1143. Longest Common Subsequence - Leetcode Solution

LeetCode problem

Solution code

💡 Step-by-Step Thought Process

Understand the problem: Find the length of the longest common subsequence (LCS) between two strings, where a subsequence is formed by deleting some characters without changing the order.
Get the lengths m and n of text1 and text2, respectively.
Create a 2D array dp of size (m+1) x (n+1) initialized with zeros to store the LCS lengths for prefixes of text1 and text2.
Iterate through indices i from 1 to m and j from 1 to n to fill the dp array.
For each (i, j), if text1[i-1] equals text2[j-1], set dp[i][j] to dp[i-1][j-1] + 1, including the matching character.
Otherwise, set dp[i][j] to the maximum of dp[i-1][j] (excluding text1[i-1]) and dp[i][j-1] (excluding text2[j-1]).
Return dp[m][n] as the length of the longest common subsequence.

Code Solution

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
	      # Bottom Up DP (Tabulation)
	      # Time: O(m*n)
        # Space: O(m*n)

        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]

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int m = text1.size();
        int n = text2.size();
        
        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
        
        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] = max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        
        return dp[m][n];
    }
};

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.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];
    }
}

var longestCommonSubsequence = function(text1, text2) {
    const m = text1.length;
    const n = text2.length;
    
    const dp = Array.from({ length: m + 1 }, () => 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];
};

Detailed Explanation

What is the Longest Common Subsequence (LCS)?

The Longest Common Subsequence problem is a classic example of dynamic programming. It asks us to find the longest sequence of characters that appear left-to-right (not necessarily contiguously) in both input strings. For example, the LCS of "abcde" and "ace" is "ace" with a length of 3. This problem is fundamental in text comparison, DNA sequence alignment, and file differencing tools.

Dynamic Programming Approach to LCS

To solve the LCS problem efficiently, we use a 2D dynamic programming table. Suppose the input strings are text1 and text2 of lengths m and n respectively. We create a table dp where dp[i][j] stores the length of the LCS between text1[0..i-1] and text2[0..j-1].

We initialize a matrix of size (m+1) x (n+1) with zeros. Then we iterate through the characters of both strings. If the characters at current positions match, we extend the LCS by 1; otherwise, we take the maximum LCS length from the previous row or column.

Time and Space Complexity

Time Complexity: O(m × n), where m and n are the lengths of the two input strings. We fill a matrix of size m×n.
Space Complexity: O(m × n), due to the 2D array storing intermediate LCS lengths. This can be optimized to O(min(m, n)) using a rolling array.

Optimization Tip

If memory is a concern, you can reduce the space usage by only storing the previous row of the DP matrix. Since each cell dp[i][j] only depends on dp[i-1][j], dp[i][j-1], and dp[i-1][j-1], a single row or two rows suffice.

Conclusion

The Longest Common Subsequence is a cornerstone problem in computer science education. Mastering its dynamic programming solution provides a strong foundation for tackling more complex string processing challenges. Whether you're preparing for coding interviews or implementing text comparison tools, the LCS algorithm is essential knowledge.

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