[Algorithms] Longest Common Subsequence

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

Given two sequences s and t, find the length of the longest sequence r, which is a subsequence of both s and t.

Do you know the difference between substring and subequence? Well, substring is a contiguous series of characters while subsequence is not necessarily. For example, "abc" is a both a substring and a subseqeunce of "abcde" while "ade" is only a subsequence.

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 subsequence ends at i of s and j of t. Then the state equations are

P[i][j] = max(P[i][j - 1], P[i - 1][j]) 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 subsequence. The code is as follows.

1 int longestCommonSubsequence(string s, string t) {
2     int m = s.length(), n = t.length();
3     vector<vector<int> > dp(m + 1, vector<int> (n + 1, 0));
4     for (int i = 1; i <= m; i++)
5         for (int j = 1; j <= n; j++)
6             dp[i][j] = (s[i - 1] == t[j - 1] ? dp[i - 1][j - 1] + 1 : max(dp[i - 1][j], dp[i][j - 1]));
7     return dp[m][n];
8 }

Well, this code has both time and space complexity of O(m*n). Note that when we update dp[i][j], we only need dp[i - 1][j - 1], dp[i - 1][j] and dp[i][j - 1]. So we simply need to maintain two columns for them. The code is as follows.

 1 int longestCommonSubsequenceSpaceEfficient(string s, string t) {
 2     int m = s.length(), n = t.length();
 3     int maxlen = 0;
 4     vector<int> pre(m, 0);
 5     vector<int> cur(m, 0);
 6     pre[0] = (s[0] == t[0]);
 7     maxlen = max(maxlen, pre[0]);
 8     for (int i = 1; i < m; i++) {
 9         if (s[i] == t[0] || pre[i - 1] == 1) pre[i] = 1;
10         maxlen = max(maxlen, pre[i]);
11     }
12     for (int j = 1; j < n; j++) {
13         if (s[0] == t[j] || pre[0] == 1) cur[0] = 1;
14         maxlen = max(maxlen, cur[0]);
15         for (int i = 1; i < m; i++) {
16             if (s[i] == t[j]) cur[i] = pre[i - 1] + 1;
17             else cur[i] = max(cur[i - 1], pre[i]);
18             maxlen = max(maxlen, cur[i]);
19         }
20         swap(pre, cur);
21         fill(cur.begin(), cur.end(), 0);
22     }
23     return maxlen;
24 }

Well, keeping two columns is just for retriving pre[i - 1], we can maintain a single variable for it and keep only one column. The code becomes more efficient and also shorter. However, you may need to run some examples to see how it achieves the things done by the two-column version.

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

Now you may try this problem on UVa Online Judge and get Accepted:)

Of course, the above code only returns the length of the longest common subsequence. If you want to print the lcs itself, you need to visit the 2-d table from bottom-right to top-left. The detailed algorithm is clearly explained here. The code is as follows.

 1 int longestCommonSubsequence(string s, string t) {
 2     int m = s.length(), n = t.length();
 3     vector<vector<int> > dp(m + 1, vector<int> (n + 1, 0));
 4     for (int i = 1; i <= m; i++)
 5         for (int j = 1; j <= n; j++)
 6             dp[i][j] = (s[i - 1] == t[j - 1] ? dp[i - 1][j - 1] + 1 : max(dp[i - 1][j], dp[i][j - 1]));
 7     int len = dp[m][n];
 8     // Print out the longest common subsequence
 9     string lcs(len, ' ');
10     for (int i = m, j = n, index = len - 1; i > 0 && j > 0;) {
11         if (s[i - 1] == t[j - 1]) {
12             lcs[index--] = s[i - 1];
13             i--;
14             j--;
15         }
16         else if (dp[i - 1][j] > dp[i][j - 1]) i--;
17         else j--;
18     }
19     printf("%s\n", lcs.c_str());
20     return len;
21 }