Medium

You are given a list of songs where the ith song has a duration of time[i] seconds. Return the number of pairs of songs for which their total duration in seconds is divisible by 60. Formally, we want the number of indices i, j such that i < j with (time[i] + time[j]) % 60 == 0. Example 1: Input: time = [30,20,150,100,40] Output: 3 Explanation: Three pairs have a total duration divisible by 60: (time[0] = 30, time[2] = 150): total duration 180 (time[1] = 20, time[3] = 100): total duration 120 (time[1] = 20, time[4] = 40): total duration 60 Example 2: Input: time = [60,60,60] Output: 3 Explanation: All three pairs have a total duration of 120, which is divisible by 60. Constraints: ...

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree. According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).” Example 1: Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1 Output: 3 Explanation: The LCA of nodes 5 and 1 is 3. ...

An a N x N grid composed of 1 x 1 squares, each 1 x 1 square consists of a /, , or blank space. These characters divide the square into contiguous regions. (Note that backslash characters are escaped, so a \ is represented as “\”.) Return the number of regions. Example 1: Input: [ " /", "/ " ] Output: 2 Explanation: The 2x2 grid is as follows: ...

Given a string s, find the length of the longest substring without repeating characters. Example 1: Input: s = "abcabcbb" Output: 3 Explanation: The answer is "abc", with the length of 3. Example 2: Input: s = "bbbbb" Output: 1 Explanation: The answer is "b", with the length of 1. Example 3: Input: s = "pwwkew" Output: 3 Explanation: The answer is "wke", with the length of 3. Notice that the answer must be a substring, "pwke" is a subsequence and not a substring. Example 4: ...

Implement pow(x, n), which calculates x raised to the power n (i.e. xn). Example 1: Input: x = 2.00000, n = 10 Output: 1024.00000 Example 2: Input: x = 2.10000, n = 3 Output: 9.26100 Example 3: Input: x = 2.00000, n = -2 Output: 0.25000 Explanation: 2-2 = 1/22 = 1/4 = 0.25 Constraints: -100.0 < x < 100.0 -231 <= n <= 231-1 -104 <= xn <= 104 ...

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] ] Solution: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public List<List<Integer>> zigzagLevelOrder(TreeNode root) { List<List<Integer>> result = new ArrayList<>(); dfs(root, result, 0); return result; } void dfs(TreeNode node, List<List<Integer>> result, int level) { if (node == null) { return; } checkSize(result, level); if (level % 2 == 1) { result.get(level).add(0, node.val); } else { result.get(level).add(node.val); } dfs(node.left, result, level + 1); dfs(node.right, result, level + 1); } void checkSize(List<List<Integer>> result, int level) { // 0 [ [] ] // 1 [[][]] // 2 [ [][] ] if (result.size() <= level) { result.add(new ArrayList<>()); } } } Solution 2021-11-18 DFS, DFS + Stack, BFS /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public List<List<Integer>> zigzagLevelOrder(TreeNode root) { List<List<Integer>> res = new ArrayList<>(); Queue<TreeNode> q = new ArrayDeque<>(); if (root != null) { q.add(root); } int level = 0; while (q.size() > 0) { int size = q.size(); List<Integer> levelList = new ArrayList(); for (int i = 0; i < size; i++) { TreeNode node = q.poll(); if (level % 2 == 0) { levelList.add(node.val); } else { levelList.add(0, node.val); } if (node.left != null) { q.add(node.left); } if (node.right != null) { q.add(node.right); } } res.add(levelList); level++; } return res; } public List<List<Integer>> zigzagLevelOrderDfsStack(TreeNode root) { List<List<Integer>> res = new ArrayList<>(); Stack<Pair<TreeNode, Integer>> stack = new Stack<>(); if (root != null) { stack.push(new Pair<>(root, 0)); } while (stack.size() > 0) { Pair<TreeNode, Integer> curr = stack.pop(); TreeNode node = curr.getKey(); Integer level = curr.getValue(); if (res.size() == level) { res.add(new ArrayList<>()); } if (level % 2 == 0) { res.get(level).add(node.val); } else { res.get(level).add(0, node.val); } if (node.right != null) { stack.push(new Pair<>(node.right, level + 1)); } if (node.left != null) { stack.push(new Pair<>(node.left, level + 1)); } } return res; } public List<List<Integer>> zigzagLevelOrderDfsRecursive(TreeNode root) { List<List<Integer>> res = new ArrayList<>(); dfs(root, 0, res); return res; } void dfs(TreeNode node, int level, List<List<Integer>> res) { if (node == null) return; if (res.size() == level) { res.add(new ArrayList<>()); } if (level % 2 == 0) { res.get(level).add(node.val); } else { res.get(level).add(0, node.val); } dfs(node.left, level + 1, res); dfs(node.right, level + 1, res); } }

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. It’s guaranteed that the answer is unique, in other words the set of the top k frequent elements is unique. You can return the answer in any order. Solution: ...

A company has n employees with a unique ID for each employee from 0 to n - 1. The head of the company has is the one with headID. Each employee has one direct manager given in the manager array where manager[i] is the direct manager of the i-th employee, manager[headID] = -1. Also it’s guaranteed that the subordination relationships have a tree structure. The head of the company wants to inform all the employees of the company of an urgent piece of news. He will inform his direct subordinates and they will inform their subordinates and so on until all employees know about the urgent news. ...

For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees. A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations. Given the roots of two binary trees root1 and root2, return true if the two trees are flip equivelent or false otherwise. ...

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below. For example, given the following triangle [ [2], [3,4], [6,5,7], [4,1,8,3] ] The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11). Note: Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle. ...