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 precedeng 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, p):
         """
         :type s: str
         :type p: str
         :rtype: bool
         """
         dp = [[False] * (len(p) + 1) for _ in range(len(s) + 1)]
         dp[-1][-1] = True
         
         for i in range(len(s), -1, -1):
             for j in range(len(p) - 1, -1, -1):
                 first_match = i < len(s) and p[j] in {s[i], '.'}
                 if j + 1 < len(p) and p[j+1] is '*':
                     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, p):
         """
         :type s: str
         :type p: str
         :rtype: bool
         """
         dp = [[False] * (len(p) + 1) for _ in range(len(s) + 1)]
         dp[-1][-1] = True
         
         for i in range(len(s), -1, -1):
             for j in range(len(p) - 1, -1, -1):
                 first_match = i < len(s) and p[j] in {s[i], '.'}
                 if j + 1 < len(p) and p[j+1] is '*':
                     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]
                     
                         
                     
                     
                     
                 
