Leetcode: Longest Common Subsequence

Given two strings text1 and text2, return the length of their longest common subsequence.

A subsequence of a string is a new string generated from the original string with some characters(can be none) deleted without changing the relative order of the remaining characters. (eg, "ace" is a subsequence of "abcde" while "aec" is not). A common subsequence of two strings is a subsequence that is common to both strings.

If there is no common subsequence, return 0.

Example 1:

Input: text1 = "abcde", text2 = "ace"
Output: 3
Explanation: The longest common subsequence is "ace" and its length is 3.
Example 2:

Input: text1 = "abc", text2 = "abc"
Output: 3
Explanation: The longest common subsequence is "abc" and its length is 3.
Example 3:

Input: text1 = "abc", text2 = "def"
Output: 0
Explanation: There is no such common subsequence, so the result is 0.

DP with a 2D array

Time & space: O(m * n)

 class Solution {
     public int longestCommonSubsequence(String text1, String text2) {
         if (text1 == null || text1.length() == 0 || text2 == null || text2.length() == 0)
             return 0;
         int[][] dp = new int[text1.length() + 1][text2.length() + 1];
         for (int i = 1; i <= text1.length(); i ++) {
             for (int j = 1; j <= text2.length(); j ++) {
                 int val = Math.max(dp[i - 1][j], dp[i][j - 1]);
                 if (text1.charAt(i - 1) == text2.charAt(j - 1)) {
                     val = Math.max(val, dp[i - 1][j - 1] + 1);
                 }
                 dp[i][j] = val;
             }
         }
         return dp[text1.length()][text2.length()];
     }
 }

(Skim through) memory optimization, referencing: https://leetcode.com/problems/longest-common-subsequence/discuss/351689/Java-Two-DP-codes-of-O(mn)-and-O(min(m-n))-spaces-w-picture-and-analysis

Obviously, the code in method 1 only needs information of previous row to update current row. So we just use a two-row 2D array to save and update the matching results for chars in s1 and s2.

Note: use k ^ 1 and k ^= 1 to switch between dp[0] (row 0) and dp[1] (row 1).

     public int longestCommonSubsequence(String s1, String s2) {
         int m = s1.length(), n = s2.length();
         if (m < n)  return longestCommonSubsequence(s2, s1);
         int[][] dp = new int[2][n + 1];
         for (int i = 0, k = 1; i < m; ++i, k ^= 1)
             for (int j = 0; j < n; ++j)
                 if (s1.charAt(i) == s2.charAt(j)) dp[k][j + 1] = 1 + dp[k ^ 1][j];
                 else dp[k][j + 1] = Math.max(dp[k ^ 1][j + 1], dp[k][j]);
         return dp[m % 2][n];
     }