[Algorithms] Longest Common Substring

The Longest Common Substring (LCS) problem is as follows:

Given two strings s and t, find the length of the longest string r, which is a substring of both s and t.

This problem is a classic application of Dynamic Programming. Let's define the sub-problem (state) P[i][j] to be the length of the longest substring ends at i of s and j of t. Then the state equations are

P[i][j] = 0 if s[i] != t[j];
P[i][j] = P[i - 1][j - 1] + 1 if s[i] == t[j].

This algorithm gives the length of the longest common substring. If we want the substring itself, we simply find the largest P[i][j] and return s.substr(i - P[i][j] + 1, P[i][j]) or t.substr(j - P[i][j] + 1, P[i][j]).

Then we have the following code.

 1 string longestCommonSubstring(string s, string t) {
 2     int m = s.length(), n = t.length();
 3     vector<vector<int> > dp(m, vector<int> (n, 0));
 4     int start = 0, len = 0;
 5     for (int i = 0; i < m; i++) {
 6         for (int j = 0; j < n; j++) {
 7             if (i == 0 || j == 0) dp[i][j] = (s[i] == t[j]);
 8             else dp[i][j] = (s[i] == t[j] ? dp[i - 1][j - 1] + 1: 0);
 9             if (dp[i][j] > len) {
10                 len = dp[i][j];
11                 start = i - len + 1;
12             }
13         }
14     }
15     return s.substr(start, len);
16 }

The above code costs O(m*n) time complexity and O(m*n) space complexity. In fact, it can be optimized to O(min(m, n)) space complexity. The observations is that each time we update dp[i][j], we only need dp[i - 1][j - 1], which is simply the value of the above grid before updates.

Now we will have the following code.

 1 string longestCommonSubstringSpaceEfficient(string s, string t) {
 2     int m = s.length(), n = t.length();
 3     vector<int> cur(m, 0);
 4     int start = 0, len = 0, pre = 0;
 5     for (int j = 0; j < n; j++) {
 6         for (int i = 0; i < m; i++) {
 7             int temp = cur[i];
 8             cur[i] = (s[i] == t[j] ? pre + 1 : 0);
 9             if (cur[i] > len) {
10                 len = cur[i];
11                 start = i - len + 1;
12             }
13             pre = temp;
14         }
15     }
16     return s.substr(start, len);
17 }

In fact, the code above is of O(m) space complexity. You may choose the small size for cur and repeat the same code using if..else.. to save more spaces :)