Open In App

Minimum adjacent swaps required to Sort Binary array

Last Updated : 18 Feb, 2023
Improve
Improve
Like Article
Like
Save
Share
Report

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.

Steps to solve this problem:

1. Declare an integer array noOfZeroes of size n and initialize it with zeros using memset function.

2. Initialize a variable count to keep track of the total number of swaps required.

3. Initialize another variable i to traverse the array from the end.

4. Calculate the number of zeroes on the right side of every one and store it in the noOfZeroes array. This is done using a loop which starts from the second last element of the input array and goes till the first element. For every iteration, the value of noOfZeroes[i] is updated with the value of noOfZeroes[i+1]. If the current element of the input array is 0, the value of noOfZeroes[i] is incremented.

5. A loop is run over the input array to count the total number of swaps required. For every iteration, if the current element of the input array is 1, the value of count is incremented by noOfZeroes[i].

6. The function returns the value of count, which represents the minimum number of swaps required to make all ones together.

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 contributed 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.

Implementation:

C++




#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;
        if(arr[index] == 1)
            count += num_unplaced_zeros;
    }
    return count;
}
 
// Driver Code
int main()
{
    int arr[] = {0, 0, 1, 0, 1, 0, 1, 1};
    cout<<minswaps(arr, 8);
    return 0;
}
 
// this code is contributed by Manu Pathria


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, 8));
    }
}
 
// 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))


C#




using System;
class GFG {
     
    static int minswaps(int[] arr, int n)
    {
        int count = 0;
        int num_unplaced_zeros = 0;
  
        for (int index = n - 2; index >= 0; index--)
        {
            if (arr[index] == 0)
                num_unplaced_zeros += 1;
            else
                count += num_unplaced_zeros;
        }
        return count;
    }
     
  static void Main() {
    int[] arr = { 0, 0, 1, 0, 1, 0, 1, 1 };
    Console.WriteLine(minswaps(arr, 8));
  }
}
 
// This code is contributed by divyeshrabadiya07.


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, 8));
 
// This code is contributed by itsok.
</script>


Output

3

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



Like Article
Suggest improvement
Previous
Next
Share your thoughts in the comments

Similar Reads