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Count number of bits to be flipped to convert A to B
• Difficulty Level : Easy
• Last Updated : 25 Mar, 2021

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 ``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)<

## 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

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## Javascript

 ``

Output :

`4`

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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