Description

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:

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

Example 2:
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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:
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Input: s = "ab", p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Constraints:

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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.

Approach

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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[0][0] = True

        for j in range(2, n + 1):
            if p[j - 1] == '*':
                dp[0][j] = dp[0][j - 2]

        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if p[j - 1] == '.' or p[j - 1] == s[i - 1]:
                    dp[i][j] = dp[i - 1][j - 1]
                elif p[j - 1] == '*':
                    <!-- match zero times -->
                    dp[i][j] = dp[i][j - 2]
                    if p[j - 2] == '.' or p[j - 2] == s[i - 1]:
                        dp[i][j] |= dp[i - 1][j]
        
        return dp[m][n]