Leetcode 10. Regular Expression Matching

Seawolf

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

'.' Matches any single character.
'*' Matches zero or more of the preceding element.
The matching should cover the entire input string (not partial).

Note:

s could be empty and contains only lowercase letters a-z.
p could be empty and contains only lowercase letters a-z, and characters like . or *.
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 (.)".
Example 4:

Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore, it matches "aab".
Example 5:

Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        # X X X X X
        # X X X *
        m = len(s)
        n = len(p)
        dp = [[False for _ in range(n + 1)] for _ in range(m + 1)]
        
        s = '#' + s
        p = '#' + p
        
        dp[0][0] = True
        for j in range(2, n + 1):
            dp[0][j] = (p[j] == '*' and dp[0][j - 2])
        
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if p[j].isalpha():
                    dp[i][j] = (s[i] == p[j] and dp[i - 1][j - 1])
                    
                elif p[j] == '.':
                    dp[i][j] = dp[i - 1][j - 1]
                
                elif p[j] == '*':
                    dp[i][j] = ((s[i] == p[j - 1] or p[j - 1] == '.') and dp[i - 1][j]) or dp[i][j - 2]
        
        return dp[m][n]