Minimum Swaps required to group all 1’s together

Given an array of 0’s and 1’s, we need to write a program to find the minimum number of swaps required to group all 1’s present in the array together.

Examples:

Input : arr[] = {1, 0, 1, 0, 1}
Output : 1
Explanation: Only 1 swap is required to 
group all 1's together. Swapping index 1
and 4 will give arr[] = {1, 1, 1, 0, 0}

Input : arr[] = {1, 0, 1, 0, 1, 1}
Output : 1

A simple solution is to first count total number of 1’s in the array. Suppose this count is x, now we need to find the subarray of length x of this array with maximum number of 1’s. And minimum swaps required will be the number of 0’s in the subarray of length x with maximum number of 1’s. 
Time Complexity: O(n²)

An efficient solution is to optimize the brute force technique of finding the subarray in above approach using the concept of sliding window technique. We can maintain a preCount array to find number of 1’s present in a subarray in O(1) time complexity.

Below is the implementation of above idea: 

C++

// C++ program to find minimum swaps
// required to group all 1's together
#include <iostream>
#include <limits.h>

using namespace std;

// Function to find minimum swaps
// to group all 1's together
int minSwaps(int arr[], int n) {

  int noOfOnes = 0;

  // find total number of all in the array
  for (int i = 0; i < n; i++) {
    if (arr[i] == 1)
      noOfOnes++;
  }

  // length of subarray to check for
  int x = noOfOnes;

  int maxOnes = INT_MIN;

  // array to store number of 1's upto
  // ith index
  int preCompute[n] = {0};

  // calculate number of 1's upto ith
  // index and store in the array preCompute[]
  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];
  }

  // using sliding window technique to find
  // max number of ones in subarray of length x
  for (int i = x - 1; i < n; i++) {
    if (i == (x - 1))
      noOfOnes = preCompute[i];
    else
      noOfOnes = preCompute[i] - preCompute[i - x];

    if (maxOnes < noOfOnes)
      maxOnes = noOfOnes;
  }

  // calculate number of zeros in subarray
  // of length x with maximum number of 1's
  int noOfZeroes = x - maxOnes;

  return noOfZeroes;
}

// Driver Code
int main() {
  int a[] = {1, 0, 1, 0, 1};
  int n = sizeof(a) / sizeof(a[0]);
  cout << minSwaps(a, n);
  return 0;
}

Java

// Java  program to find minimum swaps
// required to group all 1's together

import java.io.*;

class GFG {

// Function to find minimum swaps
// to group all 1's together
 static int minSwaps(int arr[], int n) {

int noOfOnes = 0;

// find total number of all in the array
for (int i = 0; i < n; i++) {
    if (arr[i] == 1)
    noOfOnes++;
}

// length of subarray to check for
int x = noOfOnes;

int maxOnes = Integer.MIN_VALUE;

// array to store number of 1's upto
// ith index
int preCompute[] = new int[n];

// calculate number of 1's upto ith
// index and store in the array preCompute[]
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];
}

// using sliding window technique to find
// max number of ones in subarray of length x
for (int i = x - 1; i < n; i++) {
    if (i == (x - 1))
    noOfOnes = preCompute[i];
    else
    noOfOnes = preCompute[i] - preCompute[i - x];

    if (maxOnes < noOfOnes)
    maxOnes = noOfOnes;
}

// calculate number of zeros in subarray
// of length x with maximum number of 1's
int noOfZeroes = x - maxOnes;

return noOfZeroes;
}

// Driver Code
public static void main (String[] args) {
int a[] = {1, 0, 1, 0, 1};
int n = a.length;
System.out.println( minSwaps(a, n));

    }
}

// This code is contributed by vt_m.

Python3

# Python program to
# find minimum swaps
# required to group
# all 1's together

# Function to find minimum swaps
# to group all 1's together
def minSwaps(arr,n):

    noOfOnes = 0

    # find total number of
    # all in the array
    for i in range(n):
        if (arr[i] == 1):
            noOfOnes=noOfOnes+1


    # length of subarray to check for
    x = noOfOnes

    maxOnes = -2147483648

    # array to store number of 1's upto
    # ith index
    preCompute={}

    # calculate number of 1's upto ith
    # index and store in the
    # array preCompute[]
    if (arr[0] == 1):
        preCompute[0] = 1
    for i in range(1,n):
        if (arr[i] == 1):
            preCompute[i] = preCompute[i - 1] + 1
        else:
            preCompute[i] = preCompute[i - 1]


    # using sliding window
    # technique to find
    # max number of ones in
    # subarray of length x
    for i in range(x-1,n):
        if (i == (x - 1)):
            noOfOnes = preCompute[i]
        else:
            noOfOnes = preCompute[i] - preCompute[i - x]

        if (maxOnes < noOfOnes):
            maxOnes = noOfOnes


    # calculate number of zeros in subarray
    # of length x with maximum number of 1's
    noOfZeroes = x - maxOnes

    return noOfZeroes

# Driver code

a = [1, 0, 1, 0, 1]
n = len(a)

print(minSwaps(a, n))

# This code is contributed
# by Anant Agarwal.

C#

// C# program to find minimum swaps
// required to group all 1's together

using System;

class GFG {

    // Function to find minimum swaps
    // to group all 1's together
    static int minSwaps(int []arr, int n) {

        int noOfOnes = 0;

        // find total number of all in the array
        for (int i = 0; i < n; i++) {
            if (arr[i] == 1)
            noOfOnes++;
        }

        // length of subarray to check for
        int x = noOfOnes;

        int maxOnes = int.MinValue;

        // array to store number of 1's upto
        // ith index
        int []preCompute = new int[n];

        // calculate number of 1's upto ith
        // index and store in the array preCompute[]
        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];
        }

        // using sliding window technique to find
        // max number of ones in subarray of length x
        for (int i = x - 1; i < n; i++) {
            if (i == (x - 1))
            noOfOnes = preCompute[i];
            else
            noOfOnes = preCompute[i] - preCompute[i - x];

            if (maxOnes < noOfOnes)
            maxOnes = noOfOnes;
        }

        // calculate number of zeros in subarray
        // of length x with maximum number of 1's
        int noOfZeroes = x - maxOnes;

        return noOfZeroes;
    }

    // Driver Code
    public static void Main ()
    {
        int []a = {1, 0, 1, 0, 1};
        int n = a.Length;
        Console.WriteLine( minSwaps(a, n));

    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to find minimum swaps
// required to group all 1's together

// Function to find minimum swaps
// to group all 1's together
function minSwaps($arr, $n)
{

        $noOfOnes = 0;

        // find total number of
        // all in the array
        for($i = 0; $i < $n; $i++)
        {
            if ($arr[$i] == 1)
            $noOfOnes++;
        }

        // length of subarray
        // to check for
        $x = $noOfOnes;

        $maxOnes = PHP_INT_MIN;

        // array to store number of
        // 1's upto ith index
        $preCompute = array();

        // calculate number of 1's
        // upto ith index and store
        // in the array preCompute[]
        if ($arr[0] == 1)
            $preCompute[0] = 1;
        for($i = 1; $i < $n; $i++)
        {
            if ($arr[$i] == 1)
            {
                $preCompute[$i] = $preCompute[$i - 1] + 1;
            }
            else
                $preCompute[$i] = $preCompute[$i - 1];
        }

        // using sliding window
        // technique to find
        // max number of ones in
        // subarray of length x
        for ( $i = $x - 1; $i < $n; $i++)
        {
            if ($i == ($x - 1))
                $noOfOnes = $preCompute[$i];
            else
                $noOfOnes = $preCompute[$i] -
                            $preCompute[$i - $x];

            if ($maxOnes < $noOfOnes)
                $maxOnes = $noOfOnes;
        }

        // calculate number of zeros in subarray
        // of length x with maximum number of 1's
        $noOfZeroes = $x - $maxOnes;

        return $noOfZeroes;
}

// Driver Code
$a = array(1, 0, 1, 0, 10);
$n = count($a);
echo minSwaps($a, $n);

// This code is contributed by anuj_67.
?>

Javascript

<script>
    // Javascript program to find minimum swaps
    // required to group all 1's together

    // Function to find minimum swaps
    // to group all 1's together
    function minSwaps(arr, n) {

        let noOfOnes = 0;

        // find total number of all in the array
        for (let i = 0; i < n; i++) {
            if (arr[i] == 1)
                noOfOnes++;
        }

        // length of subarray to check for
        let x = noOfOnes;

        let maxOnes = Number.MIN_VALUE;

        // array to store number of 1's upto
        // ith index
        let preCompute = new Array(n);
        preCompute.fill(0);

        // calculate number of 1's upto ith
        // index and store in the array preCompute[]
        if (arr[0] == 1)
            preCompute[0] = 1;
        for (let i = 1; i < n; i++) {
            if (arr[i] == 1) {
                preCompute[i] = preCompute[i - 1] + 1;
            } else
                preCompute[i] = preCompute[i - 1];
        }

        // using sliding window technique to find
        // max number of ones in subarray of length x
        for (let i = x - 1; i < n; i++) {
            if (i == (x - 1))
                noOfOnes = preCompute[i];
            else
                noOfOnes = preCompute[i] - preCompute[i - x];

            if (maxOnes < noOfOnes)
                maxOnes = noOfOnes;
        }

        // calculate number of zeros in subarray
        // of length x with maximum number of 1's
        let noOfZeroes = x - maxOnes;

        return noOfZeroes;
    }

    let a = [1, 0, 1, 0, 1];
    let n = a.length;
    document.write( minSwaps(a, n));

</script>

Output

1

Time Complexity: O(n) 
Auxiliary Space: O(n)

Another efficient approach :

First count total number of 1’s in the array. Suppose this count is x, now find the subarray of length x of this array with maximum number of 1’s using the concept of window-sliding technique. Maintain a variable to find number of 1’s present in a subarray in O(1) extra space and for each sub array maintain maxOnes Variable and at last Return numberOfZeros (numberOfZeroes = x – maxOnes).

Implementation:

C++

// C++ code for minimum swaps
// required to group all 1's together
#include <iostream>
#include <limits.h>

using namespace std;

// Function to find minimum swaps
// to group all 1's together
int minSwaps(int arr[], int n)
{

int numberOfOnes = 0;

// find total number of all 1's in the array
for (int i = 0; i < n; i++) {
    if (arr[i] == 1)
    numberOfOnes++;
}

// length of subarray to check for
int x = numberOfOnes;

int count_ones = 0, maxOnes;

// Find 1's for first subarray of length x
for(int i = 0; i < x; i++){
    if(arr[i] == 1)
    count_ones++;
}

maxOnes = count_ones;

// using sliding window technique to find
// max number of ones in subarray of length x
for (int i = 1; i <= n-x; i++) {

    // first remove leading element and check
    // if it is equal to 1 then decrement the
    // value of count_ones by 1
    if (arr[i-1] == 1)
    count_ones--;

    // Now add trailing element and check
    // if it is equal to 1 Then increment
    // the value of count_ones by 1
    if(arr[i+x-1] == 1)
    count_ones++;

    if (maxOnes < count_ones)
    maxOnes = count_ones;
}

// calculate number of zeros in subarray
// of length x with maximum number of 1's
int numberOfZeroes = x - maxOnes;

return numberOfZeroes;
}

// Driver Code
int main() {

int a[] = {0, 0, 1, 0, 1, 1, 0, 0, 1};
int n = sizeof(a) / sizeof(a[0]);

cout << minSwaps(a, n);

return 0;

}

Java

// java program to find largest number
// smaller than equal to n with m set
// bits then m-1 0 bits.
public class GFG {

    // Function to find minimum swaps
    // to group all 1's together
    static int minSwaps(int arr[], int n)
    {

        int numberOfOnes = 0;

        // find total number of all 1's
        // in the array
        for (int i = 0; i < n; i++) {
            if (arr[i] == 1)
                numberOfOnes++;
        }

        // length of subarray to check for
        int x = numberOfOnes;

        int count_ones = 0, maxOnes;

        // Find 1's for first subarray
        // of length x
        for(int i = 0; i < x; i++){
            if(arr[i] == 1)
                count_ones++;
        }

        maxOnes = count_ones;

        // using sliding window technique
        // to find max number of ones in
        // subarray of length x
        for (int i = 1; i <= n-x; i++) {

            // first remove leading element
            // and check if it is equal to
            // 1 then decrement the
            // value of count_ones by 1
            if (arr[i - 1] == 1)
                count_ones--;

            // Now add trailing element
            // and check if it is equal
            // to 1 Then increment the
            // value of count_ones by 1
            if(arr[i + x - 1] == 1)
                count_ones++;

            if (maxOnes < count_ones)
            maxOnes = count_ones;
        }

        // calculate number of zeros in
        // subarray of length x with
        // maximum number of 1's
        int numberOfZeroes = x - maxOnes;

        return numberOfZeroes;
    }

    // Driver code
    public static void main(String args[])
    {
        int a[] = new int[]{0, 0, 1, 0,
                            1, 1, 0, 0, 1};
        int n = a.length;

        System.out.println(minSwaps(a, n));
    }
}

// This code is contributed by Sam007

Python3

# Python code for minimum
# swaps required to group
# all 1's together

# Function to find minimum
# swaps to group all 1's
# together
def minSwaps(arr, n) :

    numberOfOnes = 0

    # find total number of
    # all 1's in the array
    for i in range(0, n) :

        if (arr[i] == 1) :
            numberOfOnes = numberOfOnes + 1

    # length of subarray
    # to check for
    x = numberOfOnes

    count_ones = 0
    maxOnes = 0

    # Find 1's for first
    # subarray of length x
    for i in range(0, x) :

        if(arr[i] == 1) :
            count_ones = count_ones + 1

    maxOnes = count_ones

    # using sliding window
    # technique to find
    # max number of ones in
    # subarray of length x
    for i in range(1, (n - x + 1)) :

        # first remove leading
        # element and check
        # if it is equal to 1
        # then decrement the
        # value of count_ones by 1
        if (arr[i - 1] == 1) :
            count_ones = count_ones - 1

        # Now add trailing
        # element and check
        # if it is equal to 1
        # Then increment
        # the value of count_ones by 1
        if(arr[i + x - 1] == 1) :
            count_ones = count_ones + 1

        if (maxOnes < count_ones) :
                maxOnes = count_ones

    # calculate number of
    # zeros in subarray
    # of length x with
    # maximum number of 1's
    numberOfZeroes = x - maxOnes

    return numberOfZeroes

# Driver Code
a = [0, 0, 1, 0, 1, 1, 0, 0, 1]
n = 9
print (minSwaps(a, n))

# This code is contributed
# by Manish Shaw(manishshaw1)

C#

// C# code for minimum swaps
// required to group all 1's together
using System;

class GFG{

    // Function to find minimum swaps
    // to group all 1's together
    static int minSwaps(int []arr, int n)
    {

        int numberOfOnes = 0;

        // find total number of all 1's in the array
        for (int i = 0; i < n; i++) {
            if (arr[i] == 1)
            numberOfOnes++;
        }

        // length of subarray to check for
        int x = numberOfOnes;

        int count_ones = 0, maxOnes;

        // Find 1's for first subarray of length x
        for(int i = 0; i < x; i++){
            if(arr[i] == 1)
            count_ones++;
        }

        maxOnes = count_ones;

        // using sliding window technique to find
        // max number of ones in subarray of length x
        for (int i = 1; i <= n-x; i++) {

            // first remove leading element and check
            // if it is equal to 1 then decrement the
            // value of count_ones by 1
            if (arr[i - 1] == 1)
            count_ones--;

            // Now add trailing element and check
            // if it is equal to 1 Then increment
            // the value of count_ones by 1
            if(arr[i + x - 1] == 1)
            count_ones++;

            if (maxOnes < count_ones)
            maxOnes = count_ones;
        }

        // calculate number of zeros in subarray
        // of length x with maximum number of 1's
        int numberOfZeroes = x - maxOnes;

        return numberOfZeroes;
    }

    // Driver Code
    static public void Main ()
    {
        int []a = {0, 0, 1, 0, 1, 1, 0, 0, 1};
        int n = a.Length;

        Console.WriteLine(minSwaps(a, n));

    }
}
// This code is contributed by vt_m.

PHP

<?php
// PHP code for minimum swaps
// required to group all 1's together

// Function to find minimum swaps
// to group all 1's together
function minSwaps($arr, $n)
{

    $numberOfOnes = 0;

    // find total number of
    // all 1's in the array
    for ($i = 0; $i < $n; $i++)
    {
        if ($arr[$i] == 1)
            $numberOfOnes++;
    }

    // length of subarray to check for
    $x = $numberOfOnes;

    $count_ones = 0;
    $maxOnes;

    // Find 1's for first
    // subarray of length x
    for($i = 0; $i < $x; $i++)
    {
        if($arr[$i] == 1)
            $count_ones++;
    }

    $maxOnes = $count_ones;

    // using sliding window
    // technique to find
    // max number of ones in
    // subarray of length x
    for ($i = 1; $i <= $n - $x; $i++)
    {

        // first remove leading
        // element and check
        // if it is equal to 1
        // then decrement the
        // value of count_ones by 1
        if ($arr[$i - 1] === 1)
            $count_ones--;

        // Now add trailing
        // element and check
        // if it is equal to 1
        // Then increment
        // the value of count_ones by 1
        if($arr[$i + $x - 1] === 1)
                $count_ones++;

        if ($maxOnes < $count_ones)
                $maxOnes = $count_ones;
    }

    // calculate number of zeros in subarray
    // of length x with maximum number of 1's
    $numberOfZeroes = $x - $maxOnes;

    return $numberOfZeroes;
}

// Driver Code
$a = array(0, 0, 1, 0, 1, 1, 0, 0, 1);
$n = 9;
echo minSwaps($a, $n);

// This code is contributed by Anuj_67
?>

Javascript

<script>
    // Javascript code for minimum swaps
    // required to group all 1's together

    // Function to find minimum swaps
    // to group all 1's together
    function minSwaps(arr, n)
    {

        let numberOfOnes = 0;

        // find total number of all 1's in the array
        for (let i = 0; i < n; i++) {
            if (arr[i] == 1)
            numberOfOnes++;
        }

        // length of subarray to check for
        let x = numberOfOnes;

        let count_ones = 0, maxOnes;

        // Find 1's for first subarray of length x
        for(let i = 0; i < x; i++){
            if(arr[i] == 1)
            count_ones++;
        }

        maxOnes = count_ones;

        // using sliding window technique to find
        // max number of ones in subarray of length x
        for (let i = 1; i <= n-x; i++) {

            // first remove leading element and check
            // if it is equal to 1 then decrement the
            // value of count_ones by 1
            if (arr[i - 1] == 1)
                count_ones--;

            // Now add trailing element and check
            // if it is equal to 1 Then increment
            // the value of count_ones by 1
            if(arr[i + x - 1] == 1)
                count_ones++;

            if (maxOnes < count_ones)
                maxOnes = count_ones;
        }

        // calculate number of zeros in subarray
        // of length x with maximum number of 1's
        let numberOfZeroes = x - maxOnes;

        return numberOfZeroes;
    }

    let a = [0, 0, 1, 0, 1, 1, 0, 0, 1];
    let n = a.length;

    document.write(minSwaps(a, n));

</script>

Output

1

Time Complexity : O(n) 
Auxiliary Space : O(1)

Thanks to Mr. Gera for suggesting this approach.

Sliding window Easy to understand version

Algorithm:

1. Store the total no of ones in a variable say count. This will be the window size.
2. Initialise a variable to store maximum no of ones out of all the sub arrays of size count and a variable to store count of ones in current window.
3. Now iterate over the array and as soon as you hit the window size compare the no of ones  in that window with the maximum count of ones in all the windows so far and update max count of ones if count of ones in the current window is more. If the first element of window is 1 then decrease the current count.
4. Answer will be total count of ones – max count of ones out of all window.

Implementation:

C++

#include <bits/stdc++.h>

using namespace std;

int minSwaps(int arr[], int n)
{
  int totalCount = 0; // To store total number of ones
  // count total no of ones
  for (int i = 0; i < n; i++)
    totalCount += arr[i];

  int currCount = 0; // To store count of ones in current window
  int maxCount = 0;  // To store maximum count ones out of all windows
  int i = 0;         // start of window
  int j = 0;         // end of window

  while (j < n)
  {
    currCount += arr[j];

    // update maxCount when reach window size i.e. total count of ones in array
    if ((j - i + 1) == totalCount)
    {
      maxCount = max(maxCount, currCount);
      if (arr[i] == 1)
        currCount--; // decrease current count if first element of window is 1
      i++;           // slide window
    }
    j++;
  }

  return totalCount - maxCount; // return total no of ones in array - maximum count of ones out of all windows
}

// Driver Code
int main()
{

  int a[] = {1, 0, 1, 0, 1, 1};
  int n = sizeof(a) / sizeof(a[0]);

  cout << minSwaps(a, n);

  return 0;
}

Java

import java.io.*;
import java.util.*;

class GFG{

static int minSwaps(int[] arr, int n)
{

    // To store total number of ones
    int totalCount = 0;

    // Count total no of ones
    int i;
    for(i = 0; i < n; i++)
        totalCount += arr[i];

    int currCount = 0; // To store count of ones in current window
    int maxCount = 0;  // To store maximum count ones out
                       // of all windows

    // start of window
    i = 0;

    // end of window
    int j = 0;

    while (j < n)
    {
        currCount += arr[j];

        // update maxCount when reach window size i.e.
        // total count of ones in array
        if ((j - i + 1) == totalCount)
        {
            maxCount = Math.max(maxCount, currCount);
            if (arr[i] == 1)
                currCount--; // decrease current count
                             // if first element of
                             // window is 1

            // slide window
            i++;
        }
        j++;
    }

    return totalCount - maxCount; // return total no of ones in array
                                  // - maximum count of ones out of
                                  // all windows
}

// Driver Code
public static void main(String args[])
    {
    int[] a = { 1, 0, 1, 0, 1, 1 };
    int n = a.length;

    System.out.println(minSwaps(a, n));
}
}

// This code is contributed by shivanisinghss2110

Python

def minSwaps(arr, n):

    # To store total number of ones
    totalCount = 0

    # count total no of ones
    for i in range(0,n):
        totalCount += arr[i]

    currCount = 0 # To store count of ones in current window
    maxCount = 0  # To store maximum count ones out of all windows
    i = 0         # start of window
    j = 0         # end of window

    while (j < n):
        currCount += arr[j]

        # update maxCount when reach window size i.e. total count of ones in array
        if ((j - i + 1) == totalCount):
            maxCount = max(maxCount, currCount)
            if (arr[i] == 1):
                currCount -= 1

                # decrease current count if first element of window is 1
            i += 1           # slide window
        j += 1

    return totalCount - maxCount # return total no of ones in array - maximum count of ones out of all windows

# Driver Code
a = [1, 0, 1, 0, 1, 1]
n = len(a)
print(minSwaps(a, n))

# this code is contributed by shivanisighss2110

C#

using System;

class GFG{

static int minSwaps(int[] arr, int n)
{

    // To store total number of ones
    int totalCount = 0;

    // Count total no of ones
    int i;
    for(i = 0; i < n; i++)
        totalCount += arr[i];

    int currCount = 0; // To store count of ones in current window
    int maxCount = 0;  // To store maximum count ones out
                       // of all windows

    // start of window
    i = 0;

    // end of window
    int j = 0;

    while (j < n)
    {
        currCount += arr[j];

        // update maxCount when reach window size i.e.
        // total count of ones in array
        if ((j - i + 1) == totalCount)
        {
            maxCount = Math.Max(maxCount, currCount);
            if (arr[i] == 1)
                currCount--; // decrease current count
                             // if first element of
                             // window is 1

            // slide window
            i++;
        }
        j++;
    }

    return totalCount - maxCount; // return total no of ones in array
                                  // - maximum count of ones out of
                                  // all windows
}

// Driver Code
public static void Main()
{
    int[] a = { 1, 0, 1, 0, 1, 1 };
    int n = a.Length;

    Console.WriteLine(minSwaps(a, n));
}
}

// This code is contributed by ukasp

Javascript

<script>

function minSwaps(arr, n)
{

    // To store total number of ones
    let totalCount = 0;

    // Count total no of ones
    let i;
    for(i = 0; i < n; i++)
        totalCount += arr[i];

    let currCount = 0; // To store count of ones in current window
    let maxCount = 0;  // To store maximum count ones out
                       // of all windows

    // start of window
    i = 0;

    // end of window
    let j = 0;

    while (j < n)
    {
        currCount += arr[j];

        // update maxCount when reach window size i.e.
        // total count of ones in array
        if ((j - i + 1) == totalCount)
        {
            maxCount = Math.max(maxCount, currCount);
            if (arr[i] == 1)
                currCount--; // decrease current count
                             // if first element of
                             // window is 1

            // slide window
            i++;
        }
        j++;
    }

    return totalCount - maxCount; // return total no of ones in array
                                  // - maximum count of ones out of
                                  // all windows
}

// Driver Code

    let a = [ 1, 0, 1, 0, 1, 1 ];
    let n = a.length;

    document.write(minSwaps(a, n));

// This code is contributed by shivanisinghss2110

</script>

Output

1

Complexities: 

• Time Complexities: O(n)
• Space Complexities: O(1)