# Minimum Swaps required to group all 1’s together in C++

## Problem statement

Given an array of 0’s and 1’s. The task is to find the minimum number of swaps required to group all 1’s present in the array together.

## Example

If input array = {1, 0, 1, 1, 0, 1} then 1 swap is required. i.e. swap first 0 with last 1.

## Algorithm

• Count total number of 1’s in the array
• If count is x, then we need to find the subarray of length x of this array with maximum number of 1’s
• Minimum swaps required will be the number of 0’s in the subarray of length x with maximum number of 1’s

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
int getMinSwaps(int *arr, int n) {
int oneCnt = 0;
for (int i = 0; i < n; ++i) {
if (arr[i] == 1) {
++oneCnt;
}
}
int x = oneCnt;
int maxOnes = INT_MIN;
int preCompute[n] = {0};
if (arr[0] == 1) {
preCompute[0] = 1;
}
for (int i = 1; i < n; ++i) {
if (arr[i] == 1) {
preCompute[i] = preCompute[i - 1] + 1;
} else {
preCompute[i] = preCompute[i - 1];
}
}
for (int i = x - 1; i < n; ++i) {
if (i == (x - 1)) {
oneCnt = preCompute[i];
} else {
oneCnt = preCompute[i] - preCompute[i - x];
} if (maxOnes < oneCnt) {
maxOnes = oneCnt;
}
}
int swapCnt = x - oneCnt;
return swapCnt;
}
int main() {
int arr[] = {1, 0, 1, 1, 0, 1};
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Minimum swap count = " << getMinSwaps(arr, n) << endl;
return 0;
}

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

## Output

Minimum swap count = 1