Count number of bits to be flipped to convert A to B

Given two numbers ‘a’ and b’. Write a program to count number of bits needed to be flipped to convert ‘a’ to ‘b’. 
Example : 

Input : a = 10, b = 20
Output : 4
Binary representation of a is 00001010
Binary representation of b is 00010100
We need to flip highlighted four bits in a
to make it b.

Input : a = 7, b = 10
Output : 3
Binary representation of a is 00000111
Binary representation of b is 00001010
We need to flip highlighted three bits in a
to make it b.

  1. Calculate XOR of A and B.      
        a_xor_b = A ^ B
  2. Count the set bits in the above 
     calculated XOR result.
        countSetBits(a_xor_b)

XOR of two number will have set bits only at those places where A differs from B. 

C++

// Count number of bits to be flipped
// to convert A into B
#include <iostream>
using namespace std;

// Function that count set bits
int countSetBits(int n)
{
    int count = 0;
    while (n > 0) {
        count++;
        n &= (n - 1);
    }
    return count;
}

// Function that return count of
// flipped number
int FlippedCount(int a, int b)
{
    // Return count of set bits in
    // a XOR b
    return countSetBits(a ^ b);
}

// Driver code
int main()
{
    int a = 10;
    int b = 20;
    cout << FlippedCount(a, b) << endl;
    return 0;
}

// This code is contributed by Sania Kumari Gupta (kriSania804)

C

// Count number of bits to be flipped to convert A into B
#include <stdio.h>

// Function that count set bits
int countSetBits(int n)
{
    int count = 0;
    while (n > 0) {
        count++;
        n &= (n - 1);
    }
    return count;
}

// Function that return count of flipped number
int FlippedCount(int a, int b)
{
    // Return count of set bits in a XOR b
    return countSetBits(a ^ b);
}

// Driver code
int main()
{
    int a = 10;
    int b = 20;
    printf("%d\n", FlippedCount(a, b));
    return 0;
}

// This code is contributed by Sania Kumari Gupta (kriSania804)

Java

// Count number of bits to be flipped
// to convert A into B
import java.util.*;

class Count {

    // Function that count set bits
    public static int countSetBits(int n)
    {
        int count = 0;
        while (n != 0) {
            count++;
            n &=(n-1);
        }
        return count;
    }

    // Function that return count of
    // flipped number
    public static int FlippedCount(int a, int b)
    {
        // Return count of set bits in
        // a XOR b
        return countSetBits(a ^ b);
    }

    // Driver code
    public static void main(String[] args)
    {
        int a = 10;
        int b = 20;
        System.out.print(FlippedCount(a, b));
    }
}

// This code is contributed by rishabh_jain

Python3

# Count number of bits to be flipped
# to convert A into B

# Function that count set bits
def countSetBits( n ):
    count = 0
    while n:
        count += 1
        n &= (n-1)
    return count

# Function that return count of
# flipped number
def FlippedCount(a , b):

    # Return count of set bits in
    # a XOR b
    return countSetBits(a^b)

# Driver code
a = 10
b = 20
print(FlippedCount(a, b))

# This code is contributed by "Sharad_Bhardwaj".

C#

// Count number of bits to be
// flipped to convert A into B
using System;

class Count {

    // Function that count set bits
    public static int countSetBits(int n)
    {
        int count = 0;
        while (n != 0) {
            count++;
            n &= (n-1);
        }
        return count;
    }

    // Function that return
    // count of flipped number
    public static int FlippedCount(int a, int b)
    {
    // Return count of set
    // bits in a XOR b
        return countSetBits(a ^ b);
    }

    // Driver code
    public static void Main()
    {
        int a = 10;
        int b = 20;
        Console.WriteLine(FlippedCount(a, b));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// Count number of bits to be
// flipped to convert A into B

// Function that count set bits
function countSetBits($n)
{
    $count = 0;
    while($n)
    {
        $count += 1;
        $n &= (n-1);
    }
    return $count;
}

// Function that return
// count of flipped number
function FlippedCount($a, $b)
{
    // Return count of set
    // bits in a XOR b
    return countSetBits($a ^ $b);
}

// Driver code
$a = 10;
$b = 20;
echo FlippedCount($a, $b);

// This code is contributed by mits
?>

Javascript

<script>
// Count number of bits to be flipped
// to convert A into Bclass Count {

    // Function that count set bits
    function countSetBits(n) {
        var count = 0;
        while (n != 0) {
            count++;
            n &= (n - 1);
        }
        return count;
    }

    // Function that return count of
    // flipped number
    function FlippedCount(a , b) {
        // Return count of set bits in
        // a XOR b
        return countSetBits(a ^ b);
    }

    // Driver code
        var a = 10;
        var b = 20;
        document.write(FlippedCount(a, b));

// This code is contributed by shikhasingrajput
</script>

Output

4

Another approach : 

C++

// C++ program
#include <iostream>
using namespace std;

int countFlips(int a, int b)
{

  // initially flips is equal to 0
  int flips = 0;

  // & each bits of a && b with 1
  // and store them if t1 and t2
  // if t1 != t2 then we will flip that bit

  while(a > 0 || b > 0){

    int t1 = (a&1);
    int t2 = (b&1);

    if(t1!=t2){
      flips++;
    }
    // right shifting a and b
    a>>=1;
    b>>=1;
  }

  return flips;
}

int main () {
  int a = 10;
  int b = 20;
  cout <<countFlips(a, b);
}

// this code is contributed by shivanisinghss2110

Java

/*package whatever //do not write package name here */

// CONTRIBUTED BY PRAVEEN VISHWAKARMA

import java.io.*;

class GFG {

      public static int countFlips(int a, int b){
          // initially flips is equal to 0
        int flips = 0;

          // & each bits of a && b with 1
          // and store them if t1 and t2
        // if t1 != t2 then we will flip that bit

          while(a>0 || b>0){

            int t1 = (a&1);
              int t2 = (b&1);

              if(t1!=t2){
                flips++;
            }
              // right shifting a and b
              a>>>=1;
              b>>>=1;
        }

      return flips;
    }

    public static void main (String[] args) {
        int a = 10;
          int b = 20;
          System.out.println(countFlips(a, b));
    }

}

Python3

def countFlips(a, b):

    # initially flips is equal to 0
    flips = 0

    # & each bits of a && b with 1
    # and store them if t1 and t2
    # if t1 != t2 then we will flip that bit
    while(a > 0 or b > 0):
        t1 = (a & 1)
        t2 = (b & 1)
        if(t1 != t2):
            flips += 1

        # right shifting a and b
        a>>=1
        b>>=1

    return flips

a = 10
b = 20
print(countFlips(a, b))

# This code is contributed by shivanisinghss2110

C#

/*package whatever //do not write package name here */
using System;

class GFG {

      public static int countFlips(int a, int b)
      {

          // initially flips is equal to 0
        int flips = 0;

          // & each bits of a && b with 1
          // and store them if t1 and t2
        // if t1 != t2 then we will flip that bit
          while(a > 0 || b > 0){

            int t1 = (a&1);
              int t2 = (b&1);

              if(t1 != t2){
                flips++;
            }
              // right shifting a and b
              a>>=1;
              b>>=1;
        }

      return flips;
    }

  // Driver code
    public static void Main (String[] args) {
        int a = 10;
          int b = 20;
          Console.Write(countFlips(a, b));
    }

}

// This code is contributed by shivanisinghss2110

Javascript

<script>
/*package whatever //do not write package name here */

function countFlips(a, b){
          // initially flips is equal to 0
        var flips = 0;

          // & each bits of a && b with 1
          // and store them if t1 and t2
        // if t1 != t2 then we will flip that bit

          while(a>0 || b>0){

            var t1 = (a&1);
              var t2 = (b&1);

              if(t1!=t2){
                flips++;
            }
              // right shifting a and b
              a>>>=1;
              b>>>=1;
        }

      return flips;
    }

          var a = 10;
          var b = 20;
          document.write(countFlips(a, b));

// This code is contributed by shivanisinghss2110
</script>

Output

4

Thanks to Sahil Rajput for providing above implementation.

Another approach :

 In this approach we will directly count the set bits of XOR of the two numbers using __builtin__popcount() which we can easily do in GCC and thus we can avoid a separate function for counting set bits. 

C++

// C++ program to Count number of bits to be flipped
// to convert A into B
#include <iostream>
using namespace std;

// Driver code
int main()
{
    int a = 10;
    int b = 20;
    cout <<__builtin_popcount(a^b) << endl;
    return 0;
}

// This code is contributed by Suruchi Kumari

C

//C program to Count number of bits to be flipped
// to convert A into B
#include <stdio.h>

// Driver code
int main()
{
    int a = 10;
    int b = 20;
    printf("%d\n",__builtin_popcount(a^b));
    return 0;
}

// This code is contributed by Suruchi Kumari

Output :

4

Time complexity : O(1)

Auxiliary space and space complexity : O(1)

To get the set bit count please see this post: Count set bits in an integer
If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.