You are given two strings of equal length, you have to find the Hamming Distance between these string.

Where the Hamming distance between two strings of equal length is the number of positions at which the corresponding character is different. **Examples: **

Input : str1[] = "geeksforgeeks", str2[] = "geeksandgeeks" Output : 3 Explanation : The corresponding character mismatch are highlighted. "geeksgeeks" and "geeksforgeeks" Input : str1[] = "1011101", str2[] = "1001001" Output : 2 Explanation : The corresponding character mismatch are highlighted. "10and1101" and "1011001"0

This problem can be solved with a simple approach in which we traverse the strings and count the mismatch at the corresponding position. The extended form of this problem is edit distance.**Algorithm : **

int hammingDist(char str1[], char str2[]) { int i = 0, count = 0; while(str1[i]!='\0') { if (str1[i] != str2[i]) count++; i++; } return count; }

Below is the implementation of two strings.

## C++

`// C++ program to find hamming distance b/w ` `// two string` `#include<bits/stdc++.h>` `using` `namespace` `std;` `// function to calculate Hamming distance` `int` `hammingDist(` `char` `*str1, ` `char` `*str2)` `{` ` ` `int` `i = 0, count = 0;` ` ` `while` `(str1[i] != ` `'\0'` `)` ` ` `{` ` ` `if` `(str1[i] != str2[i])` ` ` `count++;` ` ` `i++;` ` ` `}` ` ` `return` `count;` `}` `// driver code` `int` `main()` `{` ` ` `char` `str1[] = ` `"geekspractice"` `;` ` ` `char` `str2[] = ` `"nerdspractise"` `;` ` ` `// function call` ` ` `cout << hammingDist (str1, str2);` ` ` `return` `0;` `} ` |

## Java

`// Java program to find hamming distance` `// b/w two string` `class` `GFG` `{` `// function to calculate Hamming distance` `static` `int` `hammingDist(String str1, String str2)` `{` ` ` `int` `i = ` `0` `, count = ` `0` `;` ` ` `while` `(i < str1.length())` ` ` `{` ` ` `if` `(str1.charAt(i) != str2.charAt(i))` ` ` `count++;` ` ` `i++;` ` ` `}` ` ` `return` `count;` `} ` `// Driver code ` `public` `static` `void` `main (String[] args)` `{` ` ` `String str1 = ` `"geekspractice"` `;` ` ` `String str2 = ` `"nerdspractise"` `;` ` ` `// function call` ` ` `System.out.println(hammingDist (str1, str2));` `}` `}` `// This code is contributed by Anant Agarwal.` |

## Python3

`# Python3 program to find ` `# hamming distance b/w two ` `# string ` `# Function to calculate` `# Hamming distance ` `def` `hammingDist(str1, str2):` ` ` `i ` `=` `0` ` ` `count ` `=` `0` ` ` `while` `(i < ` `len` `(str1)):` ` ` `if` `(str1[i] !` `=` `str2[i]):` ` ` `count ` `+` `=` `1` ` ` `i ` `+` `=` `1` ` ` `return` `count` `# Driver code ` `str1 ` `=` `"geekspractice"` `str2 ` `=` `"nerdspractise"` `# function call ` `print` `(hammingDist(str1, str2))` `# This code is contributed by avanitrachhadiya2155` |

## C#

`// C# program to find hamming ` `// distance b/w two string` `using` `System;` `class` `GFG {` ` ` `// function to calculate ` `// Hamming distance` `static` `int` `hammingDist(String str1, ` ` ` `String str2)` `{` ` ` `int` `i = 0, count = 0;` ` ` `while` `(i < str1.Length)` ` ` `{` ` ` `if` `(str1[i] != str2[i])` ` ` `count++;` ` ` `i++;` ` ` `}` ` ` `return` `count;` `} ` `// Driver code ` `public` `static` `void` `Main ()` `{` ` ` `String str1 = ` `"geekspractice"` `;` ` ` `String str2 = ` `"nerdspractise"` `;` ` ` `// function call` ` ` `Console.Write(hammingDist(str1, str2));` `}` `}` `// This code is contributed by nitin mittal` |

## PHP

`<?php` `// PHP program to find hamming distance b/w ` `// two string` `// function to calculate` `// Hamming distance` `function` `hammingDist(` `$str1` `, ` `$str2` `)` `{` ` ` `$i` `= 0; ` `$count` `= 0;` ` ` `while` `(isset(` `$str1` `[` `$i` `]) != ` `''` `)` ` ` `{` ` ` `if` `(` `$str1` `[` `$i` `] != ` `$str2` `[` `$i` `])` ` ` `$count` `++;` ` ` `$i` `++;` ` ` `}` ` ` `return` `$count` `;` `}` ` ` `// Driver Code` ` ` `$str1` `= ` `"geekspractice"` `;` ` ` `$str2` `= ` `"nerdspractise"` `;` ` ` `// function call` ` ` `echo` `hammingDist (` `$str1` `, ` `$str2` `);` ` ` `// This code is contributed by nitin mittal.` `?>` |

**Output: **

4

Time complexity: O(n)**Note: **For Hamming distance of two binary numbers, we can simply return a count of set bits in XOR of two numbers.

This article is contributed by **Shivam Pradhan (anuj_charm)**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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