# Number of mismatching bits in the binary representation of two integers

Given two integers(less than 2^31) A and B. The task is to find the number of bits that are different in their binary representation.

Examples:

```Input :  A = 12, B = 15
Output : Number of different bits : 2
Explanation: The binary representation of
12 is 1100 and 15 is 1111.
So, the number of different bits are 2.

Input : A = 3, B = 16
Output : Number of different bits : 3
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Run a loop from ‘0’ to ’31’ and right shift the bits of A and B by ‘i’ places, then check whether the bit at the ‘0th’ position is different.
• If the bit is different then increase the count.
• As the numbers are less than 2^31, we only have to run the loop ’32’ times i.e. from ‘0’ to ’31’.
• We can get the 1st bit if we bitwise AND the number by 1.
• At the end of the loop display the count.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// compute number of different bits ` `void` `solve(``int` `A, ``int` `B) ` `{ ` `    ``int` `count = 0; ` ` `  `    ``// since, the numbers are less than 2^31 ` `    ``// run the loop from '0' to '31' only ` `    ``for` `(``int` `i = 0; i < 32; i++) { ` ` `  `        ``// right shift both the numbers by 'i' and ` `        ``// check if the bit at the 0th position is different ` `        ``if` `(((A >> i) & 1) != ((B >> i) & 1)) { ` `            ``count++; ` `        ``} ` `    ``} ` ` `  `    ``cout << ``"Number of different bits : "` `<< count << endl; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `A = 12, B = 15; ` ` `  `    ``// find number of different bits ` `    ``solve(A, B); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` ` `  `import` `java.io.*; ` ` `  `class` `GFG { ` ` `  `// compute number of different bits ` `static` `void` `solve(``int` `A, ``int` `B) ` `{ ` `    ``int` `count = ``0``; ` ` `  `    ``// since, the numbers are less than 2^31 ` `    ``// run the loop from '0' to '31' only ` `    ``for` `(``int` `i = ``0``; i < ``32``; i++) { ` ` `  `        ``// right shift both the numbers by 'i' and ` `        ``// check if the bit at the 0th position is different ` `        ``if` `(((A >> i) & ``1``) != ((B >> i) & ``1``)) { ` `            ``count++; ` `        ``} ` `    ``} ` ` `  `    ``System.out.println(``"Number of different bits : "` `+ count); ` `} ` ` `  `// Driver code ` ` `  ` `  `    ``public` `static` `void` `main (String[] args) { ` `        ``int` `A = ``12``, B = ``15``; ` ` `  `    ``// find number of different bits ` `    ``solve(A, B); ` ` `  `    ``} ` `} ` `// this code is contributed by anuj_67.. `

## Python3

 `# Python3 implementation of the approach ` ` `  `# compute number of different bits ` `def` `solve( A,  B): ` `  `  `    ``count ``=` `0`  ` `  `    ``# since, the numbers are less than 2^31 ` `    ``# run the loop from '0' to '31' only ` `    ``for` `i ``in` `range``(``0``,``32``): ` ` `  `        ``# right shift both the numbers by 'i' and ` `        ``# check if the bit at the 0th position is different ` `        ``if` `((( A >>  i) & ``1``) !``=` `(( B >>  i) & ``1``)):  ` `             ``count``=``count``+``1` `          `  `      `  ` `  `    ``print``(``"Number of different bits :"``,count)  ` `  `  ` `  `# Driver code ` `A ``=` `12`  `B ``=` `15`  ` `  `# find number of different bits ` `solve( A,  B)  ` ` `  ` `  `# This code is contributed by ihritik ` ` `

## C#

 `// C# implementation of the approach ` ` `  `using` `System; ` `class` `GFG ` `{ ` `    ``// compute number of different bits ` `    ``static` `void` `solve(``int` `A, ``int` `B) ` `    ``{ ` `        ``int` `count = 0; ` `     `  `        ``// since, the numbers are less than 2^31 ` `        ``// run the loop from '0' to '31' only ` `        ``for` `(``int` `i = 0; i < 32; i++) { ` `     `  `            ``// right shift both the numbers by 'i' and ` `            ``// check if the bit at the 0th position is different ` `            ``if` `(((A >> i) & 1) != ((B >> i) & 1)) { ` `                ``count++; ` `            ``} ` `        ``} ` `     `  `        ``Console.WriteLine(``"Number of different bits : "` `+ count); ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void`  `Main() ` `    ``{ ` `        ``int` `A = 12, B = 15; ` `     `  `        ``// find number of different bits ` `        ``solve(A, B); ` `     `  `    ``} ` ` `  `} ` ` `  `// This code is contributed by ihritik `

## PHP

 `> ``\$i``) & 1) != ((``\$B` `>> ``\$i``) & 1)) { ` `            ``\$count``++; ` `        ``} ` `    ``} ` ` `  `    ``echo` `"Number of different bits : \$count"``; ` `} ` ` `  `// Driver code ` `\$A` `= 12; ` `\$B` `= 15; ` ` `  `// find number of different bits ` `solve(``\$A``, ``\$B``); ` ` `  `// This code is contributed by ihritik ` `?> `

Output:

```Number of different bits : 2
```

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