Number of ways to change the XOR of two numbers by swapping the bits
Last Updated :
08 Jun, 2022
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: 4
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:
- Count the number of 1’s and 0’s in s1.
- 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.
- For the first case, the number of ways of replacement will be the number of ones-already used 1’s.
- For the second case, the number of ways of replacement will be the number of zeros-already used 0’s.
- 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;
int countWays(string s1, string s2)
{
int c1 = 0, c0 = 0;
int n = s1.length();
for ( int i = 0; i < n; i++) {
if (s1[i] == '1' )
c1++;
else
c0++;
}
int used1 = 0, used0 = 0;
int ways = 0;
for ( int i = 0; i < n; i++) {
if (s1[i] == '0' and s2[i] == '0' ) {
ways += c1;
ways -= used1;
used0++;
}
else if (s1[i] == '1' and s2[i] == '0' ) {
ways += c0;
ways -= used0;
used1++;
}
}
return ways;
}
int main()
{
string s1 = "01011" ;
string s2 = "11001" ;
cout << countWays(s1, s2);
return 0;
}
|
Java
class GFG
{
static int countWays(String s1,
String s2)
{
int c1 = 0 , c0 = 0 ;
int n = s1.length();
for ( int i = 0 ; i < n; i++)
{
if (s1.charAt(i) == '1' )
c1++;
else
c0++;
}
int used1 = 0 , used0 = 0 ;
int ways = 0 ;
for ( int i = 0 ; i < n; i++)
{
if (s1.charAt(i) == '0' &&
s2.charAt(i) == '0' )
{
ways += c1;
ways -= used1;
used0++;
}
else if (s1.charAt(i) == '1' &&
s2.charAt(i) == '0' )
{
ways += c0;
ways -= used0;
used1++;
}
}
return ways;
}
public static void main(String[] args)
{
String s1 = "01011" ;
String s2 = "11001" ;
System.out.println(countWays(s1, s2));
}
}
|
Python3
def countWays(s1, s2):
c1 = 0
c0 = 0
n = len (s1)
for i in range ( 0 ,n) :
if (s1[i] = = '1' ):
c1 + = 1
else :
c0 + = 1
used1 = 0
used0 = 0
ways = 0
for i in range ( 0 ,n) :
if (s1[i] = = '0' and s2[i] = = '0' ) :
ways + = c1
ways - = used1
used0 + = 1
elif (s1[i] = = '1' and s2[i] = = '0' ) :
ways + = c0
ways - = used0
used1 + = 1
return ways
if __name__ = = '__main__' :
s1 = "01011"
s2 = "11001"
print (countWays(s1, s2))
|
C#
using System;
class GFG
{
static int countWays(String s1,
String s2)
{
int c1 = 0, c0 = 0;
int n = s1.Length;
for ( int i = 0; i < n; i++)
{
if (s1[i] == '1' )
c1++;
else
c0++;
}
int used1 = 0, used0 = 0;
int ways = 0;
for ( int i = 0; i < n; i++)
{
if (s1[i] == '0' &&
s2[i] == '0' )
{
ways += c1;
ways -= used1;
used0++;
}
else if (s1[i] == '1' &&
s2[i] == '0' )
{
ways += c0;
ways -= used0;
used1++;
}
}
return ways;
}
public static void Main(String[] args)
{
String s1 = "01011" ;
String s2 = "11001" ;
Console.WriteLine(countWays(s1, s2));
}
}
|
PHP
<?php
function countWays( $s1 , $s2 )
{
$c1 = 0;
$c0 = 0;
$n = strlen ( $s1 );
for ( $i = 0; $i < $n ; $i ++)
{
if ( $s1 [ $i ] == '1' )
$c1 ++;
else
$c0 ++;
}
$used1 = 0;
$used0 = 0;
$ways = 0;
for ( $i = 0; $i < $n ; $i ++)
{
if ( $s1 [ $i ] == '0' and
$s2 [ $i ] == '0' )
{
$ways += $c1 ;
$ways -= $used1 ;
$used0 ++;
}
else if ( $s1 [ $i ] == '1' and
$s2 [ $i ] == '0' )
{
$ways += $c0 ;
$ways -= $used0 ;
$used1 ++;
}
}
return $ways ;
}
$s1 = "01011" ;
$s2 = "11001" ;
echo countWays( $s1 , $s2 );
?>
|
Javascript
<script>
function countWays(s1, s2)
{
let c1 = 0, c0 = 0;
let n = s1.length;
for (let i = 0; i < n; i++) {
if (s1[i] == '1' )
c1++;
else
c0++;
}
let used1 = 0, used0 = 0;
let ways = 0;
for (let i = 0; i < n; i++) {
if (s1[i] == '0' && s2[i] == '0' ) {
ways += c1;
ways -= used1;
used0++;
}
else if (s1[i] == '1' && s2[i] == '0' ) {
ways += c0;
ways -= used0;
used1++;
}
}
return ways;
}
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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