
import re
"""
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
"""

"""
Runtime: 32 ms, faster than 99.40% of Python3 online submissions for Regular Expression Matching.
Memory Usage: 14.1 MB, less than 10.16% of Python3 online submissions for Regular Expression Matching.
"""


class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        return re.match(p+"$", s) != None


class Solution(object):
    def isMatch(self, text, pattern):
        dp = [[False] * (len(pattern) + 1) for _ in range(len(text) + 1)]

        dp[-1][-1] = True
        for i in range(len(text), -1, -1):
            for j in range(len(pattern) - 1, -1, -1):
                first_match = i < len(text) and pattern[j] in {text[i], '.'}
                if j+1 < len(pattern) and pattern[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]


class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        allMatch = True
        sRe = re.search("[a-z]", s)
        pRe = re.search("[.*]", p)
        if sRe == None or pRe == None:
            allMatch = False
        return allMatch
