Skip to content
Related Articles

Related Articles

Improve Article

Count number of bits to be flipped to convert A to B | Set-2

  • Last Updated : 26 Apr, 2021

Given two integers A and B, the task is to count the number of bits needed to be flipped to convert A to B.
Examples: 
 

Input: A = 10, B = 7 
Output:
binary(10) = 1010 
binary(7) = 0111 
1010 
0111 
3 bits need to be flipped.
Input: A = 8, B = 7 
Output:
 

 

Approach: An approach to solve this problem has already been discussed here. Here, the count of bits that need to be flipped can be found by matching all the bits in both the integers one by one. If the bit under consideration differs then increment the count.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count of bits
// to be flipped to convert a to b
int countBits(int a, int b)
{
 
    // To store the required count
    int count = 0;
 
    // Loop until both of them become zero
    while (a || b) {
 
        // Store the last bits in a
        // as well as b
        int last_bit_a = a & 1;
        int last_bit_b = b & 1;
 
        // If the current bit is not same
        // in both the integers
        if (last_bit_a != last_bit_b)
            count++;
 
        // Right shift both the integers by 1
        a = a >> 1;
        b = b >> 1;
    }
 
    // Return the count
    return count;
}
 
// Driver code
int main()
{
    int a = 10, b = 7;
 
    cout << countBits(a, b);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
class GFG
{
 
// Function to return the count of bits
// to be flipped to convert a to b
static int countBits(int a, int b)
{
 
    // To store the required count
    int count = 0;
 
    // Loop until both of them become zero
    while (a > 0 || b > 0)
    {
 
        // Store the last bits in a
        // as well as b
        int last_bit_a = a & 1;
        int last_bit_b = b & 1;
 
        // If the current bit is not same
        // in both the integers
        if (last_bit_a != last_bit_b)
            count++;
 
        // Right shift both the integers by 1
        a = a >> 1;
        b = b >> 1;
    }
 
    // Return the count
    return count;
}
 
// Driver code
public static void main(String[] args)
{
    int a = 10, b = 7;
 
    System.out.println(countBits(a, b));
}
}
 
// This code is contributed by Princi Singh

Python3




# Python3 implementation of the approach
 
# Function to return the count of bits
# to be flipped to convert a to b
def countBits(a, b):
 
    # To store the required count
    count = 0
 
    # Loop until both of them become zero
    while (a or b):
 
        # Store the last bits in a
        # as well as b
        last_bit_a = a & 1
        last_bit_b = b & 1
 
        # If the current bit is not same
        # in both the integers
        if (last_bit_a != last_bit_b):
            count += 1
 
        # Right shift both the integers by 1
        a = a >> 1
        b = b >> 1
 
    # Return the count
    return count
 
# Driver code
a = 10
b = 7
 
print(countBits(a, b))
 
# This code is contributed by Mohit Kumar

C#




// C# implementation of the above approach
using System;
 
class GFG
{
 
// Function to return the count of bits
// to be flipped to convert a to b
static int countBits(int a, int b)
{
 
    // To store the required count
    int count = 0;
 
    // Loop until both of them become zero
    while (a > 0 || b > 0)
    {
 
        // Store the last bits in a
        // as well as b
        int last_bit_a = a & 1;
        int last_bit_b = b & 1;
 
        // If the current bit is not same
        // in both the integers
        if (last_bit_a != last_bit_b)
            count++;
 
        // Right shift both the integers by 1
        a = a >> 1;
        b = b >> 1;
    }
 
    // Return the count
    return count;
}
 
// Driver code
public static void Main(String[] args)
{
    int a = 10, b = 7;
 
    Console.WriteLine(countBits(a, b));
}
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the count of bits
// to be flipped to convert a to b
function countBits(a, b)
{
 
    // To store the required count
    var count = 0;
 
    // Loop until both of them become zero
    while (a || b) {
 
        // Store the last bits in a
        // as well as b
        var last_bit_a = a & 1;
        var last_bit_b = b & 1;
 
        // If the current bit is not same
        // in both the integers
        if (last_bit_a != last_bit_b)
            count++;
 
        // Right shift both the integers by 1
        a = a >> 1;
        b = b >> 1;
    }
 
    // Return the count
    return count;
}
 
// Driver code
    var a = 10, b = 7;
    document.write(countBits(a, b));
 
</script>
Output: 
3

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :