Intercambios mínimos necesarios para convertir una string binaria en otra

Dadas dos strings binarias M y N de igual longitud, la tarea es encontrar un número mínimo de operaciones (intercambios) necesarias para convertir la string N en M.
Ejemplos: 
 

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

Enfoque: inicialice el contador e itere sobre la M de modo que si se encuentran elementos no iguales en ambas strings binarias, incremente el contador. Al final, si el contador es par, imprima el resultado/2 porque para un intercambio, dos elementos no son idénticos. 
Supongamos que S1 = «10» y S2 = «01» , por lo que dos pares no son idénticos, el conteo = 2 y como el conteo es par, el número de intercambios es conteo/2, es decir, 1. El conteo par determina que hay posibilidades de intercambiar los elementos.
A continuación se muestra la implementación del enfoque anterior: 
 

// 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 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);
    }
}

# 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# 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 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
?>

<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>

3

Complejidad de tiempo: O(n)