1143 - Longest Common Subsequence (Medium)

Problem Link

https://leetcode.com/problems/longest-common-subsequence

Problem Statement

Given two strings text1 and text2, return the length of their longest common subsequence. If there is no common subsequence, return 0.

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.

• For example, "ace" is a subsequence of "abcde".

A common subsequence of two strings is a subsequence that is common to both strings.

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.

Constraints:

• 1 <= text1.length, text2.length <= 1000
• text1 and text2 consist of only lowercase English characters.

Approach 1: DP

LCS is a classic problem. Let $dp[i][j]$ be the LCS for string $text1$ ends at index $i$ and string $text2$ ends at index $j$. If $text1[i] == text2[j]$, then $dp[i][j]$ would be $1 + dp[i - 1][j - 1]$. Otherwise, we target the largest LCS if we skip one character from either $text1$ or $text2$, i.e. $dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])$.

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int n = text1.size(), m = text2.size();
        vector<vector<int>> dp(n + 1, vector<int>(m + 1));
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                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[n][m];
    }
};