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.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

  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)
    {
        count += n & 1;
        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;
}

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 & 1;
            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 += n & 1
        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 & 1;
            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.

Output :

Thanks to Sahil Rajput for providing above implementation.

To get the set bit count please see this post: Count set bits in an integer

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Practice Tags :

Amazon

Samsung

Bit Magic