# Minimum toggles to partition a binary array so that it has first 0s then 1s in C++

## Problem statement

Given an array of n integers containing only 0 and 1. Find the minimum toggles (switch from 0 to 1 or vice-versa) required such the array the array become partitioned, i.e., it has first 0s then 1s.

## Example

If arr[] = {1, 0, 0, 1, 1, 1, 0} then 2 toggle is required i.e. toggle first one and last zero.

## Algorithm

• If we observe the question, then we will find that there will definitely exist a point from 0 to n-1 where all elements left to that point should contains all 0’s and right to point should contains all 1’s
• Those indices which don’t obey this law will have to be removed. The idea is to count all 0s from left to right.

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
int getMinToggles(int *arr, int n) {
int zeroCnt[n + 1] = {0};
for (int i = 1; i <= n; ++i) {
if (arr[i - 1] == 0) {
zeroCnt[i] = zeroCnt[i - 1] + 1;
} else {
zeroCnt[i] = zeroCnt[i - 1];
}
}
int result = n;
for (int i = 1; i <= n; ++i) {
result = min(result, i - zeroCnt[i] + zeroCnt[n] - zeroCnt[i]);
}
return result;
}
int main() {
int arr[] = {1, 0, 0, 1, 1, 1, 0};
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Minimum toggles = " << getMinToggles(arr, n) << endl;
return 0;
}

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

## Output

Minimum toggles = 2