Leetcode Regular Expression Matching

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

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

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

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

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".

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

class Solution:
    def isMatch(self, s, p):
        m, n = len(s), len(p)
        if n > 0 and p[0] == '*': return False
        
        s, p = s + '#' , p + '#'    # the hashtags mark the end of string/pattern

        prev = [False] * (m+1)
        current = [False] * (m+1)
        prev[-1] = True            # end of pattern matches end of string

        for i in range(n-1,-1,-1):
            if True not in prev: return False     # terminate early
            if p[i] == '*': continue         # ignore * and move to next char in pattern

            elif p[i] != '.' and p[i+1] == '*':
                current[-1] = prev[-1]
                for j in range(m-1,-1,-1):
                    current[j] =  prev[j] or (p[i] == s[j] and current[j+1]) 

            elif p[i] == '.' and p[i+1] == '*':
                current[-1] = prev[-1]
                for j in range(m-1,-1,-1):
                    current[j] = current[j+1] or prev[j]
            elif p[i] == '.':
                for j in range(m-1,-1,-1):
                    current[j] = prev[j+1]
            else:
                for j in range(m-1,-1,-1):
                    current[j] = prev[j+1] and p[i] == s[j]

            prev = current
            current = [False] * (m+1)

        return prev[0]