Recommendation System Practice — Chapter 35 Media Design💗💗💗💗

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        @cache
        def dfs(i, j):
            if j >= n:
                return i == m
            if j + 1 < n and p[j + 1] == '*':
                return dfs(i, j + 2) or (
                    i < m and (s[i] == p[j] or p[j] == '.') and dfs(i + 1, j)
                )
            return i < m and (s[i] == p[j] or p[j] == '.') and dfs(i + 1, j + 1)

        m, n = len(s), len(p)
        return dfs(0, 0)

Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where:

• '.' Matches any single character.​​​​
• '*' Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

Example 1:

Input: s = "aa", p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input: s = "aa", p = "a*"
Output: true
Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input: s = "ab", p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Constraints:

• 1 <= s.length <= 20
• 1 <= p.length <= 20
• s contains only lowercase English letters.
• p contains only lowercase English letters, '.', and '*'.
• It is guaranteed for each appearance of the character '*', there will be a previous valid character to match.

Solution 1: Memoization Search

We design a function dfs(i,j), which indicates whether the i-th character of s matches the j-th character of p. The answer is dfs(0,0).

The calculation process of the function dfs(i,j) is as follows:

• If j has reached the end of p, then if i has also reached the end of s, the match is successful, otherwise, the match fails.
• If the next character of j is '*', we can choose to match 0 s[i] characters, which is dfs(i,j+2). If i<m and s[i] matches p[j], we can choose to match 1 s[i] character, which is dfs(i+1,j).
• If the next character of j is not '*', then if i<m and s[i] matches p[j], it is dfs(i+1,j+1). Otherwise, the match fails.

During the process, we can use memoization search to avoid repeated calculations.

The time complexity is O(m×n), and the space complexity is O(m×n). Here, m and n are the lengths of s and p respectively.