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Given two strings s and part, perform the following operation on s until all occurrences of the substring part are removed: Find the leftmost occurrence of the substring part and remove it from s. Return s after removing all occurrences of part. A substring is a contiguous sequence of characters in a string. Example 1: Input: s = "daabcbaabcbc", part = "abc" Output: "dab" Explanation: The follow..

You are given a 0-indexed integer array nums. Swaps of adjacent elements are able to be performed on nums. A valid array meets the following conditions: The largest element (any of the largest elements if there are multiple) is at the rightmost position in the array. The smallest element (any of the smallest elements if there are multiple) is at the leftmost position in the array. Return the min..

Given an m x n matrix board where each cell is a battleship 'X' or empty '.', return the number of the battleships on board. Battleships can only be placed horizontally or vertically on board. In other words, they can only be made of the shape 1 x k (1 row, k columns) or k x 1 (k rows, 1 column), where k can be of any size. At least one horizontal or vertical cell separates between two battleshi..

# Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def equalToDescendants(self, root: Optional[TreeNode]) -> int: self.count=0 def dfs(node): if not node: return 0 add=dfs(node.left)+dfs(node.right) if node.val==add: self.count+=1 return add+node.val dfs(root) return s..

A website domain "discuss.leetcode.com" consists of various subdomains. At the top level, we have "com", at the next level, we have "leetcode.com" and at the lowest level, "discuss.leetcode.com". When we visit a domain like "discuss.leetcode.com", we will also visit the parent domains "leetcode.com" and "com" implicitly. A count-paired domain is a domain that has one of the two formats "rep d1.d..

# Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def findNearestRightNode(self, root: TreeNode, u: TreeNode) -> Optional[TreeNode]: nodes=deque() nodes.append([root,0]) self.check=float('inf') while nodes: n=nodes.popleft() node=n[0] level=n[1] if node.val==u.val: se..

Given the root of a binary tree, find the maximum value v for which there exist different nodes a and b where v = |a.val - b.val| and a is an ancestor of b. A node a is an ancestor of b if either: any child of a is equal to b or any child of a is an ancestor of b. Example 1: Input: root = [8,3,10,1,6,null,14,null,null,4,7,13] Output: 7 Explanation: We have various ancestor-node differences, some..

Given 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] Constraints: m == matrix.length n == matrix[i].length 1

A string is good if there are no repeated characters. Given a string s​​​​​, return the number of good substrings of length three in s​​​​​​. Note that if there are multiple occurrences of the same substring, every occurrence should be counted. A substring is a contiguous sequence of characters in a string. Example 1: Input: s = "xyzzaz" Output: 1 Explanation: There are 4 substrings of size 3: "..

A parentheses string is valid if and only if: It is the empty string, It can be written as AB (A concatenated with B), where A and B are valid strings, or It can be written as (A), where A is a valid string. You are given a parentheses string s. In one move, you can insert a parenthesis at any position of the string. For example, if s = "()))", you can insert an opening parenthesis to be "(()))"..