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

• Difficulty Level : Basic
• Last Updated : 17 Nov, 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 ``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

 ``

Output:

`3`

Time Complexity: O(min(log a, log b))

Auxiliary Space: O(1)

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