Longest Common Subsequence

Problem Statement

Given two strings, find the length of their longest common subsequence (not necessarily contiguous). This medium-level DP puzzle is a treasure hunt—dig out the longest shared thread between two texts!

Example

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

Output: 3 ("ace")

Input: text1 = "abc", text2 = "abc"

Output: 3 ("abc")

Input: text1 = "abc", text2 = "def"

Output: 0 (no common subsequence)

Code

public 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];
    }
}
            

def longest_common_subsequence(text1, text2):
    m, n = len(text1), len(text2)
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if text1[i-1] == text2[j-1]:
                dp[i][j] = dp[i-1][j-1] + 1
            else:
                dp[i][j] = max(dp[i-1][j], dp[i][j-1])
    return dp[m][n]

# Example usage
print(longest_common_subsequence("abcde", "ace"))  # 3
print(longest_common_subsequence("abc", "def"))  # 0
            

function longestCommonSubsequence(text1, text2) {
    let m = text1.length, n = text2.length;
    let dp = Array(m + 1).fill().map(() => Array(n + 1).fill(0));
    for (let i = 1; i <= m; i++) {
        for (let j = 1; j <= n; j++) {
            if (text1[i-1] === text2[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];
}

// Example usage
console.log(longestCommonSubsequence("abcde", "ace"));  // 3
console.log(longestCommonSubsequence("abc", "def"));  // 0
            

Explanation

• DP Insight: Build a 2D table where dp[i][j] = LCS length up to i,j.
• Steps: If chars match, add 1 to diagonal; else, max of left or up.
• Flow Example ("abc","ace"): dp[1][1]=1 (a), dp[2][2]=1, dp[3][3]=2 (ac), final=3 (ace).
• Why It Works: Captures all possible subsequences efficiently.

Note