Longest common substring: Difference between revisions

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Intermedate results:
Intermedate results:


C shows which characters are in common between the two strings.
C shows which characters are in common between the two strings.
M marks the length of the longest common substring ending at that position in the right argument
M marks the length of the longest common substring ending at that position in the right argument
N is the length of the longest common substring
N is the length of the longest common substring


Example use:
Example use:

Revision as of 03:29, 23 February 2015

Longest common substring is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Write a program that prompts the user to enter two different strings, and then output their longest common substring. Note that substrings are consecutive characters within a string. This distinguishes them from subsequences, which is any sequence of characters within a string, even if there are extraneous characters in between them. Hence, the longest common subsequence between "thisisatest" and "testing123testing" is "tsitest", whereas the longest common substring is just "test".

References:

C#

<lang csharp>using System; namespace LongestCommonSubstring {

   class Program
   {
       static void Main(string[] args)
       {
           Console.Write("Enter word 1: ");
           string word1 = Console.ReadLine();
           Console.Write("Enter word 2: ");
           string word2 = Console.ReadLine();
           Console.WriteLine(lcs(word1, word2));
           Console.ReadKey(true);
       }
       //This algorithm for finding the longest commmon substring uses dynamic programming.
       public static string lcs(string a, string b)
       {
           var lengths = new int[a.Length, b.Length];
           int greatestLength = 0;
           string output = "";
           //Compare each consecutive character in string a with each consecutive character in string b.
           for (int i = 0; i < a.Length; i++)
           {
               for (int j = 0; j < b.Length; j++)
               {
                   //Keep a running count of the number of consecutive characters in each that are the same.
                   if (a[i] == b[j])
                   {
                       if (i == 0 || j == 0)
                       {
                           lengths[i, j] = 1;
                       }
                       else
                       {
                           lengths[i, j] = lengths[i - 1, j - 1] + 1;
                       }
                       //When you find the greatest count so far, note the substring of which it is. 
                       if (lengths[i, j] > greatestLength)
                       {
                           greatestLength = lengths[i, j];
                           output = a.Substring(i - greatestLength + 1, greatestLength);
                       }
                   }
                   else
                   {
                       lengths[i, j] = 0;
                   }
               }
           }
           //Output the substring you've noted.
           return output;
       }
   }

}</lang>

Go

Translation of: C#

<lang go>package main

import "fmt"

func lcs(a, b string) (output string) {

   lengths := make([]int, len(a)*len(b))
   greatestLength := 0
   for i, x := range a {
       for j, y := range b {
           if x == y {
               if i == 0 || j == 0 {
                   lengths[i*len(b)+j] = 1
               } else {
                   lengths[i*len(b)+j] = lengths[(i-1)*len(b)+j-1] + 1
               }
               if lengths[i*len(b)+j] > greatestLength {
                   greatestLength = lengths[i*len(b)+j]
                   output = a[i-greatestLength+1 : i+1]
               }
           }
       }
   }
   return

}

func main() {

   fmt.Println(lcs("thisisatest", "testing123testing"))

}</lang>

Output:
test

J

This algorithm starts by comparing each character in the one string to each character in the other, and then iterates on this result until it finds the length of the longest common substring. So if Lx is the length of one argument string, Ly is the length of the other argument string, and Lr is the length of the result string, this algorithm uses space on the order of Lx*Ly and time on the order of Lx*Ly*Lr.

In other words: this can be suitable for small problems, but you might want something better if you're comparing gigabyte length strings with high commonality.

<lang J>lcstr=:4 :0

 C=. ({.~ 1+$) x=/y
 M=. >./ (* * * >. * + (_1&|.)@:|:^:2)^:_  C
 N=. >./ M
 y {~ (M i. N)-i.-N

)</lang>

Intermedate results:

   C shows which characters are in common between the two strings.
   M marks the length of the longest common substring ending at that position in the right argument
   N is the length of the longest common substring

Example use:

<lang J> 'thisisatest' lcs 'testing123testing' test</lang>