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# Find minimum operations needed to make an Array beautiful in C++

In this problem, we are given a binary array bin[] consisting n binary values which happen to be 0 and 1. Our task is to find minimum operations needed to make an Array beautiful.

Beautiful array is a special type of binary array which consists of a pattern of alternate 0’s and 1’s.

**Problem Description** − We need to find the number operations that are required to make the array beautiful. An operations consists of these steps −

**Step 1** − Cut the array into two halves.

**Step 2** − Reverse any one of the two halves.

**Step 3** − Join then halves back.

We will be counting the number of operations that are required to make sure that the array becomes a beautiful array.

**Let’s take an example to understand the problem,**

## Input

bin[] = {1, 0, 1, 0, 0, 1}

## Output

1

## Explanation

We will cut the array, make a subarray bin[4, 5], reverse it and join it back.

## Solution Approach

The solution to the problem is based on finding the minimum number of switch operations which is equal to the number of consecutive zeros. The base cases are −

If the size of the array is 1, it is a beautiful array. If the size of the array is an odd number then it can never be a beautiful array.

For all even lengths, we will check the total number of consecutive zeros or ones which will be the number of operations to be performed.

## Algorithm

Initialise − zeroCount , oneCount, consZero = 0

**Step 1** − if ( n = 1), return 0

**Step 2** − if (n%2 != 0), return -1.

**Step 3** − Loop for i -> 0 to n-1.

**Step 3.1** − if bin[i] == bin[i+1] == 0, consZero++.

**Step 4** − if bin[n-1] == bin[0] == 0, consZero++.

**Step 5** − return consZero.

**Program to illustrate the working of our solution,**

## Example

#include <iostream> using namespace std; int minOperations(int bin[], int n) { if(n == 1) return 0; if(n%2 != 0) return -1; int consZero = 0; for (int i = 0; i < n; ++i) { if (i + 1 < n) { if (bin[i] == 0 && bin[i + 1] == 0) consZero++; } } if (bin[0] == bin[n - 1] && bin[0] == 0) consZero++; return consZero; } int main() { int bin[] = { 1, 0, 1, 0, 0, 1}; int n = sizeof(bin) / sizeof(bin[0]); cout<<"The minimum operations needed to make an Array beautiful is "<<minOperations(bin, n); return 0; }

## Output

The minimum operations needed to make an Array beautiful is 1

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