Longest Common Subsequence

Problem

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.

Solution

Approach: 2D Dynamic Programming

The key insight is that if characters match, we extend the LCS. If they don’t match, we take the maximum LCS from either skipping a character in text1 or text2.

Implementation

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        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]:
                    # Characters match - extend LCS
                    dp[i][j] = dp[i-1][j-1] + 1
                else:
                    # Take max of skipping char in text1 or text2
                    dp[i][j] = max(dp[i-1][j], dp[i][j-1])

        return dp[m][n]

Approach 2: Space-Optimized

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

        # Use shorter string for space optimization
        if m < n:
            text1, text2 = text2, text1
            m, n = n, m

        prev = [0] * (n + 1)

        for i in range(1, m + 1):
            curr = [0] * (n + 1)
            for j in range(1, n + 1):
                if text1[i-1] == text2[j-1]:
                    curr[j] = prev[j-1] + 1
                else:
                    curr[j] = max(prev[j], curr[j-1])
            prev = curr

        return prev[n]

Approach 3: Recursion with Memoization

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        memo = {}

        def dp(i, j):
            if i == len(text1) or j == len(text2):
                return 0

            if (i, j) in memo:
                return memo[(i, j)]

            if text1[i] == text2[j]:
                result = 1 + dp(i + 1, j + 1)
            else:
                result = max(dp(i + 1, j), dp(i, j + 1))

            memo[(i, j)] = result
            return result

        return dp(0, 0)

Complexity Analysis

2D DP:

Time Complexity: O(m * n), where m and n are lengths of the two strings.
Space Complexity: O(m * n) for the dp table.

Space-Optimized:

Time Complexity: O(m * n).
Space Complexity: O(min(m, n)), only storing two rows.

Key Insights

DP Table Meaning: dp[i][j] = length of LCS of text1[0:i] and text2[0:j].
Character Match: If text1[i-1] == text2[j-1], we found a common character. Add 1 to LCS of previous positions.
Character Mismatch: Take maximum of:
- LCS without current char from text1: dp[i-1][j]
- LCS without current char from text2: dp[i][j-1]
Base Cases: Empty string with any string has LCS of 0 (first row and column are 0).
Building Up: We build the solution from smaller subproblems (shorter prefixes).
Subsequence vs Substring: Unlike substring, subsequence allows gaps, so we check all combinations.