331. Longest Common Subsequence

Compare the two strings prefix by prefix. When the last characters match, extend the diagonal LCS; otherwise take the better of dropping one character from either string.

Definition

dp[i][j] = length of the LCS of the prefixes text1[:i] and text2[:j].

Transition

if text1[i-1] == text2[j-1]:
    dp[i][j] = dp[i-1][j-1] + 1        # matched one more character
else:
    dp[i][j] = max(dp[i-1][j], dp[i][j-1])   # drop a char from one side

Row 0 and column 0 are all zeros (an empty string shares nothing).

Worked example

text1 = "abcde", text2 = "ace":

• a matches a → 1, c matches c → 2, e matches e → 3.
• Non-matching cells inherit the best neighbour.
• dp[5][3] = 3 → LCS is "ace".

Complexity

• Time: O(m·n).
• Space: O(m·n) (reducible to O(min(m,n))).

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