LeetCode - Number of Good Pairs Given an array of integers nums. A pair (i,j) is called goodifnums[i] == nums[j] and i <j. Return the number of good pairs. Example1: Input: nums= [1,2,3,1,1,3] Output:4Explanation: There are4 good pairs (0,3), (0,4), (3,4), (...
Given an array of integers nums, return the number of good pairs. A pair (i, j) is called good if nums[i] == nums[j] and i < j. Example 1: Input: nums = [1,2,3,1,1,3] Output: 4 Explanation: There are 4 good pairs (0,3), (0,4), (3,4), (2,5) 0-indexed. E...
Can you solve this real interview question? Number of Good Pairs - Given an array of integers nums, return the number of good pairs. A pair (i, j) is called good if nums[i] == nums[j] and i < j. Example 1: Input: nums = [1,2,3,1,1,3] Output: 4 Exp
Runtime:1 ms, faster than81.21%of Java online submissions for Number of Good Pairs. Memory Usage:39.7 MB, less than14.56%of Java online submissions for Number of Good Pairs. 整体表现不太好。只能说一般。
1512. Number of Good Pairs # 题目 # Given an array of integers nums. A pair (i,j) is called good if nums[i] == nums[j] and i < j. Return the number of good pairs. Example 1: Input: nums = [1,2,3,1,1,3] Output: 4 Explanation: There are 4 good pairs (0,3
Leetcode solution 1512. Number of Good Pairs Blogger: https://blog.baozitraining.org/2020/07/leetcode-solution-1512-number-of-good.html Youtube: https://youtu.be/dvnjvOLh88k 博客园:https://www.cnblogs.com/baozitraining/p/13338525.html ...
Can you solve this real interview question? Number of Good Leaf Nodes Pairs - You are given the root of a binary tree and an integer distance. A pair of two different leaf nodes of a binary tree is said to be good if the length of the shortest path betwe
LeetCode——1128. 等价多米诺骨牌对的数量[Number of Equivalent Domino Pairs]——分析及代码[Java] 一、题目 二、分析及代码 1. 记录等价对数 (1)思路 (2)代码 (3)结果 三、其他 一、题目 给你一个由一些多米诺骨牌组成的列表 dominoes。 如果其中某一张多米诺骨牌可以通过旋转 0 ...5130. Number of Eq...
class Solution { public int numOfPairs(String[] nums, String target) { int cnt=0; int n=nums.length; for (int i = 0; i < n-1; i++) { for (int j = i+1; j < n; j++) { String temp1=nums[i]+nums[j]; String temp2=nums[j]+nums[i]; if(target.equals(temp1)){ cnt...
to this problem perfectly. The O(N^2) brute force solution is naive. We can also use a Map data structure (key is the number, value is the occurrence count) thus O(N). We can also sort the array and use this simple formula (also leetcode's hint) to calculate the good number ...