Longest Common Subsequence

LeetCode 1250 | Difficulty: Medium

Medium

Problem Description

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.

Topics: String, Dynamic Programming

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Approach

Dynamic Programming

Break the problem into overlapping subproblems. Define a state (what information do you need?), a recurrence (how does state[i] depend on smaller states?), and a base case. Consider both top-down (memoization) and bottom-up (tabulation) approaches.

When to use

Optimal substructure + overlapping subproblems (counting ways, min/max cost, feasibility).

String Processing

Consider character frequency counts, two-pointer approaches, or building strings efficiently. For pattern matching, think about KMP or rolling hash. For palindromes, expand from center or use DP.

When to use

Anagram detection, palindrome checking, string transformation, pattern matching.

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Solutions

Solution 1: C# (Best: 84 ms)

Metric	Value
Runtime	84 ms
Memory	25.7 MB
Date	2020-01-12

Solution

public class Solution {
    public int LongestCommonSubsequence(string text1, string text2) {
        int m = text1.Length;
    int 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[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];
    }
}

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Complexity Analysis

Approach	Time	Space
DP (2D)	$O(n × m)$	$O(n × m)$

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Interview Tips

Key Points

• Discuss the brute force approach first, then optimize. Explain your thought process.
• Define the DP state clearly. Ask: "What is the minimum information I need to make a decision at each step?"
• Consider if you can reduce space by only keeping the last row/few values.
• LeetCode provides 2 hint(s) for this problem — try solving without them first.

💡 Hints

Hint 1: Try dynamic programming. DP[i][j] represents the longest common subsequence of text1[0 ... i] & text2[0 ... j].

Hint 2: DP[i][j] = DP[i - 1][j - 1] + 1 , if text1[i] == text2[j] DP[i][j] = max(DP[i - 1][j], DP[i][j - 1]) , otherwise