# Minimum number of moves to make all elements equal using C++.

## Problem statement

Given an array of N elements and an integer K., It is allowed to perform the following operation any number of times on the given array −

• Insert the Kth element at the end of the array and delete the first element of the array.

The task is to find the minimum number of moves needed to make all elements of the array equal. Print -1 if it is not possible

If arr[] = {1, 2, 3, 4, 5, 6} and k = 6 then minimum 5 moves are
required:
Move-1: {2, 3, 4, 5, 6, 6}
Move-2: {3, 4, 5, 6, 6, 6}
Move-3: {4, 5, 6, 6, 6, 6}
Move-4: {5, 6, 6, 6, 6, 6}
Move-5: {6, 6, 6, 6, 6, 6}

## Algorithm

1. First we copy a[k] to the end, then a[k+1] and so on
2. To make sure that we only copy equal elements, all elements in the range K to N should be equal. We need to remove all elements in range 1 to K that are not equal to a[k]
3. Keep applying operations until we reach the rightmost term in range 1 to K that is not equal to a[k].

## Example

#include <iostream>
#define SIZE(arr) (sizeof(arr) / sizeof(arr[0]))
using namespace std;
int getMinMoves(int *arr, int n, int k){
int i;
for (i = k - 1; i < n; ++i) {
if (arr[i] != arr[k - 1]) {
return -1;
}
}
for (i = k - 1; i >= 0; --i) {
if (arr[i] != arr[k - 1]) {
return i + 1;
}
}
return 0;
}
int main(){
int arr[] = {1, 2, 3, 4, 5, 6};
int k = 6;
cout << "Minimum moves required = " << getMinMoves(arr, SIZE(arr), k) << endl;
return 0;
}

## Output

When you compile and execute the above program. It generates the following output −

Minimum moves required = 5

Updated on: 31-Oct-2019

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