Program to find xor of two numbers in python

Given two integers, find XOR of them without using the XOR operator, i.e., without using ^ in C/C++.

Examples :  

Input:  x = 1, y = 2
Output: 3

Input:  x = 3, y = 5
Output: 6

A Simple Solution is to traverse all bits one by one. For every pair of bits, check if both are the same, set the corresponding bit like 0 in output, otherwise set it as 1. 

C++

#include

using namespace std;

int myXOR(int x, int y)

{

    int res = 0;

    for (int i = 31; i >= 0; i--)                    

    {

       bool b1 = x & (1 << i);

       bool b2 = y & (1 << i);

        bool xoredBit = (b1 & b2) ? 0 : (b1 | b2);         

        res <<= 1;

        res |= xoredBit;

    }

    return res;

}

int main()

{

   int x = 3, y = 5;

   cout << "XOR is " << myXOR(x, y);

   return 0;

}

C

#include

#include //to use bool

int myXOR(int x, int y)

{

    int res = 0;

    for (int i = 31; i >= 0; i--)

    {

        bool b1 = x & (1 << i);

        bool b2 = y & (1 << i);

        bool xoredBit = (b1 & b2) ? 0 : (b1 | b2);

        res <<= 1;

        res |= xoredBit;

    }

    return res;

}

int main()

{

    int x = 3, y = 5;

    printf("XOR is %d\n", myXOR(x, y));

    return 0;

}

Java

import java.io.*;

class GFG{

static int myXOR(int x, int y)

{

    int res = 0;

    for(int i = 31; i >= 0; i--)                    

    {

        int b1 = ((x & (1 << i)) == 0 ) ? 0 : 1; 

        int b2 = ((y & (1 << i)) == 0 ) ? 0 : 1; 

        int xoredBit = ((b1 & b2) != 0) ? 0 : (b1 | b2);         

        res <<= 1;

        res |= xoredBit;

    }

    return res;

}

public static void main (String[] args)

{

    int x = 3, y = 5;

    System.out.println("XOR is " + myXOR(x, y));

}

}

Python3

def myXOR(x, y):

    res = 0

    for i in range(31, -1, -1):

        b1 = x & (1 << i)

        b2 = y & (1 << i)

        b1 = min(b1, 1)

        b2 = min(b2, 1)

        xoredBit = 0

        if (b1 & b2):

            xoredBit = 0

        else:

            xoredBit = (b1 | b2)

        res <<= 1;

        res |= xoredBit

    return res

x = 3

y = 5

print("XOR is", myXOR(x, y))

C#

using System;

class GFG{

static int myXOR(int x,

                 int y)

{   

  int res = 0;

  for(int i = 31; i >= 0; i--)                    

  {

    int b1 = ((x & (1 << i)) == 0 ) ?

               0 : 1; 

    int b2 = ((y & (1 << i)) == 0 ) ?

               0 : 1; 

    int xoredBit = ((b1 & b2) != 0) ?

                     0 : (b1 | b2);         

    res <<= 1;

    res |= xoredBit;

  }

  return res;

}

public static void Main(String[] args)

{

  int x = 3, y = 5;

  Console.WriteLine("XOR is " +

                     myXOR(x, y));

}

}

Javascript

Time Complexity: O(num), where num is the number of bits in the maximum of the two numbers.

Space Complexity: O(1)

Thanks to Utkarsh Trivedi for suggesting this solution.
 
A Better Solution can find XOR without using a loop. 
1) Find bitwise OR of x and y (Result has set bits where either x has set or y has set bit). OR of x = 3 (011) and y = 5 (101) is 7 (111)
2) To remove extra set bits find places where both x and y have set bits. The value of the expression “~x | ~y” has 0 bits wherever x and y both have set bits.
3) bitwise AND of “(x | y)” and “~x | ~y” produces the required result.

Below is the implementation. 

C++

#include

using namespace std;

int myXOR(int x, int y)

{

   return (x | y) & (~x | ~y);

}

int main()

{

   int x = 3, y = 5;

   cout << "XOR is " << myXOR(x, y);

   return 0;

}

Java

import java.io.*;

class GFG

{

static int myXOR(int x, int y)

{

    return (x | y) &

           (~x | ~y);

}

public static void main (String[] args)

{

    int x = 3, y = 5;

    System.out.println("XOR is "+

                      (myXOR(x, y)));

}

}

Python3

def myXOR(x, y):

    return ((x | y) &

            (~x | ~y))

x = 3

y = 5

print("XOR is" ,

       myXOR(x, y))

C#

using System;

class GFG

{

static int myXOR(int x, int y)

{

    return (x | y) &

           (~x | ~y);

}

static public void Main ()

{

    int x = 3, y = 5;

    Console.WriteLine("XOR is "+

                     (myXOR(x, y)));

}

}

PHP

function myXOR($x, $y)

{

    return ($x | $y) & (~$x | ~$y);

}

$x = 3;

$y = 5;

echo "XOR is " , myXOR($x, $y);

?>

Javascript

Time Complexity: O(1)

Space Complexity: O(1)

Thanks to jitu_the_best for suggesting this solution. 

Alternate Solution : 

C++

#include

using namespace std;

int myXOR(int x, int y)

{

   return (x & (~y)) | ((~x )& y);

}

int main()

{

   int x = 3, y = 5;

   cout << "XOR is " << myXOR(x, y);

   return 0;

}

Java

import java.io.*;

class GFG

{

static int myXOR(int x, int y)

{

return (x & (~y)) | ((~x )& y);

}

public static void main (String[] args)

{

int x = 3, y = 5;

System.out.println("XOR is "+

                      (myXOR(x, y)));

}

}

Python3

def myXOR(x, y):

    return (x & (~y)) | ((~x )& y)

x = 3

y = 5

print("XOR is" ,

    myXOR(x, y))

C#

using System;

class GFG{

static int myXOR(int x, int y)

{

    return (x & (~y)) | ((~x )& y);