Medium — Dynamic Programming | String | Two Pointers

The Problem

Find the length of the longest common subsequence (LCS) between two strings, where a subsequence maintains the relative order of characters but doesn’t need to be contiguous.

Approach

Use dynamic programming with a 2D table where dp[i][j] represents the LCS length of the first i characters of text1 and first j characters of text2. If characters match, add 1 to the diagonal value; otherwise, take the maximum of the left or top cell.

Time: O(mn) · Space: O(mn)

Code

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        m, n = len(text1), len(text2)

        # Create a 2D dp array of size (m+1) x (n+1) and initialize it with zeros.
        dp = [[0] * (n + 1) for _ in range(m + 1)]

        # Fill in the dp array using dynamic programming.
        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])

        # The value in dp[m][n] represents the length of the LCS.
        return dp[m][n]

Watch It Run

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