int[] timestamps;
+ private int[] hits;
+
+ /**
+ * Initialize your data structure here.
+ */
+ public DesignHitCounter() {
+ this.timestamps = new int[300];
+ this.hits = new int[300];
+ }
+
+ /**
+ * Record a hit.
+ *
+ * @param timestamp - The current timestamp (in seconds granularity).
+ */
+ public void hit(int timestamp) {
+ int bucket = timestamp % 300;
+ if (timestamps[bucket] == timestamp) {
+ hits[bucket]++;
+ } else {
+ timestamps[bucket] = timestamp;
+ hits[bucket] = 1;
+ }
+ }
+
+ /**
+ * Return the number of hits in the past 5 minutes.
+ *
+ * @param timestamp - The current timestamp (in seconds granularity).
+ */
+ public int getHits(int timestamp) {
+ int totalHits = 0;
+ for (int i = 0; i < 300; i++) {
+ if (timestamp - 300 < timestamps[i]) {
+ totalHits += hits[i];
+ }
+ }
+ return totalHits;
+ }
+
+ public static void main(String[] args) {
+ DesignHitCounter counter = new DesignHitCounter();
+
+ // hit at timestamp 1.
+ counter.hit(1);
+
+ // hit at timestamp 2.
+ counter.hit(2);
+
+ // hit at timestamp 3.
+ counter.hit(3);
+
+ // get hits at timestamp 4, should return 3.
+ assertEquals(3, counter.getHits(4));
+
+ // hit at timestamp 300.
+ counter.hit(300);
+
+ // get hits at timestamp 300, should return 4.
+ assertEquals(4, counter.getHits(300));
+
+ // get hits at timestamp 301, should return 3.
+ assertEquals(3, counter.getHits(301));
+ }
+}
diff --git a/src/main/java/com/leetcode/design/InsertDeleteGetRandom.java b/src/main/java/com/leetcode/design/InsertDeleteGetRandom.java
new file mode 100644
index 00000000..a6c2f376
--- /dev/null
+++ b/src/main/java/com/leetcode/design/InsertDeleteGetRandom.java
@@ -0,0 +1,106 @@
+package com.leetcode.design;
+
+import java.util.*;
+
+import static org.junit.jupiter.api.Assertions.*;
+
+/**
+ * Level: Medium
+ * Link: https://leetcode.com/problems/insert-delete-getrandom-o1/
+ * Description:
+ * Design a data structure that supports all following operations in average O(1) time.
+ *
+ * insert(val): Inserts an item val to the set if not already present.
+ * remove(val): Removes an item val from the set if present.
+ * getRandom: Returns a random element from current set of elements. Each element must have the same probability of
+ * being returned.
+ *
+ * Example:
+ *
+ * // Init an empty set.
+ * RandomizedSet randomSet = new RandomizedSet();
+ *
+ * // Inserts 1 to the set. Returns true as 1 was inserted successfully.
+ * randomSet.insert(1);
+ *
+ * // Returns false as 2 does not exist in the set.
+ * randomSet.remove(2);
+ *
+ * // Inserts 2 to the set, returns true. Set now contains [1,2].
+ * randomSet.insert(2);
+ *
+ * // getRandom should return either 1 or 2 randomly.
+ * randomSet.getRandom();
+ *
+ * // Removes 1 from the set, returns true. Set now contains [2].
+ * randomSet.remove(1);
+ *
+ * // 2 was already in the set, so return false.
+ * randomSet.insert(2);
+ *
+ * // Since 2 is the only number in the set, getRandom always return 2.
+ * randomSet.getRandom();
+ *
+ * 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/dynamicprogramming/PaintHouse.java b/src/main/java/com/leetcode/dynamicprogramming/PaintHouse.java
new file mode 100644
index 00000000..5692b0d8
--- /dev/null
+++ b/src/main/java/com/leetcode/dynamicprogramming/PaintHouse.java
@@ -0,0 +1,52 @@
+package com.leetcode.dynamicprogramming;
+
+/**
+ * Level: Easy
+ * Problem Link: https://leetcode.com/problems/paint-house/
+ * Problem Description:
+ * There are a row of n houses, each house can be painted with one of the three colors: red, blue or green. The cost
+ * of painting each house with a certain color is different. You have to paint all the houses such that no two adjacent
+ * houses have the same color. The cost of painting each house with a certain color is represented by a n x 3 cost matrix.
+ *
+ * For example, costs[0][0] is the cost of painting house 0 with color red; costs[1][2] is the cost of painting
+ * house 1 with color green, and so on... Find the minimum cost to paint all houses.
+ *
+ * Companies: LinkedIn.
+ * Related: {@link PaintHouseII}.
+ *
+ * @author rampatra
+ * @since 2019-07-23
+ */
+public class PaintHouse {
+
+ /**
+ * Runtime: 1 ms.
+ *
+ * @param costs of coloring the houses with red, blue, and green respectively. 1st row represents house 1, 2nd row
+ * house 2 and so on
+ * @return the minimum cost to paint all houses such that no two adjacent houses are of same color
+ */
+ public static int minCost(int[][] costs) {
+ if (costs == null || costs.length == 0) return 0;
+
+ for (int i = 1; i < costs.length; i++) {
+ costs[i][0] += Math.min(costs[i - 1][1], costs[i - 1][2]);
+ costs[i][1] += Math.min(costs[i - 1][0], costs[i - 1][2]);
+ costs[i][2] += Math.min(costs[i - 1][0], costs[i - 1][1]);
+ }
+
+ int lastRow = costs.length - 1;
+ return Math.min(Math.min(costs[lastRow][0], costs[lastRow][1]), costs[lastRow][2]);
+ }
+
+ public static void main(String[] args) {
+ System.out.println(minCost(new int[][]{
+ }));
+
+ System.out.println(minCost(new int[][]{
+ {2, 3, 4},
+ {5, 7, 6},
+ {8, 7, 2}
+ }));
+ }
+}
\ No newline at end of file
diff --git a/src/main/java/com/leetcode/dynamicprogramming/PaintHouseII.java b/src/main/java/com/leetcode/dynamicprogramming/PaintHouseII.java
new file mode 100644
index 00000000..431ce612
--- /dev/null
+++ b/src/main/java/com/leetcode/dynamicprogramming/PaintHouseII.java
@@ -0,0 +1,120 @@
+package com.leetcode.dynamicprogramming;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/**
+ * Level: Hard
+ * Problem Link: https://leetcode.com/problems/paint-house-ii/
+ * Problem Description:
+ * There are a row of n houses, each house can be painted with one of the m colors.
+ * The cost of painting each house with a certain color is different.
+ * You have to paint all the houses such that no two adjacent houses have the same color.
+ *
+ * The cost of painting each house with a certain color is represented by a n x m cost matrix.
+ *
+ * For example, costs[0][0] is the cost of painting house 0 with color 0;
+ * costs[1][2] is the cost of painting house 1 with color 2, and so on...
+ * Find the minimum cost to paint all houses.
+ *
+ * Note: All costs are positive integers.
+ *
+ * Follow up: Could you solve it in O(n * m) runtime? // TODO
+ *
+ * Companies: LinkedIn.
+ * Related: {@link PaintHouse}.
+ *
+ * @author rampatra
+ * @since 2019-07-24
+ */
+public class PaintHouseII {
+
+ /**
+ * The approach is similar to {@link PaintHouse} but slightly complex as the number of colors are arbitrary
+ * instead of the 3 fixed colors. So, we use two additional for loops to cycle through all the colors.
+ * Time Complexity: O(n * m * m)
+ * Space Complexity: O(1)
+ *
+ * @param costs the costs of coloring the house, each row represents the cost of coloring a particular
+ * house with different colors
+ * @return the minimum cost to paint all houses such that no two adjacent houses are of same color
+ */
+ public static int minCost(int[][] costs) {
+ if (costs == null || costs.length == 0) return 0;
+
+ for (int i = 1; i < costs.length; i++) {
+ for (int j = 0; j < costs[0].length; j++) {
+ int min = Integer.MAX_VALUE;
+ // loop through all colors for the previous house except the color of the current house
+ for (int k = 0; k < costs[0].length; k++) {
+ if (k != j) {
+ min = Math.min(costs[i - 1][k], min);
+ }
+ }
+ costs[i][j] += min;
+ }
+ }
+
+ int minCost = Integer.MAX_VALUE;
+ for (int i = 0; i < costs[0].length; i++) {
+ minCost = Math.min(costs[costs.length - 1][i], minCost);
+ }
+
+ return minCost;
+ }
+
+ public static void main(String[] args) {
+ assertEquals(0, minCost(new int[][]{}));
+
+ assertEquals(10, minCost(new int[][]{
+ {2, 3, 4},
+ {5, 7, 6},
+ {8, 7, 2}
+ }));
+
+ assertEquals(10, minCost(new int[][]{{10, 30, 20}}));
+
+ assertEquals(3, minCost(new int[][]{{1, 1, 1},
+ {1, 1, 1},
+ {1, 1, 1}}));
+
+ assertEquals(5, minCost(new int[][]{{1, 2, 3},
+ {3, 2, 1},
+ {2, 2, 2},
+ {3, 1, 2}}));
+
+ assertEquals(10, minCost(new int[][]{{17, 2, 17},
+ {16, 16, 5},
+ {14, 3, 19}}));
+
+ assertEquals(5, minCost(new int[][]{{1, 5, 3},
+ {2, 9, 4}}));
+
+ assertEquals(8, minCost(new int[][]{{8}}));
+
+ assertEquals(143, minCost(new int[][]{{12, 1, 19},
+ {15, 1, 10},
+ {3, 11, 10},
+ {9, 3, 10},
+ {4, 8, 7},
+ {4, 18, 2},
+ {16, 6, 6},
+ {3, 3, 6},
+ {10, 18, 16},
+ {5, 4, 8},
+ {5, 3, 16},
+ {11, 8, 19},
+ {18, 15, 18},
+ {16, 4, 15},
+ {10, 7, 13},
+ {11, 10, 14},
+ {3, 9, 8},
+ {5, 2, 2},
+ {3, 2, 5},
+ {2, 19, 14},
+ {17, 3, 6},
+ {6, 4, 17},
+ {5, 15, 19},
+ {2, 14, 14},
+ {19, 4, 16}}));
+ }
+}
diff --git a/src/main/java/com/leetcode/graphs/GraphValidTree.java b/src/main/java/com/leetcode/graphs/GraphValidTree.java
new file mode 100644
index 00000000..15950143
--- /dev/null
+++ b/src/main/java/com/leetcode/graphs/GraphValidTree.java
@@ -0,0 +1,131 @@
+package com.leetcode.graphs;
+
+import java.util.ArrayList;
+import java.util.List;
+
+import static org.junit.jupiter.api.Assertions.assertFalse;
+import static org.junit.jupiter.api.Assertions.assertTrue;
+
+/**
+ * Level: Medium
+ * Problem Link: https://leetcode.com/problems/graph-valid-tree/
+ * Problem Description:
+ * Given n nodes labeled from 0 to n-1 and a list of undirected edges (each edge is a pair of nodes), write a function
+ * to check whether these edges make up a valid tree.
+ *
+ * Example 1:
+ * Input: n = 5, and edges = [[0,1], [0,2], [0,3], [1,4]]
+ * Output: true
+ *
+ * Example 2:
+ * Input: n = 5, and edges = [[0,1], [1,2], [2,3], [1,3], [1,4]]
+ * Output: false
+ *
+ * Note: you can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0,1] is the
+ * same as [1,0] and thus will not appear together in edges.
+ *
+ * @author rampatra
+ * @since 2019-08-05
+ */
+public class GraphValidTree {
+
+ /**
+ *
+ * @param n
+ * @param edges
+ * @return
+ */
+ public static boolean isValidTree(int n, int[][] edges) {
+ List> adjacencyList = new ArrayList<>(n);
+
+ for (int i = 0; i < n; i++) {
+ adjacencyList.add(new ArrayList<>());
+ }
+
+ for (int i = 0; i < edges.length; i++) {
+ adjacencyList.get(edges[i][0]).add(edges[i][1]);
+ }
+
+ boolean[] visited = new boolean[n];
+
+ if (hasCycle(adjacencyList, 0, -1, visited)) {
+ return false;
+ }
+
+ for (int i = 0; i < n; i++) {
+ if (!visited[i]) {
+ return false;
+ }
+ }
+
+ return true;
+ }
+
+ private static boolean hasCycle(List> adjacencyList, int node1, int exclude, boolean[] visited) {
+ visited[node1] = true;
+
+ for (int i = 0; i < adjacencyList.get(node1).size(); i++) {
+ int node2 = adjacencyList.get(node1).get(i);
+
+ if ((visited[node2] && exclude != node2) || (!visited[node2] && hasCycle(adjacencyList, node2, node1, visited))) {
+ return true;
+ }
+ }
+
+ return false;
+ }
+
+
+ /**
+ * Union-find algorithm: We keep all connected nodes in one set in the union operation and in find operation we
+ * check whether two nodes belong to the same set. If yes then there's a cycle and if not then no cycle.
+ *
+ * Good articles on union-find:
+ * - https://www.hackerearth.com/practice/notes/disjoint-set-union-union-find/
+ * - https://www.youtube.com/watch?v=wU6udHRIkcc
+ *
+ * @param n
+ * @param edges
+ * @return
+ */
+ public static boolean isValidTreeUsingUnionFind(int n, int[][] edges) {
+ int[] roots = new int[n];
+
+ for (int i = 0; i < n; i++) {
+ roots[i] = i;
+ }
+
+ for (int i = 0; i < edges.length; i++) {
+ // find operation
+ if (roots[edges[i][0]] == roots[edges[i][1]]) {
+ return false;
+ }
+ // union operation
+ roots[edges[i][1]] = findRoot(roots, roots[edges[i][0]]); // note: we can optimize this even further by
+ // considering size of each side and then join the side with smaller size to the one with a larger size (weighted union).
+ // We can use another array called size to keep count of the size or we can use the same root array with
+ // negative values, i.e, negative resembles that the node is pointing to itself and the number will represent
+ // the size. For example, roots = [-2, -1, -1, 0] means that node 3 is pointing to node 0 and node 0 is pointing
+ // to itself and is has 2 nodes under it including itself.
+ }
+
+ return edges.length == n - 1;
+ }
+
+ private static int findRoot(int[] roots, int node) {
+ while (roots[node] != node) {
+ node = roots[node];
+ }
+ return node;
+ }
+
+ public static void main(String[] args) {
+ assertTrue(isValidTree(5, new int[][]{{0, 1}, {0, 2}, {0, 3}, {1, 4}}));
+ assertFalse(isValidTree(5, new int[][]{{0, 1}, {1, 2}, {2, 3}, {1, 3}, {1, 4}}));
+ assertFalse(isValidTree(3, new int[][]{{0, 1}, {1, 2}, {2, 0}}));
+
+ assertTrue(isValidTreeUsingUnionFind(5, new int[][]{{0, 1}, {0, 2}, {0, 3}, {1, 4}}));
+ assertFalse(isValidTreeUsingUnionFind(5, new int[][]{{0, 1}, {1, 2}, {2, 3}, {1, 3}, {1, 4}}));
+ assertFalse(isValidTreeUsingUnionFind(3, new int[][]{{0, 1}, {1, 2}, {2, 0}}));
+ }
+}
\ No newline at end of file
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/hashtables/IsomorphicStrings.java b/src/main/java/com/leetcode/hashtables/IsomorphicStrings.java
new file mode 100644
index 00000000..f9a7f4eb
--- /dev/null
+++ b/src/main/java/com/leetcode/hashtables/IsomorphicStrings.java
@@ -0,0 +1,104 @@
+package com.leetcode.hashtables;
+
+import java.util.HashMap;
+import java.util.Map;
+
+import static org.junit.jupiter.api.Assertions.assertFalse;
+import static org.junit.jupiter.api.Assertions.assertTrue;
+
+/**
+ * Level: Easy
+ * Link: https://leetcode.com/problems/isomorphic-strings/
+ * Description:
+ * Given two strings s and t, determine if they are isomorphic.
+ *
+ * Two strings are isomorphic if the characters in s can be replaced to get t.
+ *
+ * All occurrences of a character must be replaced with another character while preserving the order of characters. No
+ * two characters may map to the same character but a character may map to itself.
+ *
+ * Example 1:
+ * Input: s = "egg", t = "add"
+ * Output: true
+ *
+ * Example 2:
+ * Input: s = "foo", t = "bar"
+ * Output: false
+ *
+ * Example 3:
+ * Input: s = "paper", t = "title"
+ * Output: true
+ *
+ * Note:
+ * You may assume both s and t have the same length.
+ *
+ * @author rampatra
+ * @since 2019-08-11
+ */
+public class IsomorphicStrings {
+
+ /**
+ * Time Complexity:
+ * Space Complexity:
+ * Runtime: 8 ms.
+ *
+ * @param s
+ * @param t
+ * @return
+ */
+ public static boolean isIsomorphic(String s, String t) {
+
+ Map sToTCharMap = new HashMap<>();
+ Map tToSCharMap = new HashMap<>();
+
+ for (int i = 0; i < s.length(); i++) {
+ char chFromS = s.charAt(i);
+ char chFromT = t.charAt(i);
+ if (sToTCharMap.get(chFromS) == null && tToSCharMap.get(chFromT) == null) {
+ sToTCharMap.put(chFromS, chFromT);
+ tToSCharMap.put(chFromT, chFromS);
+ }
+ Character mappedChFromSToT = sToTCharMap.get(chFromS);
+ if (mappedChFromSToT == null || mappedChFromSToT != chFromT) {
+ return false;
+ }
+ }
+
+ return true;
+ }
+
+ /**
+ * Time Complexity:
+ * Space Complexity:
+ * Runtime: 3 ms.
+ *
+ * @param s
+ * @param t
+ * @return
+ */
+ public static boolean isIsomorphicWithoutMaps(String s, String t) {
+ int[] charMap = new int[512];
+ for (int i = 0; i < s.length(); i++) {
+ char chFromS = s.charAt(i);
+ char chFromT = t.charAt(i);
+ if (charMap[chFromS] != charMap[chFromT + 256]) {
+ return false;
+ }
+ charMap[chFromS] = charMap[chFromT + 256] = i + 1;
+ }
+
+ return true;
+ }
+
+ public static void main(String[] args) {
+ assertTrue(isIsomorphic("egg", "add"));
+ assertFalse(isIsomorphic("foo", "bar"));
+ assertTrue(isIsomorphic("paper", "title"));
+ assertFalse(isIsomorphic("ab", "aa"));
+
+ assertTrue(isIsomorphicWithoutMaps("egg", "add"));
+ assertFalse(isIsomorphicWithoutMaps("foo", "bar"));
+ assertTrue(isIsomorphicWithoutMaps("paper", "title"));
+ assertFalse(isIsomorphicWithoutMaps("ab", "aa"));
+ }
+}
diff --git a/src/main/java/com/leetcode/hashtables/RepeatedDnaSequence.java b/src/main/java/com/leetcode/hashtables/RepeatedDnaSequence.java
new file mode 100644
index 00000000..7674273d
--- /dev/null
+++ b/src/main/java/com/leetcode/hashtables/RepeatedDnaSequence.java
@@ -0,0 +1,89 @@
+package com.leetcode.hashtables;
+
+import java.util.*;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/**
+ * Level: Medium
+ * Problem Link: https://leetcode.com/problems/repeated-dna-sequences/submissions/
+ * Problem Description:
+ * All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACGAATTCCG". When
+ * studying DNA, it is sometimes useful to identify repeated sequences within the DNA.
+ *
+ * Write a function to find all the 10-letter-long sequences (substrings) that occur more than once in a DNA molecule.
+ *
+ * Example:
+ * Input: s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"
+ * Output: ["AAAAACCCCC", "CCCCCAAAAA"]
+ *
+ * TODO: Figure another method which would have a better runtime.
+ *
+ * @author rampatra
+ * @since 2019-07-29
+ */
+public class RepeatedDnaSequence {
+
+ /**
+ * Rabin-Karp Algorithm: https://brilliant.org/wiki/rabin-karp-algorithm/
+ * Following Rabin-Karp's approach let's you avoid spurious hits (worst case scenario) but once the hash matches,
+ * you will have to compare and check the string you're searching. I tried to just rely on the hash and few test
+ * cases failed for me (https://leetcode.com/submissions/detail/247342702/).
+ *
+ * Time Complexity:
+ * Space Complexity:
+ * Runtime: 38 ms.
+ *
+ * @param s
+ * @return
+ */
+ public static List findRepeatedDnaSequences(String s) {
+ if (s.length() < 10) return new ArrayList<>();
+
+ Set repeatedSequences = new HashSet<>();
+ Map> hashToStringMap = new HashMap<>();
+ long hashOfSequence = computeHash(s);
+ hashToStringMap.put(hashOfSequence, new HashSet() {{
+ add(s.substring(0, 10));
+ }});
+
+ long pow = (long) Math.pow(4, 9);
+
+ for (int i = 10; i < s.length(); i++) {
+ hashOfSequence = (hashOfSequence - (pow * (s.charAt(i - 10) - 'A'))) * 4 + (s.charAt(i) - 'A');
+ String subString = s.substring(i - 10 + 1, i + 1);
+
+ if (hashToStringMap.get(hashOfSequence) != null && hashToStringMap.get(hashOfSequence).contains(subString)) {
+ repeatedSequences.add(subString);
+ continue;
+ }
+
+ hashToStringMap.putIfAbsent(hashOfSequence, new HashSet<>());
+ hashToStringMap.get(hashOfSequence).add(subString);
+ }
+
+ return new ArrayList<>(repeatedSequences);
+ }
+
+ private static long computeHash(String s) {
+ long hash = 0;
+ for (int i = 0; i < 10; i++) {
+ hash += (Math.pow(4, i) * (s.charAt(9 - i) - 'A'));
+ }
+ return hash;
+ }
+
+ public static void main(String[] args) {
+ assertEquals(new ArrayList<>(),
+ findRepeatedDnaSequences("AAAAACCC"));
+
+ assertEquals(Arrays.asList("AAAAACCCCC", "CCCCCAAAAA"),
+ findRepeatedDnaSequences("AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"));
+
+ assertEquals(Collections.singletonList("AAAAAAAAAA"),
+ findRepeatedDnaSequences("AAAAAAAAAAAA"));
+
+ assertEquals(Collections.singletonList("BBBBBBBBBB"),
+ findRepeatedDnaSequences("BBBBBBBBBBBB"));
+ }
+}
\ No newline at end of file
diff --git a/src/main/java/com/leetcode/hashtables/ShortestWordDistanceII.java b/src/main/java/com/leetcode/hashtables/ShortestWordDistanceII.java
new file mode 100644
index 00000000..3932cebf
--- /dev/null
+++ b/src/main/java/com/leetcode/hashtables/ShortestWordDistanceII.java
@@ -0,0 +1,89 @@
+package com.leetcode.hashtables;
+
+import java.util.ArrayList;
+import java.util.HashMap;
+import java.util.List;
+import java.util.Map;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/**
+ * Level: Medium
+ * Problem Link: https://leetcode.com/problems/shortest-word-distance-ii/
+ * Problem Description:
+ * Design a class which receives a list of words in the constructor, and implements a method that takes two words
+ * word1 and word2 and return the shortest distance between these two words in the list. Your method will be called
+ * repeatedly many times with different parameters. For a simpler variant, see {@link com.leetcode.arrays.ShortestWordDistance}.
+ *
+ * Examples:
+ * Assume that words = ["practice", "makes", "perfect", "coding", "makes"].
+ *
+ * Input1: word1 = “coding”, word2 = “practice”
+ * Output1: 3
+ *
+ * Input2: word1 = "makes", word2 = "coding"
+ * Output2: 1
+ *
+ * Note: You may assume that word1 does not equal to word2, and word1 and word2 are both in the list.
+ *
+ * @author rampatra
+ * @since 2019-07-31
+ */
+public class ShortestWordDistanceII {
+
+ private String[] words;
+ private Map> wordsToIndexesMap;
+
+ ShortestWordDistanceII(String[] words) {
+ this.words = words;
+ this.wordsToIndexesMap = getWordsToIndexesMap();
+ }
+
+ /**
+ * Runtime: 65 ms.
+ *
+ * @param word1
+ * @param word2
+ * @return
+ */
+ public int findShortestDistance(String word1, String word2) {
+ return findShortestDistance(wordsToIndexesMap.get(word1), wordsToIndexesMap.get(word2));
+ }
+
+ private int findShortestDistance(List indexes1, List indexes2) {
+ int minDistance = Integer.MAX_VALUE;
+
+ for (int i = 0, j = 0; i < indexes1.size() && j < indexes2.size(); ) {
+ if (indexes1.get(i) <= indexes2.get(j)) {
+ minDistance = Math.min(minDistance, Math.abs(indexes1.get(i) - indexes2.get(j)));
+ i++;
+ } else if (indexes1.get(i) > indexes2.get(j)) {
+ minDistance = Math.min(minDistance, Math.abs(indexes1.get(i) - indexes2.get(j)));
+ j++;
+ }
+ }
+
+ return minDistance;
+ }
+
+ private Map> getWordsToIndexesMap() {
+ Map> wordsToIndexesMap = new HashMap<>();
+
+ for (int i = 0; i < words.length; i++) {
+ wordsToIndexesMap.putIfAbsent(words[i], new ArrayList<>());
+ wordsToIndexesMap.get(words[i]).add(i);
+ }
+ return wordsToIndexesMap;
+ }
+
+ public static void main(String[] args) {
+ ShortestWordDistanceII shortestWordDist = new ShortestWordDistanceII(new String[]{"practice", "makes", "perfect", "coding", "makes"});
+ assertEquals(1, shortestWordDist.findShortestDistance("coding", "makes"));
+ assertEquals(1, shortestWordDist.findShortestDistance("perfect", "makes"));
+ assertEquals(1, shortestWordDist.findShortestDistance("practice", "makes"));
+ assertEquals(1, shortestWordDist.findShortestDistance("makes", "practice"));
+ assertEquals(3, shortestWordDist.findShortestDistance("coding", "practice"));
+ assertEquals(0, shortestWordDist.findShortestDistance("coding", "coding"));
+ assertEquals(0, shortestWordDist.findShortestDistance("makes", "makes"));
+ }
+}
\ No newline at end of file
diff --git a/src/main/java/com/leetcode/hashtables/TwoSumIII.java b/src/main/java/com/leetcode/hashtables/TwoSumIII.java
new file mode 100644
index 00000000..88916db0
--- /dev/null
+++ b/src/main/java/com/leetcode/hashtables/TwoSumIII.java
@@ -0,0 +1,78 @@
+package com.leetcode.hashtables;
+
+import java.util.HashMap;
+import java.util.Map;
+
+import static org.junit.jupiter.api.Assertions.assertTrue;
+
+/**
+ * Level: Easy
+ * Problem Link: https://leetcode.com/problems/two-sum-iii-data-structure-design/
+ * Problem Description:
+ * Design and implement a TwoSum class. It should support the following operations: add and find.
+ *
+ * add - Add the number to an internal data structure.
+ * find - Find if there exists any pair of numbers which sum is equal to the value.
+ *
+ * Example 1:
+ * add(1); add(3); add(5);
+ * find(4) -> true
+ * find(7) -> false
+ *
+ * Example 2:
+ * add(3); add(1); add(2);
+ * find(3) -> true
+ * find(6) -> false
+ *
+ * @author rampatra
+ * @since 2019-08-03
+ */
+public class TwoSumIII {
+
+ Map numCount;
+
+ /**
+ * Initialize your data structure here.
+ */
+ public TwoSumIII() {
+ this.numCount = new HashMap<>();
+ }
+
+ /**
+ * Add the number to an internal data structure..
+ */
+ public void add(int number) {
+ if (numCount.containsKey(number)) {
+ numCount.put(number, 2);
+ } else {
+ numCount.put(number, 1);
+ }
+ }
+
+ /**
+ * Find if there exists any pair of numbers which sum is equal to the value.
+ */
+ public boolean find(int value) {
+ for (Map.Entry entry : numCount.entrySet()) {
+ int num1 = entry.getKey();
+ int num2 = value - num1;
+ if ((num2 == num1 && entry.getValue() == 2) || (num1 != num2 && numCount.containsKey(num2))) {
+ return true;
+ }
+ }
+ return false;
+ }
+
+ /**
+ * Runtime: 115 ms.
+ *
+ * @param args
+ */
+ public static void main(String[] args) {
+ TwoSumIII twoSumIII = new TwoSumIII();
+ twoSumIII.add(0);
+ twoSumIII.add(-1);
+ twoSumIII.add(1);
+ assertTrue(twoSumIII.find(0));
+ }
+}
\ No newline at end of file
diff --git a/src/main/java/com/leetcode/hashtables/slidingwindow/LongestSubstringWithKDistinctCharacters.java b/src/main/java/com/leetcode/hashtables/slidingwindow/LongestSubstringWithKDistinctCharacters.java
new file mode 100644
index 00000000..4b209b6e
--- /dev/null
+++ b/src/main/java/com/leetcode/hashtables/slidingwindow/LongestSubstringWithKDistinctCharacters.java
@@ -0,0 +1,83 @@
+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/longest-substring-with-at-most-k-distinct-characters/
+ * Description:
+ * Given a string, find the length of the longest substring T that contains at most k distinct characters.
+ *
+ * Example 1:
+ * Input: s = "eceba", k = 2
+ * Output: 3
+ * Explanation: T is "ece" which its length is 3.
+ *
+ * Example 2:
+ * Input: s = "aa", k = 1
+ * Output: 2
+ * Explanation: T is "aa" which its length is 2.
+ *
+ * @author rampatra
+ * @since 2019-08-09
+ */
+public class LongestSubstringWithKDistinctCharacters {
+
+ /**
+ * Time Complexity: O(n)
+ * Space Complexity: O(k), as we keep at most k characters in the hash table
+ *
+ * @param str
+ * @param k
+ * @return
+ */
+ public static int lengthOfLongestSubstringKDistinct(String str, int k) {
+ int length = 0;
+ Map letterCountInWindow = new HashMap<>();
+
+ int left = 0; // start of window
+ int right = 0; // end of window
+
+ 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 has more than one of this character
+ // - lastly, move the window forward
+ if (letterCountInWindow.keySet().size() > k) {
+ 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, right - left);
+ left++;
+ }
+ right++;
+ }
+
+ 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(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/math/BestMeetingPoint.java b/src/main/java/com/leetcode/math/BestMeetingPoint.java
new file mode 100644
index 00000000..84d96fe9
--- /dev/null
+++ b/src/main/java/com/leetcode/math/BestMeetingPoint.java
@@ -0,0 +1,141 @@
+package com.leetcode.math;
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/**
+ * Level: Hard
+ * Link: https://leetcode.com/problems/best-meeting-point/
+ * Description:
+ * A group of two or more people wants to meet and minimize the total travel distance. You are given a 2D grid
+ * of values 0 or 1, where each 1 marks the home of someone in the group. The distance is calculated using
+ * Manhattan Distance, where distance(p1, p2) = |p2.x - p1.x| + |p2.y - p1.y|.
+ *
+ * Example:
+ *
+ * Input:
+ *
+ * 1 - 0 - 0 - 0 - 1
+ * | | | | |
+ * 0 - 0 - 0 - 0 - 0
+ * | | | | |
+ * 0 - 0 - 1 - 0 - 0
+ *
+ * Output: 6
+ *
+ * Explanation: Given three people living at (0,0), (0,4), and (2,2):
+ * The point (0,2) is an ideal meeting point, as the total travel distance
+ * of 2+2+2=6 is minimal. So, return 6.
+ *
+ * @author rampatra
+ * @since 2019-08-07
+ */
+public class BestMeetingPoint {
+
+ /**
+ * Time Complexity: O(k * i * j)
+ * Space Complexity: O(1)
+ * where,
+ * k = no of homes
+ * i = rows in grid
+ * j = columns in grid
+ *
+ * So, if i = j = k then you can see that it has a O(n^3) time complexity.
+ *
+ * @param grid
+ * @return
+ */
+ public static int minTotalDistanceBrutForce(int[][] grid) {
+ int minDistance = Integer.MAX_VALUE;
+ List> homeCoordinates = new ArrayList<>();
+
+ for (int i = 0; i < grid.length; i++) {
+ for (int j = 0; j < grid[0].length; j++) {
+ if (grid[i][j] == 1) {
+ homeCoordinates.add(Arrays.asList(i, j));
+ }
+ }
+ }
+
+ for (int i = 0; i < grid.length; i++) {
+ for (int j = 0; j < grid[0].length; j++) {
+ int distance = 0;
+ for (int k = 0; k < homeCoordinates.size(); k++) {
+ distance += Math.abs(homeCoordinates.get(k).get(0) - i) + Math.abs(homeCoordinates.get(k).get(1) - j);
+ }
+ minDistance = Math.min(minDistance, distance);
+ }
+ }
+
+ return minDistance;
+ }
+
+ public static int minTotalDistance(int[][] grid) {
+ return -1; // todo
+ }
+
+ public static void main(String[] args) {
+ assertEquals(6, minTotalDistanceBrutForce(new int[][]{
+ {1,0,0,0,1},
+ {0,0,0,0,0},
+ {0,0,1,0,0}
+ }));
+
+ assertEquals(4, minTotalDistanceBrutForce(new int[][]{
+ {1,0,0,0,1},
+ {0,0,0,0,0},
+ {0,0,0,0,0}
+ }));
+
+ assertEquals(1, minTotalDistanceBrutForce(new int[][]{
+ {1,1,0,0,0},
+ {0,0,0,0,0},
+ {0,0,0,0,0}
+ }));
+
+ assertEquals(0, minTotalDistanceBrutForce(new int[][]{
+ {1,0,0,0,0},
+ {0,0,0,0,0},
+ {0,0,0,0,0}
+ }));
+
+ assertEquals(0, minTotalDistanceBrutForce(new int[][]{
+ {0,0,0,0,0},
+ {0,0,0,0,0},
+ {0,0,0,0,0}
+ }));
+
+ assertEquals(6, minTotalDistance(new int[][]{
+ {1,0,0,0,1},
+ {0,0,0,0,0},
+ {0,0,1,0,0}
+ }));
+
+ assertEquals(4, minTotalDistance(new int[][]{
+ {1,0,0,0,1},
+ {0,0,0,0,0},
+ {0,0,0,0,0}
+ }));
+
+ assertEquals(1, minTotalDistance(new int[][]{
+ {1,1,0,0,0},
+ {0,0,0,0,0},
+ {0,0,0,0,0}
+ }));
+
+ assertEquals(0, minTotalDistance(new int[][]{
+ {1,0,0,0,0},
+ {0,0,0,0,0},
+ {0,0,0,0,0}
+ }));
+
+ assertEquals(0, minTotalDistance(new int[][]{
+ {0,0,0,0,0},
+ {0,0,0,0,0},
+ {0,0,0,0,0}
+ }));
+ }
+}
\ No newline at end of file
diff --git a/src/main/java/com/leetcode/recursion/FlattenNestListIterator.java b/src/main/java/com/leetcode/recursion/FlattenNestListIterator.java
new file mode 100644
index 00000000..b443e954
--- /dev/null
+++ b/src/main/java/com/leetcode/recursion/FlattenNestListIterator.java
@@ -0,0 +1,69 @@
+package com.leetcode.recursion;
+
+import java.util.ArrayList;
+import java.util.Iterator;
+import java.util.List;
+
+/**
+ * Level: Medium
+ * Link: https://leetcode.com/problems/flatten-nested-list-iterator/
+ * Description:
+ * Given a nested list of integers, implement an iterator to flatten it.
+ *
+ * Each element is either an integer, or a list -- whose elements may also be integers or other lists.
+ *
+ * Example 1:
+ * Input: [[1,1],2,[1,1]]
+ * Output: [1,1,2,1,1]
+ * Explanation: By calling next repeatedly until hasNext returns false,
+ * the order of elements returned by next should be: [1,1,2,1,1].
+ *
+ * Example 2:
+ * Input: [1,[4,[6]]]
+ * Output: [1,4,6]
+ * Explanation: By calling next repeatedly until hasNext returns false,
+ * the order of elements returned by next should be: [1,4,6].
+ *
+ * Runtime: 2 ms.
+ *
+ * @author rampatra
+ * @since 2019-08-12
+ */
+public class FlattenNestListIterator implements Iterator {
+
+ private int index;
+ private List flattenedList;
+
+ public FlattenNestListIterator(List nestedList) {
+ index = 0;
+ flattenedList = getFlattenedList(nestedList);
+ }
+
+ private List getFlattenedList(List nestedList) {
+ List flattenedList = new ArrayList<>();
+
+ for (NestedInteger nestedInteger : nestedList) {
+ if (nestedInteger.isInteger()) {
+ flattenedList.add(nestedInteger.getInteger());
+ } else {
+ flattenedList.addAll(getFlattenedList(nestedInteger.getList()));
+ }
+ }
+
+ return flattenedList;
+ }
+
+ @Override
+ public Integer next() {
+ return flattenedList.get(index++);
+ }
+
+ @Override
+ public boolean hasNext() {
+ return index < flattenedList.size();
+ }
+
+ public static void main(String[] args) {
+ // TODO add some test cases
+ }
+}
\ No newline at end of file
diff --git a/src/main/java/com/leetcode/recursion/NestedInteger.java b/src/main/java/com/leetcode/recursion/NestedInteger.java
new file mode 100644
index 00000000..1bba817e
--- /dev/null
+++ b/src/main/java/com/leetcode/recursion/NestedInteger.java
@@ -0,0 +1,46 @@
+package com.leetcode.recursion;
+
+import java.util.ArrayList;
+import java.util.List;
+
+/**
+ * Class needed for various problems like {@link NestedListWeightSumII}, {@link FlattenNestListIterator}, etc.
+ *
+ * @author rampatra
+ * @since 2019-08-12
+ */
+public class NestedInteger {
+
+ private Integer integer;
+ private List list;
+
+ public NestedInteger() {
+ this.list = new ArrayList<>();
+ }
+
+ public NestedInteger(int integer) {
+ this.integer = integer;
+ this.list = new ArrayList<>();
+ }
+
+ public boolean isInteger() {
+ return this.integer != null;
+ }
+
+ public Integer getInteger() {
+ return integer;
+ }
+
+ public void setInteger(Integer integer) {
+ this.integer = integer;
+ }
+
+ public List getList() {
+ return list;
+ }
+
+ public NestedInteger add(NestedInteger nestedInteger) {
+ this.list.add(nestedInteger);
+ return this;
+ }
+}
diff --git a/src/main/java/com/leetcode/recursion/NestedListWeightSum.java b/src/main/java/com/leetcode/recursion/NestedListWeightSum.java
new file mode 100644
index 00000000..1079b29f
--- /dev/null
+++ b/src/main/java/com/leetcode/recursion/NestedListWeightSum.java
@@ -0,0 +1,57 @@
+package com.leetcode.recursion;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/**
+ * Level: Easy
+ * Problem Link: https://leetcode.com/problems/nested-list-weight-sum/
+ * Problem Description:
+ * Given a nested list of integers, return the sum of all integers in the list weighted by their depth. Each element
+ * is either an integer, or a list – whose elements may also be integers or other lists.
+ *
+ * Example 1:
+ * Given the list [[1,1],2,[1,1]], return 10. (four 1’s at depth 2, one 2 at depth 1)
+ *
+ * Example 2:
+ * Given the list [1,[4,[6]]], return 27. (one 1 at depth 1, one 4 at depth 2, and one 6 at depth 3; 1 + 42 + 63 = 27)
+ *
+ * Note: For a more complex variant, see {@link NestedListWeightSumII}.
+ *
+ * @author rampatra
+ * @since 2019-07-27
+ */
+public class NestedListWeightSum {
+
+ /**
+ * Time Complexity: The algorithm takes O(N) time, where N is the total number of nested elements in the input
+ * list. For example, the list [ [[[[1]]]], 2 ] contains 4 nested lists and 2 nested integers (11 and 22), so N=6.
+ * Space Complexity: In terms of space, at most O(D) recursive calls are placed on the stack, where D is the
+ * maximum level of nesting in the input. For example, D=2 for the input [[1,1],2,[1,1]], and D=3 for the
+ * input [1,[4,[6]]].
+ *
+ * @param nestedList
+ * @return
+ */
+ public static long nestedSum(Object[] nestedList) {
+ return nestedSum(nestedList, 1);
+ }
+
+ private static long nestedSum(Object[] nestedList, int depth) {
+ long sum = 0;
+ for (int i = 0; i < nestedList.length; i++) {
+ if (nestedList[i] instanceof Integer) {
+ sum += ((int) nestedList[i] * depth);
+ } else {
+ sum += nestedSum((Object[]) nestedList[i], depth + 1);
+ }
+ }
+ return sum;
+ }
+
+ public static void main(String[] args) {
+ assertEquals(0, nestedSum(new Object[]{}));
+ assertEquals(0, nestedSum(new Object[]{new Object[]{}}));
+ assertEquals(10, nestedSum(new Object[]{new Object[]{1, 1}, 2, new Object[]{1, 1}}));
+ assertEquals(27, nestedSum(new Object[]{1, new Object[]{4, new Object[]{6}}}));
+ }
+}
\ No newline at end of file
diff --git a/src/main/java/com/leetcode/recursion/NestedListWeightSumII.java b/src/main/java/com/leetcode/recursion/NestedListWeightSumII.java
new file mode 100644
index 00000000..eadd121b
--- /dev/null
+++ b/src/main/java/com/leetcode/recursion/NestedListWeightSumII.java
@@ -0,0 +1,80 @@
+package com.leetcode.recursion;
+
+import java.util.*;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/**
+ * Level: Medium
+ * Problem Link: https://leetcode.com/problems/nested-list-weight-sum-ii/
+ * Problem Description:
+ * Given a nested list of integers, return the sum of all integers in the list weighted by their depth. Each element
+ * is either an integer, or a list – whose elements may also be integers or other lists.
+ *
+ * Different from {@link NestedListWeightSum} where weight is increasing from root to leaf, now the weight is defined
+ * from bottom up, i.e., the leaf level integers have weight 1, and the root level integers have the largest weight.
+ *
+ * Example 1:
+ * Given the list [[1,1],2,[1,1]], return 8. (four 1’s at depth 1, one 2 at depth 2)
+ *
+ * Example 2:
+ * Given the list [1,[4,[6]]], return 17. (one 1 at depth 3, one 4 at depth 2, and one 6 at depth 1; 13 + 42 + 6*1 = 17)
+ *
+ * Note: For a simpler variant, see {@link NestedListWeightSum}.
+ *
+ * @author rampatra
+ * @since 2019-07-27
+ */
+public class NestedListWeightSumII {
+
+ /**
+ * Time Complexity:
+ * Space Complexity:
+ * Runtime: 1 ms.
+ *
+ * @param nestedList
+ * @return
+ */
+ public static int nestedSum(List nestedList) {
+ int weightedSum = 0;
+ int unweightedSum = 0;
+
+ while (!nestedList.isEmpty()) {
+ List nextLevel = new ArrayList<>();
+
+ for (NestedInteger ni : nestedList) {
+ if (ni.isInteger()) {
+ unweightedSum += ni.getInteger();
+ } else {
+ nextLevel.addAll(ni.getList());
+ }
+ }
+
+ unweightedSum += unweightedSum; // multiplication by repetitive addition
+ weightedSum = unweightedSum;
+ nestedList = nextLevel;
+ }
+
+ return weightedSum;
+ }
+
+ public static void main(String[] args) {
+ assertEquals(0, nestedSum(Collections.singletonList(new NestedInteger())));
+
+ assertEquals(0, nestedSum(Collections.singletonList(new NestedInteger().add(new NestedInteger()))));
+
+ // TODO: fix the test cases
+ // {2, {1,1}, {1,1}}
+ NestedInteger ni = new NestedInteger(2);
+ ni.add(new NestedInteger().add(new NestedInteger(1)).add(new NestedInteger(1)));
+ ni.add(new NestedInteger().add(new NestedInteger(1)).add(new NestedInteger(1)));
+
+ assertEquals(6, nestedSum(Collections.singletonList(ni)));
+
+ // {1, {4, {6}}}
+ ni = new NestedInteger(1);
+ ni.add(new NestedInteger(4).add(new NestedInteger(6)));
+
+ assertEquals(17, nestedSum(Collections.singletonList(ni)));
+ }
+}
\ 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/stacks/ReversePolishNotation.java b/src/main/java/com/leetcode/stacks/ReversePolishNotation.java
new file mode 100644
index 00000000..f917099a
--- /dev/null
+++ b/src/main/java/com/leetcode/stacks/ReversePolishNotation.java
@@ -0,0 +1,93 @@
+package com.leetcode.stacks;
+
+import java.util.Stack;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/**
+ * Level: Medium
+ * Problem Link: https://leetcode.com/problems/evaluate-reverse-polish-notation
+ * Problem Description:
+ * Evaluate the value of an arithmetic expression in Reverse Polish Notation.
+ *
+ * Valid operators are +, -, *, /. Each operand may be an integer or another expression.
+ *
+ * Note:
+ * Division between two integers should truncate toward zero.
+ * The given RPN expression is always valid. That means the expression would always evaluate to a result and there
+ * won't be any divide by zero operation.
+ *
+ * Example 1:
+ * Input: ["2", "1", "+", "3", "*"]
+ * Output: 9
+ * Explanation: ((2 + 1) * 3) = 9
+ *
+ * Example 2:
+ * Input: ["4", "13", "5", "/", "+"]
+ * Output: 6
+ * Explanation: (4 + (13 / 5)) = 6
+ *
+ * Example 3:
+ * Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
+ * Output: 22
+ * Explanation:
+ * ((10 * (6 / ((9 + 3) * -11))) + 17) + 5
+ * = ((10 * (6 / (12 * -11))) + 17) + 5
+ * = ((10 * (6 / -132)) + 17) + 5
+ * = ((10 * 0) + 17) + 5
+ * = (0 + 17) + 5
+ * = 17 + 5
+ * = 22
+ *
+ * @author rampatra
+ * @since 2019-07-27
+ */
+public class ReversePolishNotation {
+
+ /**
+ * Time Complexity:
+ * Space Complexity:
+ * Runtime: 5 ms.
+ *
+ * @param tokens
+ * @return
+ */
+ public static int evalRPN(String[] tokens) {
+ int operand1;
+ int operand2;
+
+ Stack stack = new Stack<>();
+
+ for (String s : tokens) {
+ switch (s) {
+ case "+":
+ stack.push(stack.pop() + stack.pop());
+ break;
+ case "-":
+ operand1 = stack.pop();
+ operand2 = stack.pop();
+ stack.push(operand2 - operand1);
+ break;
+ case "*":
+ stack.push(stack.pop() * stack.pop());
+ break;
+ case "/":
+ operand1 = stack.pop();
+ operand2 = stack.pop();
+ stack.push(operand2 / operand1);
+ break;
+ default:
+ stack.push(Integer.parseInt(s));
+ }
+ }
+
+ return stack.pop();
+ }
+
+ public static void main(String[] args) {
+ assertEquals(18, evalRPN(new String[]{"18"}));
+ assertEquals(9, evalRPN(new String[]{"2", "1", "+", "3", "*"}));
+ assertEquals(6, evalRPN(new String[]{"4", "13", "5", "/", "+"}));
+ assertEquals(22, evalRPN(new String[]{"10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"}));
+ }
+}
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/BinaryTreeUpsideDown.java b/src/main/java/com/leetcode/trees/BinaryTreeUpsideDown.java
new file mode 100644
index 00000000..aa43bf50
--- /dev/null
+++ b/src/main/java/com/leetcode/trees/BinaryTreeUpsideDown.java
@@ -0,0 +1,207 @@
+package com.leetcode.trees;
+
+import java.util.Stack;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+import static org.junit.jupiter.api.Assertions.assertNull;
+
+/**
+ * Level: Medium
+ * Problem Link: https://leetcode.com/problems/binary-tree-upside-down/
+ * Problem Description:
+ * Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the
+ * same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into
+ * left leaf nodes. Return the new root.
+ *
+ * Example:
+ * Input: [1,2,3,4,5]
+ *
+ * 1
+ * / \
+ * 2 3
+ * / \
+ * 4 5
+ *
+ * Output: return the root of the binary tree [4,5,2,#,#,3,1]
+ *
+ * 4
+ * / \
+ * 5 2
+ * / \
+ * 3 1
+ *
+ * Clarification:
+ * Confused what [4,5,2,#,#,3,1] means? Read more below on how binary tree is serialized on OJ. The serialization of
+ * a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
+ *
+ * Here's an example:
+ *
+ * 1
+ * / \
+ * 2 3
+ * /
+ * 4
+ * \
+ * 5
+ *
+ * The above binary tree is serialized as [1,2,3,#,#,4,#,#,5].
+ *
+ * @author rampatra
+ * @since 2019-08-04
+ */
+public class BinaryTreeUpsideDown {
+
+ /**
+ * The solution is simple, every node (except the root) on the left of the tree would have its parent's right child
+ * as it's left child and parent as its right child. That's all you have to do to flip the tree upside down.
+ *
+ * Time Complexity: O(h)
+ * Space Complexity: O(h)
+ * where,
+ * h = height of the tree
+ *
+ * Runtime: 1 ms.
+ *
+ * @param root
+ * @return
+ */
+ public static TreeNode upsideDownBinaryTreeUsingStack(TreeNode root) {
+ if (root == null) return null;
+
+ TreeNode curr = root;
+ TreeNode currParent;
+ TreeNode newRoot = null;
+
+ // using stack to keep track of the parent node
+ Stack stack = new Stack<>();
+
+ while (curr != null) {
+ stack.add(curr);
+ curr = curr.left;
+ }
+
+ while (!stack.empty()) {
+ curr = stack.pop();
+ currParent = stack.empty() ? null : stack.peek();
+
+ if (newRoot == null) newRoot = curr;
+
+ if (currParent != null) {
+ curr.left = currParent.right;
+ curr.right = currParent;
+ } else {
+ curr.left = null;
+ curr.right = null;
+ }
+ }
+
+ return newRoot;
+ }
+
+ /**
+ * The solution is simple, every node (except the root) on the left of the tree would have its parent's right child
+ * as it's left child and parent as its right child. That's all you have to do to flip the tree upside down.
+ *
+ * Time Complexity: O(h)
+ * Space Complexity: O(h)
+ * where,
+ * h = height of the tree
+ *
+ * Runtime: 0 ms.
+ *
+ * @param node
+ * @return
+ */
+ public static TreeNode upsideDownBinaryTree(TreeNode node) {
+ if (node == null || node.left == null) return node;
+
+ // go to the last node on the extreme left branch
+ TreeNode newRoot = upsideDownBinaryTree(node.left);
+
+ // do the node changes as you backtrack
+ node.left.left = node.right;
+ node.left.right = node;
+
+ // clean up
+ node.left = null;
+ node.right = null;
+
+ return newRoot;
+ }
+
+ public static void main(String[] args) {
+ /*
+ 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);
+
+ /*
+ Upside Down Binary Tree
+
+ 4
+ / \
+ 5 2
+ / \
+ 3 1
+ */
+ TreeNode upsideDownTree = upsideDownBinaryTreeUsingStack(tree);
+ assertEquals(4, upsideDownTree.val);
+ assertEquals(5, upsideDownTree.left.val);
+ assertEquals(2, upsideDownTree.right.val);
+ assertEquals(1, upsideDownTree.right.right.val);
+ assertEquals(3, upsideDownTree.right.left.val);
+ assertNull(upsideDownTree.right.right.left);
+ assertNull(upsideDownTree.right.right.right);
+
+
+
+ /******************************
+ *
+ * Test for the recursive method
+ *
+ ******************************/
+
+ /*
+ Binary Tree
+
+ 1
+ / \
+ 2 3
+ / \
+ 4 5
+ */
+ 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);
+
+ /*
+ Upside Down Binary Tree
+
+ 4
+ / \
+ 5 2
+ / \
+ 3 1
+ */
+ upsideDownTree = upsideDownBinaryTree(tree);
+ assertEquals(4, upsideDownTree.val);
+ assertEquals(5, upsideDownTree.left.val);
+ assertEquals(2, upsideDownTree.right.val);
+ assertEquals(1, upsideDownTree.right.right.val);
+ assertEquals(3, upsideDownTree.right.left.val);
+ assertNull(upsideDownTree.right.right.right);
+ assertNull(upsideDownTree.right.right.left);
+ }
+}
\ No newline at end of file
diff --git a/src/main/java/com/leetcode/trees/BinaryTreeZigZagLevelOrderTraversal.java b/src/main/java/com/leetcode/trees/BinaryTreeZigZagLevelOrderTraversal.java
new file mode 100644
index 00000000..1a54d952
--- /dev/null
+++ b/src/main/java/com/leetcode/trees/BinaryTreeZigZagLevelOrderTraversal.java
@@ -0,0 +1,100 @@
+package com.leetcode.trees;
+
+import java.util.*;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/**
+ * Level: Medium
+ * Link: https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/
+ * Description:
+ * Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then
+ * right to left for the next level and alternate between).
+ *
+ * For example:
+ * Given binary tree [3,9,20,null,null,15,7],
+ * 3
+ * / \
+ * 9 20
+ * / \
+ * 15 7
+ * return its zigzag level order traversal as:
+ * [
+ * [3],
+ * [20,9],
+ * [15,7]
+ * ]
+ *
+ * @author rampatra
+ * @since 2019-08-11
+ */
+public class BinaryTreeZigZagLevelOrderTraversal {
+
+ /**
+ * Time Complexity:
+ * Space Complexity:
+ * Runtime: 1 ms.
+ *
+ * @param root
+ * @return
+ */
+ public static List> zigzagLevelOrder(TreeNode root) {
+
+ int levelNo = 0;
+ LinkedList currLevel = new LinkedList<>();
+ List> levelOrderTraversal = new LinkedList<>();
+
+ if (root == null) {
+ return levelOrderTraversal;
+ }
+
+ Queue queue = new LinkedList<>();
+ queue.add(root);
+ queue.add(null);
+
+ while (!queue.isEmpty()) {
+
+ TreeNode treeNode = queue.poll();
+
+ if (treeNode == null) {
+ levelOrderTraversal.add(currLevel);
+ currLevel = new LinkedList<>();
+ levelNo++;
+
+ if (queue.size() > 0) {
+ queue.add(null);
+ }
+ } else {
+ if (levelNo % 2 == 0) {
+ currLevel.add(treeNode.val);
+ } else {
+ currLevel.add(0, treeNode.val);
+ }
+ if (treeNode.left != null) queue.add(treeNode.left);
+ if (treeNode.right != null) queue.add(treeNode.right);
+ }
+ }
+
+ return levelOrderTraversal;
+ }
+
+ public static void main(String[] args) {
+ /*
+ 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);
+
+ assertEquals("[[1], [3, 2], [4, 5]]", zigzagLevelOrder(tree).toString());
+ assertEquals("[]", zigzagLevelOrder(null).toString());
+ }
+}
diff --git a/src/main/java/com/leetcode/trees/ClosestBinarySearchTreeValue.java b/src/main/java/com/leetcode/trees/ClosestBinarySearchTreeValue.java
new file mode 100644
index 00000000..ca6d94b7
--- /dev/null
+++ b/src/main/java/com/leetcode/trees/ClosestBinarySearchTreeValue.java
@@ -0,0 +1,110 @@
+package com.leetcode.trees;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/**
+ * Level: Easy
+ * Problem Link: https://leetcode.com/problems/closest-binary-search-tree-value/
+ * Problem Description:
+ * Given a non-empty binary search tree and a target value, find the value in the BST that is closest to the target.
+ *
+ * Note:
+ * - Given target value is a floating point.
+ * - You are guaranteed to have only one unique value in the BST that is closest to the target.
+ *
+ * @author rampatra
+ * @since 2019-07-31
+ */
+public class ClosestBinarySearchTreeValue {
+
+ /**
+ * Runtime: 0 ms.
+ *
+ * @param root
+ * @param target
+ * @return
+ */
+ public static int closestValue(TreeNode root, double target) {
+ if (root == null) return -1;
+
+ return closestValue(root, root, target);
+ }
+
+ private static int closestValue(TreeNode node, TreeNode closestNode, double val) {
+ if (node == null) return closestNode.val;
+
+ if (Math.abs(node.val - val) < Math.abs(closestNode.val - val)) {
+ closestNode = node;
+ }
+
+ if (node.val > val) {
+ return closestValue(node.left, closestNode, val);
+ } else {
+ return closestValue(node.right, closestNode, val);
+ }
+ }
+
+ public static void main(String[] args) {
+
+ /*
+ BST looks like:
+
+ 9
+ / \
+ 7 13
+ / \ \
+ 5 8 20
+ / \
+ 2 6
+ */
+ TreeNode root = new TreeNode(9);
+ root.left = new TreeNode(7);
+ root.right = new TreeNode(13);
+ root.left.left = new TreeNode(5);
+ root.left.right = new TreeNode(8);
+ root.left.left.right = new TreeNode(6);
+ root.left.left.left = new TreeNode(2);
+ root.right.right = new TreeNode(20);
+
+ assertEquals(13, closestValue(root, 15));
+ assertEquals(13, closestValue(root, 13));
+ assertEquals(9, closestValue(root, 9));
+ assertEquals(2, closestValue(root, 2));
+ assertEquals(2, closestValue(root, 1));
+ assertEquals(6, closestValue(root, 6));
+ assertEquals(13, closestValue(root, 11)); // tie b/w 9 and 13
+
+ /*
+ BST looks like:
+
+ 9
+ / \
+ 7 13
+ / \ / \
+ 5 8 13 20
+ */
+ root = new TreeNode(9);
+ root.left = new TreeNode(7);
+ root.right = new TreeNode(13);
+ root.left.left = new TreeNode(5);
+ root.left.right = new TreeNode(8);
+ root.right.left = new TreeNode(13);
+ root.right.right = new TreeNode(20);
+
+ assertEquals(13, closestValue(root, 15));
+
+ /*
+ BST looks like:
+
+ 1500000000
+ /
+ /
+ /
+ 1400000000
+ */
+ root = new TreeNode(1500000000);
+ root.left = new TreeNode(1400000000);
+
+ assertEquals(1400000000, closestValue(root, -1500000000.0));
+ }
+}
diff --git a/src/main/java/com/leetcode/trees/ClosestBinarySearchTreeValueII.java b/src/main/java/com/leetcode/trees/ClosestBinarySearchTreeValueII.java
new file mode 100644
index 00000000..e583723a
--- /dev/null
+++ b/src/main/java/com/leetcode/trees/ClosestBinarySearchTreeValueII.java
@@ -0,0 +1,166 @@
+package com.leetcode.trees;
+
+import java.util.LinkedList;
+import java.util.List;
+import java.util.Queue;
+import java.util.Stack;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/**
+ * Level: Hard
+ * Problem Link: https://leetcode.com/problems/closest-binary-search-tree-value-ii/
+ * Problem Description:
+ * Given a non-empty binary search tree and a target value, find k values in the BST that are closest to the target.
+ *
+ * Note:
+ * - Given target value is a floating point.
+ * - You may assume k is always valid, that is: k ≤ total nodes.
+ * - You are guaranteed to have only one unique set of k values in the BST that are closest to the target.
+ *
+ * Example:
+ * Input: root = [4,2,5,1,3], target = 3.714286, and k = 2
+ *
+ * 4
+ * / \
+ * 2 5
+ * / \
+ * 1 3
+ *
+ * Output: [4,3]
+ *
+ * Follow up:
+ * Assume that the BST is balanced, could you solve it in less than O(n) runtime (where n = total nodes)?
+ *
+ * @author rampatra
+ * @since 2019-07-31
+ */
+public class ClosestBinarySearchTreeValueII {
+
+
+ /**
+ * The idea is simple. We do the inorder traversal and keep the values less than or equal to target in a stack and
+ * the values greater than target in a queue. And finally, we compare values from both stack and queue and take
+ * whichever is the closest to target value each time.
+ *
+ * Note: We can optimize it even further in terms of space. We can get rid of the stack and queue and just fill up
+ * the result list in the recursive inOrder call. Once the result list is of size k, we can compare and remove the
+ * farthest value and insert the closer value. See {@link ClosestBinarySearchTreeValueII#closestKValuesOptimized(TreeNode, double, int)}.
+ *
+ * @param root
+ * @param target
+ * @param k
+ * @return
+ */
+ public static List closestKValues(TreeNode root, double target, int k) {
+ int count = 0;
+ List closestKValues = new LinkedList<>();
+
+ Stack predecessors = new Stack<>();
+ Queue successors = new LinkedList<>();
+ inOrder(root, predecessors, successors, target, k);
+
+ while (count < k) {
+ if (predecessors.empty()) {
+ closestKValues.add(successors.poll());
+ } else if (successors.isEmpty()) {
+ closestKValues.add(predecessors.pop());
+ } else if (Math.abs(target - predecessors.peek()) < Math.abs(target - successors.peek())) {
+ closestKValues.add(predecessors.pop());
+ } else {
+ closestKValues.add(successors.poll());
+ }
+ count++;
+ }
+
+ return closestKValues;
+ }
+
+ private static void inOrder(TreeNode root, Stack predecessors, Queue successors, double target, int k) {
+ if (root == null || successors.size() == k) return;
+ inOrder(root.left, predecessors, successors, target, k);
+ if (root.val <= target) {
+ predecessors.add(root.val);
+ } else {
+ successors.add(root.val);
+ }
+ inOrder(root.right, predecessors, successors, target, k);
+ }
+
+
+ /**
+ * This approach is similar to the above one but it doesn't use stack or queue.
+ *
+ * @param root
+ * @param target
+ * @param k
+ * @return
+ */
+ public static List closestKValuesOptimized(TreeNode root, double target, int k) {
+ LinkedList closestKValues = new LinkedList<>();
+ inOrder(root, target, k, closestKValues);
+ return closestKValues;
+ }
+
+ private static void inOrder(TreeNode root, double target, int k, LinkedList closestKValues) {
+ if (root == null) return;
+
+ inOrder(root.left, target, k, closestKValues);
+ if (closestKValues.size() == k) {
+ //if size k, add current and remove head if it's closer to target, otherwise return
+ if (Math.abs(target - root.val) < Math.abs(target - closestKValues.peekFirst()))
+ closestKValues.removeFirst();
+ else {
+ return;
+ }
+ }
+ closestKValues.add(root.val);
+ inOrder(root.right, target, k, closestKValues);
+ }
+
+ public static void main(String[] args) {
+
+ /*
+ BST looks like:
+
+ 9
+ / \
+ 7 13
+ / \ \
+ 5 8 20
+ / \
+ 2 6
+ */
+ TreeNode root = new TreeNode(9);
+ root.left = new TreeNode(7);
+ root.right = new TreeNode(13);
+ root.left.left = new TreeNode(5);
+ root.left.right = new TreeNode(8);
+ root.left.left.left = new TreeNode(2);
+ root.left.left.right = new TreeNode(6);
+ root.right.right = new TreeNode(20);
+
+ assertEquals("[9, 8, 7, 6, 5]", closestKValues(root, 8.5, 5).toString());
+ assertEquals("[5, 6, 7, 8, 9]", closestKValuesOptimized(root, 8.5, 5).toString());
+
+ /*
+ BST looks like:
+
+ 9
+ / \
+ 7 13
+ / \ / \
+ 5 8 13 20
+ */
+ root = new TreeNode(9);
+ root.left = new TreeNode(7);
+ root.right = new TreeNode(13);
+ root.left.left = new TreeNode(5);
+ root.right.left = new TreeNode(13);
+ root.left.right = new TreeNode(8);
+ root.right.right = new TreeNode(20);
+
+ assertEquals("[13, 13, 9, 20, 8]", closestKValues(root, 14, 5).toString());
+ assertEquals("[8, 9, 13, 13, 20]", closestKValuesOptimized(root, 14, 5).toString());
+ }
+}
diff --git a/src/main/java/com/leetcode/trees/LeavesOfBinaryTree.java b/src/main/java/com/leetcode/trees/LeavesOfBinaryTree.java
new file mode 100644
index 00000000..6ca15e22
--- /dev/null
+++ b/src/main/java/com/leetcode/trees/LeavesOfBinaryTree.java
@@ -0,0 +1,97 @@
+package com.leetcode.trees;
+
+import java.util.ArrayList;
+import java.util.List;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/**
+ * Level: Medium
+ * Problem Link: https://leetcode.com/problems/find-leaves-of-binary-tree/
+ * Problem Description:
+ * Given a binary tree, collect a tree's nodes as if you were doing this: Collect and remove all leaves, repeat
+ * until the tree is empty.
+ *
+ * Example:
+ * Input: [1,2,3,4,5]
+ *
+ * 1
+ * / \
+ * 2 3
+ * / \
+ * 4 5
+ *
+ * Output: [[4,5,3],[2],[1]]
+ *
+ * Explanation:
+ * 1. Removing the leaves [4,5,3] would result in this tree:
+ * 1
+ * /
+ * 2
+ *
+ * 2. Now removing the leaf [2] would result in this tree:
+ * 1
+ *
+ * 3. Now removing the leaf [1] would result in the empty tree:
+ * []
+ *
+ * @author rampatra
+ * @since 2019-08-01
+ */
+public class LeavesOfBinaryTree {
+
+ /**
+ * THe idea is to perform a DFS and backtrack. While backtracking, check the height of the node and insert
+ * the node into the list indexed by their heights.
+ * Time Complexity:
+ * Space Complexity:
+ * Runtime: 1 ms.
+ *
+ * @param root
+ * @return
+ */
+ public static List> findLeavesOfBinaryTree(TreeNode root) {
+ List> levels = new ArrayList<>();
+ findLeavesOfBinaryTree(root, levels);
+ return levels;
+ }
+
+ private static int findLeavesOfBinaryTree(TreeNode root, List> levels) {
+ if (root == null) return -1;
+
+ int leftHeight = findLeavesOfBinaryTree(root.left, levels);
+ int rightHeight = findLeavesOfBinaryTree(root.right, levels);
+ int height = Math.max(leftHeight, rightHeight) + 1;
+
+ if (height >= levels.size()) {
+ levels.add(height, new ArrayList<>());
+ }
+ levels.get(height).add(root.val);
+
+ return height;
+ }
+
+ public static void main(String[] args) {
+ /*
+ BST looks like:
+
+ 4
+ / \
+ 1 7
+ / \ \
+ 3 8 20
+ / \
+ 2 6
+ */
+ TreeNode root = new TreeNode(4);
+ root.left = new TreeNode(1);
+ root.right = new TreeNode(7);
+ root.left.left = new TreeNode(3);
+ root.left.right = new TreeNode(8);
+ root.left.left.left = new TreeNode(2);
+ root.left.left.right = new TreeNode(6);
+ root.right.right = new TreeNode(20);
+
+ assertEquals("[[2, 6, 8, 20], [3, 7], [1], [4]]", findLeavesOfBinaryTree(root).toString());
+ }
+}
\ No newline at end of file
diff --git a/src/main/java/com/leetcode/trees/SecondMinNodeInBinaryTree.java b/src/main/java/com/leetcode/trees/SecondMinNodeInBinaryTree.java
new file mode 100644
index 00000000..a8cddf74
--- /dev/null
+++ b/src/main/java/com/leetcode/trees/SecondMinNodeInBinaryTree.java
@@ -0,0 +1,108 @@
+package com.leetcode.trees;
+
+import java.util.Stack;
+
+/**
+ * Level: Easy
+ * Problem Link: https://leetcode.com/problems/second-minimum-node-in-a-binary-tree/
+ * Problem Description:
+ * Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this
+ * tree has exactly two or zero sub-node. If the node has two sub-nodes, then this node's value is the smaller value
+ * among its two sub-nodes. More formally, the property root.val = min(root.left.val, root.right.val) always holds.
+ *
+ * Given such a binary tree, you need to output the second minimum value in the set made of all the nodes' value in
+ * the whole tree.
+ *
+ * If no such second minimum value exists, output -1 instead.
+ *
+ * Example 1:
+ * Input:
+ * 2
+ * / \
+ * 2 5
+ * / \
+ * 5 7
+ *
+ * Output: 5
+ * Explanation: The smallest value is 2, the second smallest value is 5.
+ *
+ *
+ * Example 2:
+ * Input:
+ * 2
+ * / \
+ * 2 2
+ *
+ * Output: -1
+ * Explanation: The smallest value is 2, but there isn't any second smallest value.
+ *
+ * @author rampatra
+ * @since 2019-08-03
+ */
+public class SecondMinNodeInBinaryTree {
+
+ /**
+ * Time Complexity: O(n)
+ * Space Complexity: O(n)
+ * Runtime: 1 ms.
+ * @param root
+ * @return
+ */
+ public static int findSecondMinimumValueIterative(TreeNode root) {
+ if (root == null || (root.left == null && root.right == null)) return -1;
+
+ int min = root.val;
+ long secondMin = Long.MAX_VALUE;
+
+ Stack stack = new Stack<>();
+ stack.push(root);
+
+ while (!stack.empty()) {
+ TreeNode node = stack.pop();
+ if (node == null) continue;
+
+ if (node.val > min && node.val < secondMin) {
+ secondMin = node.val;
+ }
+ stack.push(node.left);
+ stack.push(node.right);
+ }
+
+ return secondMin == Long.MAX_VALUE ? -1 : (int) secondMin;
+ }
+
+
+ /**
+ * Time Complexity:
+ * Space Complexity:
+ * Runtime: 0 ms.
+ *
+ * @param root
+ * @return
+ */
+ public static int findSecondMinimumValue(TreeNode root) {
+ // passing a long as secondMin because TreeNode can have Integer.MAX_VALUE as its value
+ long ans = findSecondMinimumValue(root, root.val, Long.MAX_VALUE);
+ return ans == Long.MAX_VALUE ? -1 : (int) ans;
+ }
+
+ private static long findSecondMinimumValue(TreeNode root, int min, long secondMin) {
+ if (root == null) return Long.MAX_VALUE;
+
+ if (root.val > min && root.val < secondMin) {
+ return root.val;
+ } else {
+ return Math.min(findSecondMinimumValue(root.left, min, secondMin),
+ findSecondMinimumValue(root.right, min, secondMin));
+ }
+ }
+
+ public static void main(String[] args) {
+ System.out.println((int) 2147483647L);
+ System.out.println(Integer.MAX_VALUE);
+ // TODO: A function called buildTree which would take an array like [1,1,3,1,1,3,4,3,1,1,1,3,8,4,8,3,3,1,6,2,1]
+ // and return a Binary Tree
+ //assertEquals(2, findSecondMinimumValue(buildTree(new int[]{1,1,3,1,1,3,4,3,1,1,1,3,8,4,8,3,3,1,6,2,1})));
+ //assertEquals(2147483647, findSecondMinimumValue(buildTree(new int[]{2,2,2147483647})));
+ }
+}
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/leetcode/trees/SymmetricTree.java b/src/main/java/com/leetcode/trees/SymmetricTree.java
new file mode 100644
index 00000000..093d9a6f
--- /dev/null
+++ b/src/main/java/com/leetcode/trees/SymmetricTree.java
@@ -0,0 +1,117 @@
+package com.leetcode.trees;
+
+import java.util.LinkedList;
+import java.util.Queue;
+
+import static org.junit.jupiter.api.Assertions.assertTrue;
+
+/**
+ * Level: Easy
+ * Problem Link: https://leetcode.com/problems/symmetric-tree/
+ * Problem Description:
+ * Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
+ *
+ * For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
+ *
+ * 1
+ * / \
+ * 2 2
+ * / \ / \
+ * 3 4 4 3
+ *
+ *
+ * But the following [1,2,2,null,3,null,3] is not:
+ *
+ * 1
+ * / \
+ * 2 2
+ * \ \
+ * 3 3
+ *
+ *
+ * Note:
+ * Bonus points if you could solve it both recursively and iteratively.
+ *
+ * @author rampatra
+ * @since 2019-07-25
+ */
+public class SymmetricTree {
+
+ /**
+ * Time Complexity: O(n) Because we traverse the entire input tree once, the total run time is O(n), where n is
+ * the total number of nodes in the tree.
+ * Space Complexity: O(n) The number of recursive calls is bound by the height of the tree. In the worst case, the
+ * tree is linear and the height is in O(n). Therefore, space complexity due to recursive calls on the stack is
+ * O(n) in the worst case.
+ * Runtime: 0 ms.
+ *
+ * @param root
+ * @return
+ */
+ public static boolean isSymmetric(TreeNode root) {
+ if (root == null) {
+ return true;
+ }
+
+ return isSymmetric(root.left, root.right);
+ }
+
+ private static boolean isSymmetric(TreeNode leftRoot, TreeNode rightRoot) {
+ if (leftRoot == null && rightRoot == null) {
+ return true;
+ } else if (leftRoot == null || rightRoot == null) {
+ return false;
+ }
+
+ return isSymmetric(leftRoot.left, rightRoot.right) && isSymmetric(leftRoot.right, rightRoot.left) && leftRoot.val == rightRoot.val;
+ }
+
+ /**
+ * Time Complexity: O(n) Because we traverse the entire input tree once, the total run time is O(n), where n is the
+ * total number of nodes in the tree.
+ * Space Complexity: There is additional space required for the search queue. In the worst case, we have to
+ * insert O(n) nodes in the queue. Therefore, space complexity is O(n).
+ * Runtime: 1 ms.
+ *
+ * @param root
+ * @return
+ */
+ public static boolean isSymmetricIterative(TreeNode root) {
+ if (root == null || (root.left == null && root.right == null)) return true;
+ if (root.left == null || root.right == null) return false;
+
+ Queue queue = new LinkedList<>();
+ queue.add(root.left);
+ queue.add(root.right);
+
+ while (!queue.isEmpty()) {
+ TreeNode t1 = queue.poll();
+ TreeNode t2 = queue.poll();
+
+ if (t1 == null && t2 == null) continue;
+ if (t1 == null || t2 == null) return false;
+ if (t1.val != t2.val) return false;
+
+ // enqueue left and then right child of t1 but do the opposite for t2
+ queue.add(t1.left);
+ queue.add(t2.right);
+ queue.add(t1.right);
+ queue.add(t2.left);
+ }
+
+ return true;
+ }
+
+ public static void main(String[] args) {
+ TreeNode root = new TreeNode(1);
+ root.left = new TreeNode(2);
+ root.right = new TreeNode(2);
+ root.left.left = new TreeNode(4);
+ root.left.right = new TreeNode(3);
+ root.right.left = new TreeNode(3);
+ root.right.right = new TreeNode(4);
+
+ assertTrue(isSymmetric(root));
+ assertTrue(isSymmetricIterative(root));
+ }
+}
diff --git a/src/main/java/com/leetcode/trees/TreeNode.java b/src/main/java/com/leetcode/trees/TreeNode.java
new file mode 100644
index 00000000..4c4c2569
--- /dev/null
+++ b/src/main/java/com/leetcode/trees/TreeNode.java
@@ -0,0 +1,21 @@
+package com.leetcode.trees;
+
+/**
+ * @author rampatra
+ * @since 2019-07-25
+ */
+public class TreeNode {
+
+ int val;
+ TreeNode left;
+ TreeNode right;
+
+ public TreeNode(int val) {
+ this.val = val;
+ }
+
+ @Override
+ public String toString() {
+ return val + "";
+ }
+}
\ 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();
}
}