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Minimum swaps required to convert one binary string to another

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  • Difficulty Level : Medium
  • Last Updated : 06 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)


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