Hamming Distance between two strings


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 are different.

Examples:

Input : str1[] = "geeksforgeeks", str2[] = "geeksandgeeks"
Output : 3
Explanation : The corresponding character mismatch are highlighted.
"geeksforgeeks" and "geeksandgeeks"

Input : str1[] = "1011101", str2[] = "1001001"
Output : 2
Explanation : The corresponding character mismatch are highlighted.
"1011101" and "1001001"



This problem can be solved with a simple approach in which we traverse the strings and count the mismatch at corresponding position. 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++

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// 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;

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Java

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// 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.

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C#

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// 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

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PHP

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<?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.
?>

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Output:

4

Time complexity : O(n)

Note : For Hamming distance of two binary numbers, we can simply return 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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



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Improved By : nitin mittal



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