Return the total number of bad pairs in nums. Example 1: Input: nums = [4,1,3,3] Output: 5 Explanation: The pair (0, 1) is a bad pair since 1 - 0 != 1 - 4. The pair (0, 2) is a bad pair since 2 - 0 != 3 - 4, 2 != -1. The pair (0, 3) is a bad pair...
count number,计数; pairs,数对; difference,差; absolute difference,差值的绝对值; absolute difference K,差值的绝对值=K。 题目:Given an integer array nums and an integer k, return the number of pairs (i, j) where i < j such that |nums[i] - nums[j]| == k. The value of |x| is ...
[LeetCode] 2563. Count the Number of Fair Pairs Given a 0-indexed integer array nums of size n and two integers lower and upper, return the number of fair pairs. A pair (i, j) is fair if: 0 <= i < j < n, and lower <= nums[i] + nums[j] <= upper Example 1: Input: n...
Can you solve this real interview question? Count Number of Nice Subarrays - Given an array of integers nums and an integer k. A continuous subarray is called nice if there are k odd numbers on it. Return the number of nice sub-arrays. Example 1:
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Can you solve this real interview question? Count Number of Possible Root Nodes - Alice has an undirected tree with n nodes labeled from 0 to n - 1. The tree is represented as a 2D integer array edges of length n - 1 where edges[i] = [ai, bi] indicates t
1497. Check If Array Pairs Are Divisible by k.md 1545. Find Kth Bit in Nth Binary String.md 155. Min Stack.md 1593. Split a String Into the Max Number of Unique Substrings.md 1684. Count the Number of Consistent Strings.md 179. Largest Number.md 1942. The Number of the...
在LeetCode 1973题中,DFS算法的时间复杂度是多少? 文章目录 1. 题目 2. 解题 1. 题目 Given the root of a binary tree, return the number of nodes where the value of the node is equal to the sum of the values of its descendants. A descendant of a node x is any node that is on the...
<LeetCode OJ> 204. Count Primes Count the number of prime numbers less than a non-negative number, n. 分析: 思路首先:一个数不是合数就是素数,合数更好推断呢! 合数:不论什么一个合数都能够表现为适当个素数的乘积的形式, 所以我们仅仅用小于sqrt(number)的素数去除要推断的数就可以,...
Count Primes -- leetcode 后端开发 Description: Count the number of prime numbers less than a non-negative number, n. 基本思路:筛法 1。 2 为素数。 筛掉以2为因子的数。 即2 * 2, 2*3, 2*4,2*5 2, 寻找到下一个未被筛除的数。如3. 再筛掉以3为因子的数。