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# Hamming distance between two Integers

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;``}` `// This code is contributed by Sania Kumari Gupta (kriSania804)`

## C

 `// C implementation of above approach``#include ` `// 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;``    ``printf``(``"%d\n"``, hammingDistance(9, 14));` `    ``return` `0;``}` `// This code is contributed by Sania Kumari Gupta (kriSania804)`

## 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 counted using __builtin_popcount() function.

Time Complexity: O(log2x), where x is n1 ^ n2, this can be interpreted as O(1), since an int can hold a maximum of 32 bits

Auxiliary Space: O(1)

Approach 2:

1. Calculate the maximum of both 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;``}`

## Java

 `/*package whatever //do not write package name here */` `import` `java.io.*;` `class` `GFG {``    ``static` `int` `hammingDistance(``int` `x, ``int` `y)``{``    ``int` `ans = ``0``;``    ``int` `m = Math.max(x, y);``    ``while` `(m>``0``) {``        ``int` `c1 = x & ``1``;``        ``int` `c2 = y & ``1``;``        ``if` `(c1 != c2)``            ``ans += ``1``;``        ``m = m >> ``1``;``        ``x = x >> ``1``;``        ``y = y >> ``1``;``    ``}``    ``return` `ans;``}``    ` `// Drivers code``public` `static` `void` `main(String args[])``{``    ``int` `n1 = ``4``, n2 = ``8``;``    ``int` `hdist = hammingDistance(n1, n2);``    ``System.out.println(hdist);``}``}` `// This code is contributed by shinjanpatra`

## Python3

 `# Python3 code for the same approach``def` `hammingDistance(x, y):` `   ``ans ``=` `0``   ``m ``=` `max``(x, y)``   ``while` `(m):``      ``c1 ``=` `x & ``1``      ``c2 ``=` `y & ``1``      ``if` `(c1 !``=` `c2):``         ``ans ``+``=` `1``      ``m ``=` `m >> ``1``      ``x ``=` `x >> ``1``      ``y ``=` `y >> ``1``   ``return` `ans` `# driver program``n1 ``=` `4``n2 ``=` `8``hdist ``=` `hammingDistance(n1, n2)``print``(hdist)` `# This code is contributed by shinjanpatra`

## C#

 `// C# code to implement the approach` `using` `System;` `public` `class` `GFG``{``  ` `    ``// Function that return the hamming distance``    ``// between x and y``    ``static` `int` `hammingDistance(``int` `x, ``int` `y)``    ``{``        ``int` `ans = 0;``        ``int` `m = Math.Max(x, y);``      ` `        ``// checking the set bits``        ``while` `(m > 0) {``            ``int` `c1 = x & 1;``            ``int` `c2 = y & 1;``            ``if` `(c1 != c2)``                ``ans += 1;``            ``m = m >> 1;``            ``x = x >> 1;``            ``y = y >> 1;``        ``}``        ``return` `ans;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``int` `n1 = 4, n2 = 8;``        ``// Function call``        ``int` `hdist = hammingDistance(n1, n2);``        ``Console.WriteLine(hdist);``    ``}``}` `// This code is contributed by phasing17`

## Javascript

 ``

Output

`2`

Time Complexity: O(log2(max(n1, n2)).
Auxiliary Space: O(1)

#### Algorithm

1. Convert the two given integers to binary strings.
2. Pad the shorter binary string with leading zeros to make both strings of equal length.
3. Compare the two binary strings bit by bit and count the number of differences.
4. The count obtained in step 3 represents the Hamming distance between the two given integers.

## C++

 `// c++ code implementation``#include ``#include ``#include ` `using` `namespace` `std;` `int` `hammingDistance(``int` `n1, ``int` `n2) {``    ``string binStr1 = bitset<32>(n1).to_string();``    ``string binStr2 = bitset<32>(n2).to_string();``    ``int` `lenDiff = ``abs``(``static_cast``<``int``>(binStr1.length() - binStr2.length()));``    ``if` `(binStr1.length() < binStr2.length()) {``        ``binStr1 = string(lenDiff, ``'0'``) + binStr1;``    ``} ``else` `{``        ``binStr2 = string(lenDiff, ``'0'``) + binStr2;``    ``}``    ``int` `count = 0;``    ``for` `(``int` `i = 0; i < binStr1.length(); i++) {``        ``if` `(binStr1[i] != binStr2[i]) {``            ``count++;``        ``}``    ``}``    ``return` `count;``}` `int` `main() {``    ``int` `n1 = 4;``    ``int` `n2 = 8;``    ``cout << hammingDistance(n1, n2) << endl;``    ``return` `0;``}`

## Java

 `public` `class` `Main {` `    ``public` `static` `int` `hammingDistance(``int` `n1, ``int` `n2) {``        ``String binStr1 = Integer.toBinaryString(n1);``        ``String binStr2 = Integer.toBinaryString(n2);``        ``int` `lenDiff = Math.abs(binStr1.length() - binStr2.length());``        ``if` `(binStr1.length() < binStr2.length()) {``            ``binStr1 = ``"0"``.repeat(lenDiff) + binStr1;``        ``} ``else` `{``            ``binStr2 = ``"0"``.repeat(lenDiff) + binStr2;``        ``}``        ``int` `count = ``0``;``        ``for` `(``int` `i = ``0``; i < binStr1.length(); i++) {``            ``if` `(binStr1.charAt(i) != binStr2.charAt(i)) {``                ``count++;``            ``}``        ``}``        ``return` `count;``    ``}` `    ``public` `static` `void` `main(String[] args) {``        ``int` `n1 = ``4``;``        ``int` `n2 = ``8``;``        ``System.out.println(hammingDistance(n1, n2));``    ``}``}`

## Python3

 `def` `hamming_distance(n1, n2):``    ``bin_str1 ``=` `bin``(n1)[``2``:]``    ``bin_str2 ``=` `bin``(n2)[``2``:]``    ``len_diff ``=` `abs``(``len``(bin_str1) ``-` `len``(bin_str2))``    ``if` `len``(bin_str1) < ``len``(bin_str2):``        ``bin_str1 ``=` `'0'` `*` `len_diff ``+` `bin_str1``    ``else``:``        ``bin_str2 ``=` `'0'` `*` `len_diff ``+` `bin_str2``    ``count ``=` `0``    ``for` `i ``in` `range``(``len``(bin_str1)):``        ``if` `bin_str1[i] !``=` `bin_str2[i]:``            ``count ``+``=` `1``    ``return` `count``n1 ``=` `4``n2 ``=` `8``print``(hamming_distance(n1, n2))`

## Javascript

 `function` `hamming_distance(n1, n2) {``    ``let bin_str1 = n1.toString(2);``    ``let bin_str2 = n2.toString(2);``    ``let len_diff = Math.abs(bin_str1.length - bin_str2.length);``    ``if` `(bin_str1.length < bin_str2.length) {``        ``bin_str1 = ``'0'``.repeat(len_diff) + bin_str1;``    ``} ``else` `{``        ``bin_str2 = ``'0'``.repeat(len_diff) + bin_str2;``    ``}``    ``let count = 0;``    ``for` `(let i = 0; i < bin_str1.length; i++) {``        ``if` `(bin_str1[i] !== bin_str2[i]) {``            ``count += 1;``        ``}``    ``}``    ``return` `count;``}` `let n1 = 4;``let n2 = 8;``console.log(hamming_distance(n1, n2));`

Output

`2`

Time Complexity: O(log n), where n is the maximum of the two integers.
Space Complexity: O(log n), where n is the maximum of the two integers.

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