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.

SOLUTION:

class Solution:
    def isSame(self, a, b):
        if a == "." or b == ".":
            return True
        return a == b

    def isMatchRec(self, s, p, i, j, m, n):
        if i >= m:
            if j >= n:
                return True
            return j < n - 1 and p[j + 1] == "*" and self.isMatchRec(s, p, i, j + 2, m, n)
        if j >= n:
            return i >= m
        if j == n - 1 or p[j + 1] != "*":
            return self.isSame(s[i], p[j]) and self.isMatchRec(s, p, i + 1, j + 1, m, n)
        if p[j + 1] == "*":
            return self.isMatchRec(s, p, i, j + 2, m, n) or (self.isSame(s[i], p[j]) and self.isMatchRec(s, p, i + 1, j, m, n))
        return False

    def isMatch(self, s: str, p: str) -> bool:
        m = len(s)
        n = len(p)
        return self.isMatchRec(s, p, 0, 0, m, n)