1551. Minimum Operations to Make Array Equal You have an arrayarrof lengthnwherearr[i] = (2 * i) + 1for all valid values ofi(i.e.0 <= i < n). In one operation, you can select two indicesxandywhere0 <= x, y < nand subtract1fromarr[x]and add1toarr[y](i.e. performarr[...
Given an integern, the length of the array. Returnthe minimum number of operationsneeded to make all the elements of arr equal. Example 1: Input: n = 3 Output: 2 Explanation: arr = [1, 3, 5] First operation choose x = 2 and y = 0, this leads arr to be [2, 3, 4] In the...
每次操作可以同时将一个数字+1 另外一个数字-1,问你操作多少次可以让这个数组里面的数字都变成一样的。 那肯定是把所有数字变成最中间那个数字,最快。二. 代码 基本就是一道小学数学题,自己在纸上写出来规律,得出公式,然后用简单的代码把这个公式写了出来。直接看我的代码,那肯定看不懂意思,因为这是自己得出来...
Can you solve this real interview question? Minimum Operations to Make the Array Increasing - You are given an integer array nums (0-indexed). In one operation, you can choose an element of the array and increment it by 1. * For example, if nums = [1,2
2967.Minimum Cost to Make Array Equalindromic题目描述You are given a 0-indexed integer array nums having length n.You are allowed to perform a special move any number of times (including zero) on nu…
Return the minimum number of operations to make all elements of nums equal to 1. If it is impossible, return -1. The gcd of two integers is the greatest common divisor of the two integers. Example 1: Input: nums = [2,6,3,4] Output: 4 Explanation: We can do the following operation...
You are given an integer arraynums(0-indexed). In one operation, you can choose an element of the array and increment it by1. For example, ifnums = [1,2,3], you can choose to incrementnums[1]to makenums = [1,3,3]. Return the minimum number of operations needed to makenumsstric...
Given a non-empty integer array of size n, find the minimum number of moves required to make all array elements equal, where a move is incrementing n - 1 elements by 1. Example: Input: [1,2,3] Output: 3 Explanation: Only three moves are needed (remember each move increments two ele...
publicstaticintminOperations(int[]nums){int cnt=0;int n=nums.length;for(int i=0;i<n-2;i++){if(nums[i]==0){cnt++;nums[i+1]=1-nums[i+1];nums[i+2]=1-nums[i+2];}}if(nums[n-2]!=1||nums[n-1]!=1){return-1;}returncnt;}...
Hence, minimum operations needed = 7-3 = 4 C++ implementation: // C++ program to find the minimum operation// to make all elements equal in array#include <bits/stdc++.h>usingnamespacestd;intminimum_operation(vector<int>arr,intn)