Algorithms for Coding Interviews in Java/

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Solution: Longest Common Substring

Solution: Longest Common Substring

This review provides a detailed analysis of the different ways to solve the Longest Common Substring problem.

We'll cover the following...

Solution #1: Brute force

Let’s start by discussing the brute force solution.

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class LongestCommonSubstring
{ 
  public static int longestCommonSubstrLengthRec(String s1, String s2, int i1, int i2, int count) {
    if (i1 == s1.length() || i2 == s2.length())
      return count;
    if (s1.charAt(i1) == s2.charAt(i2))
      count = longestCommonSubstrLengthRec(s1, s2, i1 + 1, i2 + 1, count + 1);
    int c1 = longestCommonSubstrLengthRec(s1, s2, i1, i2 + 1, 0);
    int c2 = longestCommonSubstrLengthRec(s1, s2, i1 + 1, i2, 0);
    return Math.max(count, Math.max(c1, c2));
  }
  public static int longestCommonSubstrLength(String s1, String s2) 
  {
    return longestCommonSubstrLengthRec(s1, s2, 0, 0, 0);
  }
  public static void main(String args[]) 
  {
    String s1 = "0abc321";
    String s2 = "123abcdef";
    String s3 = "educative.io/expl";
    String s4 = "educative.io/edpr";
    System.out.println(longestCommonSubstrLength(s1, s2));
    //System.out.println(longestCommonSubstrLength(s3, s4));
  }
};

Explanation

This is a naive solution which finds all the substrings of the two given strings and then discovers the longest common substrings between the two.

  • If the strings have a matching character, we recursively check for the remaining length of each and keep a track of the current matching length (line 8).

  • If the characters don’t match, we start two new recursive calls by skipping those characters from each string (lines 10-11).

  • The length of the longest common substring will be the maximum number returned by the three recursive calls (line 13).

Time complexity

The time complexity of this code is exponential: O(2m+n)O(2^{m+n}).

Notice that the time complexity is so high that the function times out before it can process the longer inputs s3 and s4. Try uncommenting the call on line 28.

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