Problem of The Day: Longest Common Subsequence

Problem Statement

leetcode problem link

My Solution [Accepted]

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

        @cache
        def dp(i, j):
            if i == n or j == m:
                return 0

            both = move_text1 = move_text2 = 0
            if text1[i] == text2[j]:
                both = dp(i + 1, j + 1) + 1
            else:
                move_text1 = dp(i + 1, j)
                move_text2 = dp(i, j + 1)

            return max(both, move_text1, move_text2)

        return dp(0, 0)

Editorial

Approach 1: Memoization

from functools import lru_cache

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:

        @lru_cache(maxsize=None)
        def memo_solve(p1, p2):

            # Base case: If either string is now empty, we can't match
            # up anymore characters.
            if p1 == len(text1) or p2 == len(text2):
                return 0

            # Option 1: We don't include text1[p1] in the solution.
            option_1 = memo_solve(p1 + 1, p2)

            # Option 2: We include text1[p1] in the solution, as long as
            # a match for it in text2 at or after p2 exists.
            first_occurence = text2.find(text1[p1], p2)
            option_2 = 0
            if first_occurence != -1:
                option_2 = 1 + memo_solve(p1 + 1, first_occurence + 1)

            # Return the best option.
            return max(option_1, option_2)

        return memo_solve(0, 0)

Approach 2: Improved Memoization

from functools import lru_cache
class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:

        @lru_cache(maxsize=None)
        def memo_solve(p1, p2):

            # Base case: If either string is now empty, we can't match
            # up anymore characters.
            if p1 == len(text1) or p2 == len(text2):
                return 0

            # Recursive case 1.
            if text1[p1] == text2[p2]:
                return 1 + memo_solve(p1 + 1, p2 + 1)

            # Recursive case 2.
            else:
                return max(memo_solve(p1, p2 + 1), memo_solve(p1 + 1, p2))

        return memo_solve(0, 0)

Approach 3: Dynamic Programming

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:

        # Make a grid of 0's with len(text2) + 1 columns
        # and len(text1) + 1 rows.
        dp_grid = [[0] * (len(text2) + 1) for _ in range(len(text1) + 1)]

        # Iterate up each column, starting from the last one.
        for col in reversed(range(len(text2))):
            for row in reversed(range(len(text1))):
                # If the corresponding characters for this cell are the same...
                if text2[col] == text1[row]:
                    dp_grid[row][col] = 1 + dp_grid[row + 1][col + 1]
                # Otherwise they must be different...
                else:
                    dp_grid[row][col] = max(dp_grid[row + 1][col], dp_grid[row][col + 1])

        # The original problem's answer is in dp_grid[0][0]. Return it.
        return dp_grid[0][0]