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# Hamming distance between two Integers

• Difficulty Level : Easy
• Last Updated : 02 Aug, 2021

Given two integers, the task is to find the hamming distance between two integers. Hamming Distance between two integers is the number of bits that are different at the same position in both numbers.
Examples:

```Input: n1 = 9, n2 = 14
Output: 3
9 = 1001, 14 = 1110
No. of Different bits = 3

Input: n1 = 4, n2 = 8
Output: 2```

Approach:

1. Calculate the XOR of two numbers.
2. Count the number of set bits.

Below is the implementation of above approach:

## C++

 `// C++ implementation of above approach``#include ``using` `namespace` `std;` `// Function to calculate hamming distance``int` `hammingDistance(``int` `n1, ``int` `n2)``{``    ``int` `x = n1 ^ n2;``    ``int` `setBits = 0;` `    ``while` `(x > 0) {``        ``setBits += x & 1;``        ``x >>= 1;``    ``}` `    ``return` `setBits;``}` `// Driver code``int` `main()``{``    ``int` `n1 = 9, n2 = 14;``    ``cout << hammingDistance(9, 14) << endl;` `    ``return` `0;``}`

## Java

 `// Java implementation of above approach``class` `GFG``{` `// Function to calculate hamming distance``static` `int` `hammingDistance(``int` `n1, ``int` `n2)``{``    ``int` `x = n1 ^ n2;``    ``int` `setBits = ``0``;` `    ``while` `(x > ``0``)``    ``{``        ``setBits += x & ``1``;``        ``x >>= ``1``;``    ``}` `    ``return` `setBits;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `n1 = ``9``, n2 = ``14``;``    ``System.out.println(hammingDistance(n1, n2));``}``}` `// This code is contributed by Bilal`

## Python3

 `# Python3 implementation of above approach` `# Function to calculate hamming distance``def` `hammingDistance(n1, n2) :` `    ``x ``=` `n1 ^ n2``    ``setBits ``=` `0` `    ``while` `(x > ``0``) :``        ``setBits ``+``=` `x & ``1``        ``x >>``=` `1``    ` `    ``return` `setBits` `if` `__name__``=``=``'__main__'``:``    ``n1 ``=` `9``    ``n2 ``=` `14``    ``print``(hammingDistance(``9``, ``14``))` `# this code is contributed by Smitha Dinesh Semwal`

## C#

 `// C# implementation of above approach``class` `GFG``{` `// Function to calculate``// hamming distance``static` `int` `hammingDistance(``int` `n1, ``int` `n2)``{``    ``int` `x = n1 ^ n2;``    ``int` `setBits = 0;` `    ``while` `(x > 0)``    ``{``        ``setBits += x & 1;``        ``x >>= 1;``    ``}` `    ``return` `setBits;``}` `// Driver code``static` `void` `Main()``{``    ``int` `n1 = 9, n2 = 14;``    ``System.Console.WriteLine(hammingDistance(n1, n2));``}``}` `// This code is contributed by mits`

## PHP

 ` 0)``    ``{``        ``\$setBits` `+= ``\$x` `& 1;``        ``\$x` `>>= 1;``    ``}` `    ``return` `\$setBits``;``}` `// Driver code``\$n1` `= 9;``\$n2` `= 14;``echo``(hammingDistance(9, 14));` `// This code is contributed by Smitha``?>`

## Javascript

 ``
Output

```3
```

Note: No. of set bits can be count using __builtin_popcount() function.

Approach 2:

1. Calculate the maximum of both the numbers.

2.Check the set bits for both at each position.

## C++

 `#include ``using` `namespace` `std;``int` `hammingDistance(``int` `x, ``int` `y)``{``    ``int` `ans = 0;``    ``int` `m = max(x, y);``    ``while` `(m) {``        ``int` `c1 = x & 1;``        ``int` `c2 = y & 1;``        ``if` `(c1 != c2)``            ``ans += 1;``        ``m = m >> 1;``        ``x = x >> 1;``        ``y = y >> 1;``    ``}``    ``return` `ans;``}``int` `main()``{``    ``int` `n1 = 4, n2 = 8;``    ``int` `hdist = hammingDistance(n1, n2);``    ``cout << hdist << endl;``    ``return` `0;``}`
Output
```2
```

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