Number of mismatching bits in the binary representation of two integers

Given two integers(less than 2^31) A and B. The task is to find the number of bits that are different in their binary representation.

Examples:

Input :  A = 12, B = 15
Output : Number of different bits : 2
Explanation: The binary representation of 
12 is 1100 and 15 is 1111.
So, the number of different bits are 2.

Input : A = 3, B = 16
Output : Number of different bits : 3


Approach:

  • Run a loop from ‘0’ to ’31’ and right shift the bits of A and B by ‘i’ places, then check whether the bit at the ‘0th’ position is different.
  • If the bit is different then increase the count.
  • As the numbers are less than 2^31, we only have to run the loop ’32’ times i.e. from ‘0’ to ’31’.
  • We can get the 1st bit if we bitwise AND the number by 1.
  • At the end of the loop display the count.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// compute number of different bits
void solve(int A, int B)
{
    int count = 0;
  
    // since, the numbers are less than 2^31
    // run the loop from '0' to '31' only
    for (int i = 0; i < 32; i++) {
  
        // right shift both the numbers by 'i' and
        // check if the bit at the 0th position is different
        if (((A >> i) & 1) != ((B >> i) & 1)) {
            count++;
        }
    }
  
    cout << "Number of different bits : " << count << endl;
}
  
// Driver code
int main()
{
    int A = 12, B = 15;
  
    // find number of different bits
    solve(A, B);
  
    return 0;
}

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Java

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// Java implementation of the approach
  
import java.io.*;
  
class GFG {
  
// compute number of different bits
static void solve(int A, int B)
{
    int count = 0;
  
    // since, the numbers are less than 2^31
    // run the loop from '0' to '31' only
    for (int i = 0; i < 32; i++) {
  
        // right shift both the numbers by 'i' and
        // check if the bit at the 0th position is different
        if (((A >> i) & 1) != ((B >> i) & 1)) {
            count++;
        }
    }
  
    System.out.println("Number of different bits : " + count);
}
  
// Driver code
  
  
    public static void main (String[] args) {
        int A = 12, B = 15;
  
    // find number of different bits
    solve(A, B);
  
    }
}
// this code is contributed by anuj_67..

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Python3

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# Python3 implementation of the approach
  
# compute number of different bits
def solve( A,  B):
   
    count = 0 
  
    # since, the numbers are less than 2^31
    # run the loop from '0' to '31' only
    for i in range(0,32):
  
        # right shift both the numbers by 'i' and
        # check if the bit at the 0th position is different
        if ((( A >>  i) & 1) != (( B >>  i) & 1)): 
             count=count+1
           
       
  
    print("Number of different bits :",count) 
   
  
# Driver code
A = 12 
B = 15 
  
# find number of different bits
solve( A,  B) 
  
  
# This code is contributed by ihritik
  

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C#

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// C# implementation of the approach
  
using System;
class GFG
{
    // compute number of different bits
    static void solve(int A, int B)
    {
        int count = 0;
      
        // since, the numbers are less than 2^31
        // run the loop from '0' to '31' only
        for (int i = 0; i < 32; i++) {
      
            // right shift both the numbers by 'i' and
            // check if the bit at the 0th position is different
            if (((A >> i) & 1) != ((B >> i) & 1)) {
                count++;
            }
        }
      
        Console.WriteLine("Number of different bits : " + count);
    }
      
    // Driver code
    public static void  Main()
    {
        int A = 12, B = 15;
      
        // find number of different bits
        solve(A, B);
      
    }
  
}
  
// This code is contributed by ihritik

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PHP

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<?php
// PHP implementation of the approach
  
// compute number of different bits
function solve($A, $B)
{
    $count = 0;
  
    // since, the numbers are less than 2^31
    // run the loop from '0' to '31' only
    for ($i = 0; $i < 32; $i++) {
  
        // right shift both the numbers by 'i' and
        // check if the bit at the 0th position is different
        if ((($A >> $i) & 1) != (($B >> $i) & 1)) {
            $count++;
        }
    }
  
    echo "Number of different bits : $count";
}
  
// Driver code
$A = 12;
$B = 15;
  
// find number of different bits
solve($A, $B);
  
// This code is contributed by ihritik
?>

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Output:

Number of different bits : 2


My Personal Notes arrow_drop_up

Second year Department of Information Technology Jadavpur University

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Improved By : vt_m, ihritik