April LeetCoding Challenge 2021 — Day 6: Minimum Operations to Make Array Equal

Today, we will solve the 6th problem of the April LeetCoding Challenge 2021.

Problem Statement

You have an array arr of length n where arr[i] = (2 * i) + 1 for all valid values of i (i.e. 0 <= i < n).

In one operation, you can select two indices x and y where 0 <= x, y < n and subtract 1 from arr[x] and add 1 to arr[y] (i.e. perform arr[x] -=1 and arr[y] += 1). The goal is to make all the elements of the array equal. It is guaranteed that all the elements of the array can be made equal using some operations.

Given an integer n, the length of the array. Return the minimum number of operations needed 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 second operation choose x = 2 and y = 0 again, thus arr = [3, 3, 3].

Example 2:

Input: n = 6
Output: 9

Solution

This is a basic problem. Here we have to make all the elements of the array equal by a particular operation. We have to find the minimum number of steps to make the array elements equal.

To do this task in minimum steps, we can convert the elements into their mean value.

According to the question, the elements at i-th index is (2*i)+1 . Therefore the mean of these elements will be n ,which is given in the question.

The code is given below.

The code can be found here

sksaikia/LeetCode