XOR of two numbers after making length of their binary representations equal

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Given two numbers say a and b. Print their XOR after making the lengths of their binary representation equal by adding trailing zeros to the binary representation of smaller one.
Examples :

Input : a = 13, b = 5 
Output : 7
Explanation : Binary representation of 13 is 1101 and 
of 5 is 101. As the length of "101" is smaller,
so add a '0' to it making it "1010', to make 
the length of binary representations equal. 
XOR of 1010 and 1101 gives 0111 which is 7.

Input : a = 7, b = 5 
Output : 2
Explanation : Since the length of binary representations
of 7 i.e, 111 and 5 i.e, 101 are same, hence simply
print XOR of a and b.

Approach : Count the number of bits in binary representation of smaller number out of a and b. If the number of bits in smaller number(say a) exceeds to that of larger number(say b), then apply left shift to the smaller number by the number of exceeding bits, i.e, a = a<<(exceeding bits). After applying left shift, trailing zeroes will be added at the end of binary representation of smaller number to make the number of bits in binary representation of both the numbers equal. XOR both the binary representations to get the final result.
Below is the implementation of above method :

C++

// C++ implementation to return 
// XOR of two numbers after making
// length of their binary representation same
#include <bits/stdc++.h>
using namespace std;
 
// function to count the number 
// of bits in binary representation 
// of an integer
int count(int n)
{
    // initialize count
    int c = 0;
     
    // count till n is non zero
    while (n)
    {
        c++;
         
        // right shift by 1 
        // i.e, divide by 2
        n = n>>1;
    }
    return c;
}
 
// function to calculate the xor of
// two numbers by adding trailing 
// zeros to the number having less number 
// of bits in its binary representation.
int XOR(int a, int b)
{
    // stores the minimum and maximum 
    int c = min(a,b);
    int d = max(a,b);
     
    // left shift if the number of bits
    // are less in binary representation
    if (count(c) < count(d))
       c = c << ( count(d) - count(c) ); 
     
    return (c^d); 
}
 
// driver code to check the above function 
int main()
{   
    int a = 13, b = 5;
    cout << XOR(a,b);    
    return 0;
}

Java

// Java implementation to return 
// XOR of two numbers after making
// length of their binary representation same
import java.io.*;
 
class GFG {
     
    // function to count the number 
    // of bits in binary representation 
    // of an integer
    static int count(int n)
    {
        // initialize count
        int c = 0;
         
        // count till n is non zero
        while (n != 0)
        {
            c++;
             
            // right shift by 1 
            // i.e, divide by 2
            n = n >> 1;
        }
        return c;
    }
     
    // function to calculate the xor of
    // two numbers by adding trailing 
    // zeros to the number having less number 
    // of bits in its binary representation.
    static int XOR(int a, int b)
    {
        // stores the minimum and maximum 
        int c = Math.min(a, b);
        int d = Math.max(a, b);
         
        // left shift if the number of bits
        // are less in binary representation
        if (count(c) < count(d))
        c = c << ( count(d) - count(c) ); 
         
        return (c ^ d); 
    }
     
    // driver code to check the above function 
    public static void main(String args[])
    { 
        int a = 13, b = 5;
        System.out.println(XOR(a, b));
    }
}
 
// This code is contributed by Nikita Tiwari.

Python3

# Python 3 implementation to return XOR
# of two numbers after making length
# of their binary representation same
 
# Function to count the number of bits
# in binary representation of an integer
def count(n) :
     
    # initialize count
    c = 0
     
    # count till n is non zero
    while (n != 0) :
        c += 1
         
        # right shift by 1 
        # i.e, divide by 2
        n = n >> 1
         
    return c
     
# Function to calculate the xor of
# two numbers by adding trailing 
# zeros to the number having less number 
# of bits in its binary representation.
def XOR(a, b) :
     
    # stores the minimum and maximum 
    c = min(a, b)
    d = max(a, b)
     
    # left shift if the number of bits
    # are less in binary representation
    if (count(c) < count(d)) :
        c = c << ( count(d) - count(c) ) 
     
    return (c^d)
 
# Driver Code
a = 13; b = 5
print(XOR(a, b))
 
 
# This code is contributed by Nikita Tiwari.

C#

// C# implementation to return XOR of two 
// numbers after making length of their 
// binary representation same
using System;
 
class GFG {
     
    // function to count the number 
    // of bits in binary representation 
    // of an integer
    static int count(int n)
    {
         
        // initialize count
        int c = 0;
         
        // count till n is non zero
        while (n != 0)
        {
             
            c++;
             
            // right shift by 1 
            // i.e, divide by 2
            n = n >> 1;
        }
         
        return c;
    }
     
    // function to calculate the xor of
    // two numbers by adding trailing 
    // zeros to the number having less number 
    // of bits in its binary representation.
    static int XOR(int a, int b)
    {
         
        // stores the minimum and maximum 
        int c = Math.Min(a, b);
        int d = Math.Max(a, b);
         
        // left shift if the number of bits
        // are less in binary representation
        if (count(c) < count(d))
        c = c << ( count(d) - count(c) ); 
         
        return (c ^ d); 
    }
     
    // driver code to check the above function 
    public static void Main()
    { 
        int a = 13, b = 5;
         
        Console.WriteLine(XOR(a, b));
    }
}
 
// This code is contributed by vt_m.

PHP

<?php
// php implementation to return XOR
// of two numbers after making
// length of their binary 
// representation same
 
// function to count the number 
// of bits in binary representation 
// of an integer
function count1($n)
{
     
    // initialize count
    $c = 0;
     
    // count till n is
    // non zero
    while ($n)
    {
        $c++;
         
        // right shift by 1 
        // i.e, divide by 2
        $n = $n>>1;
    }
    return $c;
}
 
// function to calculate the xor of
// two numbers by adding trailing 
// zeros to the number having less number 
// of bits in its binary representation.
function XOR1($a, $b)
{
     
    // stores the minimum 
    // and maximum 
    $c = min($a,$b);
    $d = max($a,$b);
     
    // left shift if the number of bits
    // are less in binary representation
    if (count1($c) < count1($d))
    $c = $c << ( count1($d) - count1($c) ); 
     
    return ($c^$d); 
}
 
    // Driver Code 
    $a = 13;
    $b = 5;
    echo XOR1($a, $b); 
 
// This code is contributed by mits 
?>

Javascript

<script>
 
// JavaScript program to return 
// XOR of two numbers after making
// length of their binary representation same
 
    // function to count the number 
    // of bits in binary representation 
    // of an integer
    function count(n)
    {
        // initialize count
        let c = 0;
           
        // count till n is non zero
        while (n != 0)
        {
            c++;
               
            // right shift by 1 
            // i.e, divide by 2
            n = n >> 1;
        }
        return c;
    }
       
    // function to calculate the xor of
    // two numbers by adding trailing 
    // zeros to the number having less number 
    // of bits in its binary representation.
    function XOR(a, b)
    {
        // stores the minimum and maximum 
        let c = Math.min(a, b);
        let d = Math.max(a, b);
           
        // left shift if the number of bits
        // are less in binary representation
        if (count(c) < count(d))
        c = c << ( count(d) - count(c) ); 
           
        return (c ^ d); 
    }
  
// Driver code
 
        let a = 13, b = 5;
        document.write(XOR(a, b));
 
</script>

Output :

Time Complexity : O(log₂n)
Auxiliary Space : O(1)

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