LeetCode 1143: Longest Common Subsequence

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.

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

Examples

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 common subsequence between the two strings.

Constraints

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

Dynamic Programming Solution

We can solve this using a 2D DP table where:

Rows represent characters from text1
Columns represent characters from text2
Each cell [i][j] represents the length of LCS for text1[0…i-1] and text2[0…j-1]

The DP state transition:

If characters match: dp[i][j] = 1 + dp[i-1][j-1]
If characters don’t match: dp[i][j] = max(dp[i-1][j], dp[i][j-1])

Implementation

Here’s the Python implementation:

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        n, m = len(text1), len(text2)
        grid = [[0] * (m+1) for _ in range(n+1)]
        
        for r in range(1, n+1):
            for c in range(1, m+1):
                if text1[r-1] == text2[c-1]:
                    grid[r][c] = 1 + grid[r-1][c-1]
                else:
                    grid[r][c] = max(grid[r-1][c], grid[r][c-1])
        
        return grid[n][m]

Bonus: Retrieving the Subsequence

While the problem only asks for the length, we can also retrieve the actual subsequence:

def getLCS(self, text1: str, text2: str) -> str:
    n, m = len(text1), len(text2)
    grid = [[0] * (m+1) for _ in range(n+1)]
    
    # Fill the grid as before
    for r in range(1, n+1):
        for c in range(1, m+1):
            if text1[r-1] == text2[c-1]:
                grid[r][c] = 1 + grid[r-1][c-1]
            else:
                grid[r][c] = max(grid[r-1][c], grid[r][c-1])
    
    # Backtrack to find the subsequence
    subsequence = ""
    r, c = n, m
    while r > 0 and c > 0:
        if text1[r-1] == text2[c-1]:
            subsequence = text1[r-1] + subsequence
            r -= 1
            c -= 1
        elif grid[r-1][c] > grid[r][c-1]:
            r -= 1
        else:
            c -= 1
    
    return subsequence

Complexity Analysis

Time Complexity: $O(n × m)$ , where n and m are the lengths of the input strings
- We need to fill every cell in the DP table exactly once
- Each cell operation is O(1)
Space Complexity: $O(n × m)$
- We need a 2D table to store the intermediate results
- This can be optimized to O(min(n,m)) by using only two rows