E
EduScale
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Given the root of a binary tree, return the length of the diameter of the tree.
The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.
The length of a path between two nodes is represented by the number of edges between them.
Example 1:
Input: root = [1,2,3,4,5]
Output: 3
Explanation: 3 is the length of the path [4,2,1,3] or [5,2,1,3].
Example 2:
Input: root = [1,2]
Output: 1
[1, 10^4].-100 <= Node.val <= 100Given an m x n matrix, return all elements of the matrix in spiral order.
Example 1:
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]
Example 2:
Input: matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]
m == matrix.lengthn == matrix[i].length1 <= m, n <= 10-100 <= matrix[i][j] <= 100Write a solution to find the rank of the scores. The ranking should be calculated according to the following rules:
Return the result table ordered by score in descending order.
| Column Name | Type | | :--- | :--- | | id | int | | score | decimal |
id is the primary key for this table.
Each row of this table contains the score of a game. Score is a fractional number with two decimal places.
Input: Scores table: | id | score | | :--- | :--- | | 1 | 3.50 | | 2 | 3.65 | | 3 | 4.00 | | 4 | 3.85 | | 5 | 4.00 | | 6 | 3.65 |
Output: | score | rank | | :--- | :--- | | 4.00 | 1 | | 4.00 | 1 | | 3.85 | 2 | | 3.65 | 3 | | 3.65 | 3 | | 3.50 | 4 |
id is a unique identifier.Given an m x n matrix board containing 'X' and 'O', capture all regions that are 4-directionally surrounded by 'X'.
A region is captured by flipping all 'O's into 'X's in that surrounded region.
Example 1:
Input: board = [
["X","X","X","X"],
["X","O","O","X"],
["X","X","O","X"],
["X","O","X","X"]
]
Output: [
["X","X","X","X"],
["X","X","X","X"],
["X","X","X","X"],
["X","O","X","X"]
]
Explanation:
Surrounded regions should not be on the border, which means that any 'O' on the border of the board are not flipped to 'X'. Any 'O' that is not on the border and it is not connected to an 'O' on the border will be flipped to 'X'. Two cells are connected if they are adjacent cells connected horizontally or vertically.
Example 2:
Input: board = [["X"]]
Output: [["X"]]
m == board.lengthn == board[i].length1 <= m, n <= 200board[i][j] is 'X' or 'O'.Given strings s1, s2, and s3, find whether s3 is formed by an interleaving of s1 and s2.
An interleaving of two strings s and t is a configuration where s and t are divided into n and m substrings respectively, such that the interleaving preserves the left-to-right order within each string.
Example 1:
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true
Explanation: One valid interleaving is: aa|dbb|cb|ca|c
Example 2:
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false
Example 3:
Input: s1 = "", s2 = "", s3 = ""
Output: true
0 <= s1.length, s2.length <= 1000 <= s3.length <= 200s1, s2, and s3 consist of lowercase English letters.Given the root of a complete binary tree, return the number of the nodes in the tree.
According to Wikipedia, every level, except possibly the last, is completely filled in a complete binary tree, and all nodes in the last level are as far left as possible. It can have between 1 and 2^h nodes inclusive at the last level h.
Design an algorithm that runs in less than O(n) time complexity.
Example 1:
Input: root = [1,2,3,4,5,6]
Output: 6
Example 2:
Input: root = []
Output: 0
Example 3:
Input: root = [1]
Output: 1
[0, 5 * 10^4].0 <= Node.val <= 5 * 10^4You are given a network of n nodes, labeled from 1 to n. You are also given times, a list of travel times as directed edges times[i] = (ui, vi, wi), where ui is the source node, vi is the target node, and wi is the time it takes for a signal to travel from source to target.
We will send a signal from a given node k. Return the minimum time it takes for all the n nodes to receive the signal. If it is impossible for all the n nodes to receive the signal, return -1.
Input: times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2
Output: 2
Input: times = [[1,2,1]], n = 2, k = 1
Output: 1
Input: times = [[1,2,1]], n = 2, k = 2
Output: -1
1 <= k <= n <= 1001 <= times.length <= 6000times[i].length == 31 <= ui, vi <= nui != vi0 <= wi <= 100(ui, vi) are unique (i.e., no multiple edges).Given the root of a binary tree, determine if it is a valid binary search tree (BST).
A valid BST is defined as follows:
Example 1:
Input: root = [2,1,3]
Output: true
Example 2:
Input: root = [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.
[1, 10^4].-2^31 <= Node.val <= 2^31 - 1Design a simplified version of Twitter where users can post tweets, follow/unfollow another user, and is able to see the 10 most recent tweets in the user's news feed.
Implement the Twitter class:
Twitter() Initializes your twitter object.postTweet(int userId, int tweetId) Composes a new tweet with ID tweetId by the user userId. Each call to this function will be made with a unique tweetId.getNewsFeed(int userId) Retrieves the 10 most recent tweet IDs in the user's news feed. Each item in the news feed must be posted by users who the user followed or by the user themselves. Tweets must be ordered from most recent to least recent.follow(int followerId, int followeeId) The user with ID followerId started following the user with ID followeeId.unfollow(int followerId, int followeeId) The user with ID followerId started unfollowing the user with ID followeeId.Example 1:
Input:
["Twitter", "postTweet", "getNewsFeed", "follow", "postTweet", "getNewsFeed", "unfollow", "getNewsFeed"]
[[], [1, 5], [1], [1, 2], [2, 6], [1], [1, 2], [1]]
Output:
[null, null, [5], null, null, [6, 5], null, [5]]
Explanation:
Twitter twitter = new Twitter();
twitter.postTweet(1, 5); // User 1 posts a new tweet (id = 5).
twitter.getNewsFeed(1); // User 1's news feed should return a list with 1 tweet id -> [5]. return [5]
twitter.follow(1, 2); // User 1 follows user 2.
twitter.postTweet(2, 6); // User 2 posts a new tweet (id = 6).
twitter.getNewsFeed(1); // User 1's news feed should return a list with 2 tweet ids -> [6, 5]. Tweet id 6 should precede tweet id 5 because it is posted after tweet id 5.
twitter.unfollow(1, 2); // User 1 unfollows user 2.
twitter.getNewsFeed(1); // User 1's news feed should return a list with 1 tweet id -> [5], since user 1 is no longer following user 2.
1 <= userId, followerId, followeeId <= 500You are given an m x n grid rooms initialized with these three possible values:
-1 A wall or an obstacle.0 A gate.INF Infinity means an empty room. We use the value 2^31 - 1 = 2147483647 to represent INF as you may assume that the distance to a gate is less than 2147483647.Fill each empty room with the distance to its nearest gate. If it is impossible to reach a gate, it should be filled with INF.
Note: Modify rooms in-place. You do not need to return anything.
Example 1:
Input: rooms = [
[2147483647,-1,0,2147483647],
[2147483647,2147483647,2147483647,-1],
[2147483647,-1,2147483647,-1],
[0,-1,2147483647,2147483647]
]
Output: [
[3,-1,0,1],
[2,2,1,-1],
[1,-1,2,-1],
[0,-1,3,4]
]
Example 2:
Input: rooms = [[-1]]
Output: [[-1]]
m == rooms.lengthn == rooms[i].length1 <= m, n <= 250rooms[i][j] is -1, 0, or 2147483647.Given two strings needle and haystack, return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
Input: haystack = "sadbutsad", needle = "sad"
Output: 0
Explanation: "sad" occurs at index 0 and 6. The first occurrence is at index 0.
Input: haystack = "leetcode", needle = "leeto"
Output: -1
Explanation: "leeto" did not occur in "leetcode", so we return -1.
1 <= haystack.length, needle.length <= 10^4haystack and needle consist of only lowercase English characters.You are given an array of integers stones where stones[i] is the weight of the ith stone.
We are playing a game with the stones. On each turn, we choose the heaviest two stones and smash them together. Suppose the heaviest two stones have weights x and y with x <= y. The result of this smash is:
x == y, both stones are destroyed, andx != y, the stone of weight x is destroyed, and the stone of weight y has new weight y - x.At the end of the game, there is at most one stone left.
Return the weight of the last remaining stone. If there are no stones left, return 0.
Example 1:
Input: stones = [2,7,4,1,8,1]
Output: 1
Explanation:
We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,
we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,
we combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
we combine 1 and 1 to get 0 so the array converts to [1] then that's the value of the last stone.
Example 2:
Input: stones = [1]
Output: 1
1 <= stones.length <= 301 <= stones[i] <= 10000 <= tweetId <= 10^43 * 10^4 calls will be made to postTweet, getNewsFeed, follow, and unfollow.