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Number of ways to change the XOR of two numbers by swapping the bits

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  • Last Updated : 08 Jun, 2022
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Given two binary strings s1 and s2. The XOR of them is X, the task is to find the number of ways to swap two-bit positions in string s1 such that XOR formed between new s1 and s2 is not same as X. 

Examples: 

Input: s1 = “01011”, s2 = “11001” 
Output:
swap bits of index(1-based) (1, 4), (2, 3), (3, 4), or (3, 5) such that XOR value is changed. 
Input: s1 = “011000”, s2 = “010011” 
Output: 6

Approach:  

  1. Count the number of 1’s and 0’s in s1.
  2. Traverse in the string s1, and check for two cases: 
    • 0 and 0 in s1[i] and s2[i], as replacing 0 with 1, will change the XOR value.
    • 1 and 0 in s1[i] and s2[i], as replacing 1 with 0 will change the XOR value.
  3. For the first case, the number of ways of replacement will be the number of ones-already used 1’s.
  4. For the second case, the number of ways of replacement will be the number of zeros-already used 0’s.
  5. summation of number of ways in both the cases will be the answer.

Below is the implementation of the above approach: 

C++




#include <bits/stdc++.h>
using namespace std;
 
// Function that returns the number of
// bit swaps such that xor is different
int countWays(string s1, string s2)
{
    int c1 = 0, c0 = 0;
    int n = s1.length();
 
    // traverse and count 1's and 0's
    for (int i = 0; i < n; i++) {
        if (s1[i] == '1')
            c1++;
        else
            c0++;
    }
    int used1 = 0, used0 = 0;
    int ways = 0;
 
    // traverse in the string
    for (int i = 0; i < n; i++) {
 
        // if both positions are 0
        if (s1[i] == '0' and s2[i] == '0') {
 
            // add the number of ones as
            // it will change the XOR
            ways += c1;
 
            // subtract the number of ones already used
            ways -= used1;
 
            // zeros have been used
            used0++;
        }
 
        // when 1 and 0, to change XOR, we have to
        // replace 1 by 0
        else if (s1[i] == '1' and s2[i] == '0') {
 
            // add number of 0's
            ways += c0;
 
            // subtract number of 0's already used
            ways -= used0;
 
            // count 1's used
            used1++;
        }
    }
 
    // return the answer
    return ways;
}
 
// Driver Code
int main()
{
    string s1 = "01011";
    string s2 = "11001";
 
    cout << countWays(s1, s2);
    return 0;
}

Java




// Java Program to find Number of
// ways to change the XOR of two
// numbers by swapping the bits
class GFG
{
// Function that returns the
// number of bit swaps such
// that xor is different
static int countWays(String s1,
                     String s2)
{
    int c1 = 0, c0 = 0;
    int n = s1.length();
 
    // traverse and count 1's and 0's
    for (int i = 0; i < n; i++)
    {
        if (s1.charAt(i) == '1')
            c1++;
        else
            c0++;
    }
    int used1 = 0, used0 = 0;
    int ways = 0;
 
    // traverse in the String
    for (int i = 0; i < n; i++)
    {
 
        // if both positions are 0
        if (s1.charAt(i) == '0' &&
            s2.charAt(i) == '0')
        {
 
            // add the number of ones as
            // it will change the XOR
            ways += c1;
 
            // subtract the number of
            // ones already used
            ways -= used1;
 
            // zeros have been used
            used0++;
        }
 
        // when 1 and 0, to change XOR,
        // we have to replace 1 by 0
        else if (s1.charAt(i) == '1' &&
                 s2.charAt(i) == '0')
        {
 
            // add number of 0's
            ways += c0;
 
            // subtract number of
            // 0's already used
            ways -= used0;
 
            // count 1's used
            used1++;
        }
    }
 
    // return the answer
    return ways;
}
 
// Driver Code
public static void main(String[] args)
{
    String s1 = "01011";
    String s2 = "11001";
 
    System.out.println(countWays(s1, s2));
}
}
 
// This code is contributed
// by Arnab Kundu

Python3




# Function that returns the number of
# bit swaps such that xor is different
def countWays(s1, s2):
 
    c1 = 0
    c0 = 0
    n = len(s1)
 
    # traverse and count 1's and 0's
    for i in range(0,n) :
        if (s1[i] == '1'):
            c1+=1
        else:
            c0+=1
     
    used1 = 0
    used0 = 0
    ways = 0
 
    # traverse in the string
    for i in range(0,n) :
 
        # if both positions are 0
        if (s1[i] == '0' and s2[i] == '0') :
 
            # add the number of ones as
            # it will change the XOR
            ways += c1
 
            # subtract the number of ones already used
            ways -= used1
 
            # zeros have been used
            used0+=1
         
 
        # when 1 and 0, to change XOR, we have to
        # replace 1 by 0
        elif (s1[i] == '1' and s2[i] == '0') :
 
            # add number of 0's
            ways += c0
 
            # subtract number of 0's already used
            ways -= used0
 
            # count 1's used
            used1+=1
 
    # return the answer
    return ways
 
# Driver Code
if __name__=='__main__':
    s1 = "01011"
    s2 = "11001"
    print(countWays(s1, s2))
 
# This code is contributed by Smitha Dinesh Semwal

C#




// C# Program to find Number of
// ways to change the XOR of two
// numbers by swapping the bits
using System;
 
class GFG
{
// Function that returns the
// number of bit swaps such
// that xor is different
static int countWays(String s1,
                     String s2)
{
    int c1 = 0, c0 = 0;
    int n = s1.Length;
 
    // traverse and count 1's and 0's
    for (int i = 0; i < n; i++)
    {
        if (s1[i] == '1')
            c1++;
        else
            c0++;
    }
    int used1 = 0, used0 = 0;
    int ways = 0;
 
    // traverse in the String
    for (int i = 0; i < n; i++)
    {
 
        // if both positions are 0
        if (s1[i] == '0' &&
            s2[i] == '0')
        {
 
            // add the number of ones as
            // it will change the XOR
            ways += c1;
 
            // subtract the number of
            // ones already used
            ways -= used1;
 
            // zeros have been used
            used0++;
        }
 
        // when 1 and 0, to change XOR,
        // we have to replace 1 by 0
        else if (s1[i] == '1' &&
                 s2[i] == '0')
        {
 
            // add number of 0's
            ways += c0;
 
            // subtract number of
            // 0's already used
            ways -= used0;
 
            // count 1's used
            used1++;
        }
    }
 
    // return the answer
    return ways;
}
 
// Driver Code
public static void Main(String[] args)
{
    String s1 = "01011";
    String s2 = "11001";
 
    Console.WriteLine(countWays(s1, s2));
}
}
 
// This code is contributed
// by Subhadeep Gupta

PHP




<?php
// Function that returns the 
// number of bit swaps such
// that xor is different
function countWays($s1, $s2)
{
    $c1 = 0;
    $c0 = 0;
    $n = strlen($s1);
 
    // traverse and count 1's and 0's
    for ($i = 0; $i < $n; $i++)
    {
        if ($s1[$i] == '1')
            $c1++;
        else
            $c0++;
    }
     
    $used1 = 0;
    $used0 = 0;
    $ways = 0;
     
    // traverse in the string
    for ($i = 0; $i < $n; $i++)
    {
     
        // if both positions are 0
        if ($s1[$i] == '0' and
            $s2[$i] == '0')
        {
     
            // add the number of ones as
            // it will change the XOR
            $ways += $c1;
     
            // subtract the number of
            // ones already used
            $ways -= $used1;
     
            // zeros have been used
            $used0++;
        }
     
        // when 1 and 0, to change XOR,
        // we have to replace 1 by 0
        else if ($s1[$i] == '1' and
                 $s2[$i] == '0')
        {
     
            // add number of 0's
            $ways += $c0;
     
            // subtract number of 0's
            // already used
            $ways -= $used0;
     
            // count 1's used
            $used1++;
        }
    }
     
    // return the answer
    return $ways;
}
 
// Driver Code
$s1 = "01011";
$s2 = "11001";
 
echo countWays($s1, $s2);
 
// This code is contributed
// by Shivi_Aggarwal
?>

Javascript




<script>
 
// Function that returns the number of
// bit swaps such that xor is different
function countWays(s1, s2)
{
    let c1 = 0, c0 = 0;
    let n = s1.length;
 
    // traverse and count 1's and 0's
    for (let i = 0; i < n; i++) {
        if (s1[i] == '1')
            c1++;
        else
            c0++;
    }
    let used1 = 0, used0 = 0;
    let ways = 0;
 
    // traverse in the string
    for (let i = 0; i < n; i++) {
 
        // if both positions are 0
        if (s1[i] == '0' && s2[i] == '0') {
 
            // add the number of ones as
            // it will change the XOR
            ways += c1;
 
            // subtract the number of ones already used
            ways -= used1;
 
            // zeros have been used
            used0++;
        }
 
        // when 1 and 0, to change XOR, we have to
        // replace 1 by 0
        else if (s1[i] == '1' && s2[i] == '0') {
 
            // add number of 0's
            ways += c0;
 
            // subtract number of 0's already used
            ways -= used0;
 
            // count 1's used
            used1++;
        }
    }
 
    // return the answer
    return ways;
}
 
// Driver Code
    let s1 = "01011";
    let s2 = "11001";
 
    document.write(countWays(s1, s2));
 
</script>

Output:  

4

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


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