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Minimum swaps required to convert one binary string to another

Last Updated : 28 Sep, 2022
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Given two binary string M and N of equal length, the task is to find a minimum number of operations (swaps) required to convert string N to M.

Examples: 

Input: str1 = "1101", str2 = "1110"
Output: 1
Swap last and second last element in the binary string, 
so that it become 1101

Input: str1 = "1110000", str2 = "0001101"
Output: 3

Approach:

Initialize the counter and Iterate over the M such that if any non-equal elements found in both binary strings, increment the counter. In the end, if the counter is even then print the result/2 because for one swap two elements are non-identical. 

Suppose S1 = "10" and S2 = "01", so two pairs are non-identical, the count = 2 and as the count is even, so number of swaps are count/2, i.e. 1. Even count determines that there are chances to swap the elements.

Below is the implementation of the above approach: 

C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;

// Method to count swaps
void minSwaps(string str1, string str2)
{
    // Initialize the count
    int count = 0;

    // Iterate the loop with str1 length
    for (int i = 0; i < str1.length(); i++) {

        // If any non-equal elements are found
        // increment the counter
        if (str1[i] != str2[i])
            count++;
    }

    // If counter is even print the swap
    if (count % 2 == 0)
        cout << count / 2;
    else
        cout << "Not Possible";
}

// Driver code
int main()
{
    // Take two input
    string binaryString1 = "1110000";
    string binaryString2 = "0001101";

    // Call the method
    minSwaps(binaryString1, binaryString2);

    return 0;
}
Java
// Java Program to count minimum number of swap
// required to make string N to M
public class GFG {

    // Method to count swaps
    static void minSwaps(String str1, String str2)
    {
        // Initialize the count
        int count = 0;

        // Iterate the loop with str1 length
        for (int i = 0; i < str1.length(); i++) {

            // If any non-equal elements are found
            // increment the counter
            if (str1.charAt(i) != str2.charAt(i))
                count++;
        }

        // If counter is even print the swap
        if (count % 2 == 0)
            System.out.println(count / 2);
        else
            System.out.println("Not Possible");
    }

    // Driver Code
    public static void main(String args[])
    {
        // Take two input
        String binaryString1 = "1110000";
        String binaryString2 = "0001101";

        // Call the method
        minSwaps(binaryString1, binaryString2);
    }
}
Python 3
# Python3 implementation of 
# the above approach 

# function to count swaps
def minSwaps(str1, str2) :

    # Initialize the count
    count = 0

    # Iterate the loop with 
    # length of str1
    for i in range(len(str1)) :

        # If any non-equal elements are 
        # found increment the counter 
        if str1[i] != str2[i] :
            count += 1

    # If counter is even print 
    # the swap 
    if count % 2 == 0 :
        print(count // 2)
    else :
        print("Not Possible")


# Driver code
if __name__ == "__main__" :

    # Take two input
    binaryString1 = "1110000"
    binaryString2 = "0001101"

    # Call the function
    minSwaps( binaryString1, binaryString2)

# This code is contributed by ANKITRAI1
C#
// C# Program to count minimum number of swap
// required to make string N to M
using System;
class GFG
{

// Method to count swaps
static void minSwaps(string str1, string str2)
{
    // Initialize the count
    int count = 0;

    // Iterate the loop with str1 length
    for (int i = 0; i < str1.Length; i++) {

        // If any non-equal elements are found
        // increment the counter
        if (str1[i] != str2[i])
            count++;
    }

    // If counter is even print the swap
    if (count % 2 == 0)
        Console.WriteLine(count / 2);
    else
        Console.WriteLine("Not Possible");
}

// Driver Code
public static void Main()
{
    // Take two input
    string binaryString1 = "1110000";
    string binaryString2 = "0001101";

    // Call the method
    minSwaps(binaryString1, binaryString2);
}
}

// This code is contributed 
// by Akanksha Rai(Abby_akku)
PHP
<?php
// PHP implementation of the above 
// approach 

// Method to count swaps 
function minSwaps($str1, $str2) 
{ 
    // Initialize the count 
    $count = 0; 
    
    // Iterate the loop with str1 length 
    for ($i = 0; $i < strlen($str1); $i++)
    { 

        // If any non-equal elements are 
        // found increment the counter 
        if ($str1[$i] != $str2[$i]) 
            $count++; 
    } 

    // If counter is even print the swap 
    if ($count % 2 == 0) 
        echo ($count / 2); 
    else
        echo "Not Possible"; 
} 

// Driver code 

// Take two input 
$binaryString1 = "1110000"; 
$binaryString2 = "0001101"; 

// Call the method 
minSwaps($binaryString1, $binaryString2); 

// This code is contributed
// by Sach_Code
?>
JavaScript
<script>
      // JavaScript Program to count minimum number of swap
      // required to make string N to M
      // Method to count swaps
      function minSwaps(str1, str2) {
        // Initialize the count
        var count = 0;

        // Iterate the loop with str1 length
        for (var i = 0; i < str1.length; i++) {
          // If any non-equal elements are found
          // increment the counter
          if (str1[i] !== str2[i]) count++;
        }

        // If counter is even print the swap
        if (count % 2 === 0) document.write(count / 2);
        else document.write("Not Possible");
      }

      // Driver Code
      // Take two input
      var binaryString1 = "1110000";
      var binaryString2 = "0001101";

      // Call the method
      minSwaps(binaryString1, binaryString2);
    </script>

Output
3

Time Complexity: O(n)

Auxiliary Space: O(1) it is using constant space for variables


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