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Minimum adjacent swaps required to Sort Binary array
  • Difficulty Level : Easy
  • Last Updated : 26 Apr, 2021

Given a binary array, task is to sort this binary array using minimum swaps. We are allowed to swap only adjacent elements

Examples: 

Input : [0, 0, 1, 0, 1, 0, 1, 1]
Output : 3
1st swap : [0, 0, 1, 0, 0, 1, 1, 1]
2nd swap : [0, 0, 0, 1, 0, 1, 1, 1]
3rd swap : [0, 0, 0, 0, 1, 1, 1, 1]

Input : Array = [0, 1, 0, 1, 0]
Output : 3

Approach : 
This can be done by finding number of zeroes to the right side of every 1 and add them. In order to sort the array every one always has to perform a swap operation with every zero on its right side. So the total number of swap operations for a particular 1 in array is the number of zeroes on its right hand side. Find the number of zeroes on right side for every one i.e. the number of swaps and add them all to obtain the total number of swaps.

Implementation : 

C++




// C++ code to find minimum number of
// swaps to sort a binary array
#include <bits/stdc++.h>
 
using namespace std;
 
// Function to find minimum swaps to
// sort an array of 0s and 1s.
int findMinSwaps(int arr[], int n)
{
    // Array to store count of zeroes
    int noOfZeroes[n];
    memset(noOfZeroes, 0, sizeof(noOfZeroes));
 
    int i, count = 0;
 
    // Count number of zeroes
    // on right side of every one.
    noOfZeroes[n - 1] = 1 - arr[n - 1];
    for (i = n - 2; i >= 0; i--) {
        noOfZeroes[i] = noOfZeroes[i + 1];
        if (arr[i] == 0)
            noOfZeroes[i]++;
    }
 
    // Count total number of swaps by adding number
    // of zeroes on right side of every one.
    for (i = 0; i < n; i++) {
        if (arr[i] == 1)
            count += noOfZeroes[i];
    }
 
    return count;
}
 
// Driver code
int main()
{
    int arr[] = { 0, 0, 1, 0, 1, 0, 1, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findMinSwaps(arr, n);
    return 0;
}

Java




// Java code to find minimum number of
// swaps to sort a binary array
class gfg {
     
    static int findMinSwaps(int arr[], int n)
    {
        // Array to store count of zeroes
        int noOfZeroes[] = new int[n];
        int i, count = 0;
 
        // Count number of zeroes
        // on right side of every one.
        noOfZeroes[n - 1] = 1 - arr[n - 1];
        for (i = n - 2; i >= 0; i--)
        {
            noOfZeroes[i] = noOfZeroes[i + 1];
            if (arr[i] == 0)
                noOfZeroes[i]++;
        }
 
        // Count total number of swaps by adding number
        // of zeroes on right side of every one.
        for (i = 0; i < n; i++)
        {
            if (arr[i] == 1)
                count += noOfZeroes[i];
        }
        return count;
    }
     
    // Driver Code
    public static void main(String args[])
    {
        int ar[] = { 0, 0, 1, 0, 1, 0, 1, 1 };
        System.out.println(findMinSwaps(ar, ar.length));
    }
}
 
// This code is contributed by Niraj_Pandey.

Python3




# Python3 code to find minimum number of
# swaps to sort a binary array
 
# Function to find minimum swaps to
# sort an array of 0s and 1s.
def findMinSwaps(arr, n) :
    # Array to store count of zeroes
    noOfZeroes = [0] * n
    count = 0
     
    # Count number of zeroes
    # on right side of every one.
    noOfZeroes[n - 1] = 1 - arr[n - 1]
    for i in range(n-2, -1, -1) :
        noOfZeroes[i] = noOfZeroes[i + 1]
        if (arr[i] == 0) :
            noOfZeroes[i] = noOfZeroes[i] + 1
 
    # Count total number of swaps by adding
    # number of zeroes on right side of
    # every one.
    for i in range(0, n) :
        if (arr[i] == 1) :
            count = count + noOfZeroes[i]
 
    return count
 
# Driver code
arr = [ 0, 0, 1, 0, 1, 0, 1, 1 ]
n = len(arr)
print (findMinSwaps(arr, n))
 
# This code is contributed by Manish Shaw
# (manishshaw1)

C#




// C# code to find minimum number of
// swaps to sort a binary array
using System;
 
class GFG {
     
    static int findMinSwaps(int []arr, int n)
    {
         
        // Array to store count of zeroes
        int []noOfZeroes = new int[n];
        int i, count = 0;
 
        // Count number of zeroes
        // on right side of every one.
        noOfZeroes[n - 1] = 1 - arr[n - 1];
        for (i = n - 2; i >= 0; i--)
        {
            noOfZeroes[i] = noOfZeroes[i + 1];
            if (arr[i] == 0)
                noOfZeroes[i]++;
        }
 
        // Count total number of swaps by
        // adding number of zeroes on right
        // side of every one.
        for (i = 0; i < n; i++)
        {
            if (arr[i] == 1)
                count += noOfZeroes[i];
        }
         
        return count;
    }
     
    // Driver Code
    public static void Main()
    {
        int []ar = { 0, 0, 1, 0, 1,
                                0, 1, 1 };
                                 
        Console.WriteLine(
              findMinSwaps(ar, ar.Length));
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP code to find minimum number of
// swaps to sort a binary array
 
// Function to find minimum swaps to
// sort an array of 0s and 1s.
function findMinSwaps($arr, $n)
{
    // Array to store count of zeroes
    $noOfZeroes[$n] = array();
    $noOfZeroes = array_fill(0, $n, true);
    $count = 0;
 
    // Count number of zeroes
    // on right side of every one.
    $noOfZeroes[$n - 1] = 1 - $arr[$n - 1];
    for ($i = $n - 2; $i >= 0; $i--)
    {
        $noOfZeroes[$i] = $noOfZeroes[$i + 1];
        if ($arr[$i] == 0)
            $noOfZeroes[$i]++;
    }
 
    // Count total number of swaps by adding
    // number of zeroes on right side of every one.
    for ($i = 0; $i < $n; $i++)
    {
        if ($arr[$i] == 1)
            $count += $noOfZeroes[$i];
    }
 
    return $count;
}
 
// Driver code
$arr = array( 0, 0, 1, 0, 1, 0, 1, 1 );
$n = sizeof($arr);
echo findMinSwaps($arr, $n);
 
// This code is contributed by Sach_code
?>

Javascript




<script>
 
// JavaScript program to find minimum number of
// swaps to sort a binary array
 
    function findMinSwaps(arr, n)
    {
        // Array to store count of zeroes
        let noOfZeroes = [];
        let i, count = 0;
  
        // Count number of zeroes
        // on right side of every one.
        noOfZeroes[n - 1] = 1 - arr[n - 1];
        for (i = n - 2; i >= 0; i--)
        {
            noOfZeroes[i] = noOfZeroes[i + 1];
            if (arr[i] == 0)
                noOfZeroes[i]++;
        }
  
        // Count total number of swaps by adding number
        // of zeroes on right side of every one.
        for (i = 0; i < n; i++)
        {
            if (arr[i] == 1)
                count += noOfZeroes[i];
        }
        return count;
    }
 
// Driver Code
 
        let ar = [ 0, 0, 1, 0, 1, 0, 1, 1 ];
        document.write(findMinSwaps(ar, ar.length));
 
// This code is contribued by avijitmondal1998.
</script>
Output



3

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

Space Optimized Solution : 
An auxiliary space is not needed. We just need to start reading the list from the back and keep track of number of zeros we encounter. If we encounter a 1 the number of zeros is the number of swaps needed to put the 1 in correct place.

C++




// this code is contributed by Manu Pathria
#include <iostream>
using namespace std;
 
int minswaps(int arr[], int n)
{
    int count = 0;
    int num_unplaced_zeros = 0;
      
    for(int index=n-1;index>=0;index--)
    {
        if(arr[index] == 0)
            num_unplaced_zeros += 1;
        else
            count += num_unplaced_zeros;
    }
    return count;
}
 
// Driver Code
int main()
{
    int arr[] = {0, 0, 1, 0, 1, 0, 1, 1};
    cout<<minswaps(arr, 9);
    return 0;
}

Java




import java.io.*;
 
class GFG {
    public static int minswaps(int arr[], int n)
    {
        int count = 0;
        int num_unplaced_zeros = 0;
 
        for (int index = n - 1; index >= 0; index--)
        {
            if (arr[index] == 0)
                num_unplaced_zeros += 1;
            else
                count += num_unplaced_zeros;
        }
        return count;
    }
   
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 0, 0, 1, 0, 1, 0, 1, 1 };
        System.out.println(minswaps(arr, 9));
    }
}
 
// this code is contributed by Manu Pathria

Python3




def minswaps(arr):
    count = 0
    num_unplaced_zeros = 0
 
    for index in range(len(arr)-1, -1, -1):
        if arr[index] == 0:
            num_unplaced_zeros += 1
        else:
            count += num_unplaced_zeros
    return count
 
 
arr = [0, 0, 1, 0, 1, 0, 1, 1]
print(minswaps(arr))

Javascript




<script>
 
function minswaps(arr, n)
{
    var count = 0;
    var num_unplaced_zeros = 0;
      
    for(var index = n - 1; index >= 0; index--)
    {
        if(arr[index] == 0)
            num_unplaced_zeros += 1;
        else
            count += num_unplaced_zeros;
    }
    return count;
}
 
// Driver Code
var arr = [0, 0, 1, 0, 1, 0, 1, 1];
document.write( minswaps(arr, 9));
 
// This code is contributed by itsok.
</script>
Output
3

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

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