Number of ways to swap two bit of s1 so that bitwise OR of s1 and s2 changes

Given two binary numbers strings and of length . Find the number of ways of swapping two bits in s1(only s1 not s2) so that bit-wise OR of these two numbers s1 and s2 are changed.

Note: The length of both string must be equal, you can take leading zeros in case of different length.

Example:

Input: s1 = "01011", s2 = "11001"
Output: 4
Explanation:
You can swap the bit of s1 at indexed:
(1, 4), (2, 3), (3, 4) and (3, 5)
there are 4 ways possible.

Input: s1 = "011000", s2 = "010011"
Output: 6

Approach: Initialize a variable result as zero that stores number of ways to swap bits of s1 so that bitwise OR of s1 and s2 changes. Initialize four variables say , , , and as zero.

Traverse both strings from left side and check for the below conditions to increment values of the 
variable declared above: 

• If the current bit of both s1 and s2 are zero, increment c.
• If current bit of s2 is zero and s1 is one, increment d.
• If current bit of s2 is one and s1 is zero, increment a.
• If current bit of both s1 and s2 is one, increment b.

Update the result by (a*d) + (b*c) + (c*d) and return the result.

Below is the implementation of the above approach:  

C++

// C++ program to find no of ways
// to swap bits of s1 so that
// bitwise OR of s1 and s2 changes
#include <iostream>
using namespace std;

// Function to find number of ways
int countWays(string s1, string s2, int n)
{

    int a, b, c, d;
    a = b = c = d = 0;

    // initialise result that store
    // No. of swaps required
    int result = 0;

    // Traverse both strings and check
    // the bits as explained
    for (int i = 0; i < n; i++) {
        if (s2[i] == '0') {
            if (s1[i] == '0') {
                c++;
            }
            else {
                d++;
            }
        }
        else {
            if (s1[i] == '0') {
                a++;
            }
            else {
                b++;
            }
        }
    }

    // calculate result
    result = a * d + b * c + c * d;

    return result;
}

// Driver code
int main()
{
    int n = 5;
    string s1 = "01011";
    string s2 = "11001";

    cout << countWays(s1, s2, n);

    return 0;
}

Java

// Java program to find no of ways
// to swap bits of s1 so that
// bitwise OR of s1 and s2 changes
import java.io.*;

class GFG {



// Function to find number of ways
static int countWays(String s1, String s2, int n)
{

    int a, b, c, d;
    a = b = c = d = 0;

    // initialise result that store
    // No. of swaps required
    int result = 0;

    // Traverse both strings and check
    // the bits as explained
    for (int i = 0; i < n; i++) {
        if (s2.charAt(i) == '0') {
            if (s1.charAt(i) == '0') {
                c++;
            }
            else {
                d++;
            }
        }
        else {
            if (s1.charAt(i) == '0') {
                a++;
            }
            else {
                b++;
            }
        }
    }

    // calculate result
    result = a * d + b * c + c * d;

    return result;
}

// Driver code

    public static void main (String[] args) {
        int n = 5;
    String s1 = "01011";
    String s2 = "11001";

    System.out.println(countWays(s1, s2, n));
    }
}
// This code is contributed by shs..

Python3

# Python 3 program to find no of ways
# to swap bits of s1 so that
# bitwise OR of s1 and s2 changes

# Function to find number of ways
def countWays(s1, s2, n):
    a = b = c = d = 0

    # initialise result that store
    # No. of swaps required
    result = 0;

    # Traverse both strings and check
    # the bits as explained
    for i in range(0, n, 1):
        if (s2[i] == '0'):
            if (s1[i] == '0'):
                c += 1;

            else:
                d += 1
        else:
            if (s1[i] == '0'):
                a += 1

            else:
                b += 1

    # calculate result
    result = a * d + b * c + c * d

    return result

# Driver code
if __name__ == '__main__':
    n = 5
    s1 = "01011"
    s2 = "11001"

    print(countWays(s1, s2, n))

# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find no of ways
// to swap bits of s1 so that
// bitwise OR of s1 and s2 changes
using System;

class GFG {



// Function to find number of ways
static int countWays(string s1, string s2, int n)
{

    int a, b, c, d;
    a = b = c = d = 0;

    // initialise result that store
    // No. of swaps required
    int result = 0;

    // Traverse both strings and check
    // the bits as explained
    for (int i = 0; i < n; i++) {
        if (s2[i] == '0') {
            if (s1[i] == '0') {
                c++;
            }
            else {
                d++;
            }
        }
        else {
            if (s1[i] == '0') {
                a++;
            }
            else {
                b++;
            }
        }
    }

    // calculate result
    result = a * d + b * c + c * d;

    return result;
}

// Driver code

    public static void Main () {
        int n = 5;
    string s1 = "01011";
    string s2 = "11001";

    Console.WriteLine(countWays(s1, s2, n));
    }
}
// This code is contributed by shs..

PHP

<?php
// PHP program to find no of ways
// to swap bits of s1 so that
// bitwise OR of s1 and s2 changes

// Function to find number of ways
function countWays($s1, $s2, $n)
{
    $a = $b = $c = $d = 0;

    // initialise result that store
    // No. of swaps required
    $result = 0;

    // Traverse both strings and check
    // the bits as explained
    for ($i = 0; $i < $n; $i++)
    {
        if ($s2[$i] == '0')
        {
            if ($s1[$i] == '0')
            {
                $c++;
            }
            else
            {
                $d++;
            }
        }
        else
        {
            if ($s1[$i] == '0')
            {
                $a++;
            }
            else
            {
                $b++;
            }
        }
    }

    // calculate result
    $result = $a * $d + $b *
              $c + $c * $d;

    return $result;
}

// Driver code
$n = 5;
$s1 = "01011";
$s2 = "11001";
echo countWays($s1, $s2, $n);

// This code is contributed by ajit
?>

Javascript

<script>
// javascript program to find no of ways
// to swap bits of s1 so that
// bitwise OR of s1 and s2 changes
 // Function to find number of ways
function countWays(s1, s2 , n)
{

    var a, b, c, d;
    a = b = c = d = 0;

    // initialise result that store
    // No. of swaps required
    var result = 0;

    // Traverse both strings and check
    // the bits as explained
    for (i = 0; i < n; i++) {
        if (s2.charAt(i) == '0') {
            if (s1.charAt(i) == '0') {
                c++;
            }
            else {
                d++;
            }
        }
        else {
            if (s1.charAt(i) == '0') {
                a++;
            }
            else {
                b++;
            }
        }
    }

    // calculate result
    result = a * d + b * c + c * d;

    return result;
}

// Driver code
var n = 5;
var s1 = "01011";
var s2 = "11001";

document.write(countWays(s1, s2, n));

// This code is contributed by Princi Singh
</script>

Output

4

Complexity y Analysis:

• Time Complexity: O(N)
• Auxiliary Space: O(1)