10 Regular Expression Matching

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 <= 30
• 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.

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class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        m, n = len(s), len(p)
        dp = [[False] * (n+1) for _ in range(m+1)]
        
        dp[-1][-1] = True
        # Bottom-Up DP
        for i in range(m, -1, -1):
            for j in range(n-1, -1, -1):
                first_match = i < m and p[j] in {s[i], '.'}
                if j+1 < n and p[j+1] == '*':
                    dp[i][j] = dp[i][j+2] or first_match and dp[i+1][j]
                else:
                    dp[i][j] = first_match and dp[i+1][j+1]
        return dp[0][0]