human
Solve in Python:
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
 ... Write me a the Python program you requested?
assistant
#Here is the Python program you requested.

class Solution:
     def isMatch(self, s, p):
         """
         :type s: str
         :type p: str
         :rtype: bool
         """
         
         self.mem = {}
         self.t = s
         self.p = p
         
         return self.match(0,0)
     
     def match(self, i, j):
         
         if (i,j) not in self.mem:
             if j == len(self.p):
                 res = i == len(self.t)
                 
             else:
                 if j+1 < len(self.p) and self.p[j+1] == '*':
                     res = self.match(i, j+2) or (i < len(self.t) and self.p[j] in {self.t[i], '.'} and self.match(i+1, j))
                 else:
                     res = i < len(self.t) and self.p[j] in {self.t[i], '.'} and self.match(i+1, j+1)
                     
             self.mem[i, j] = res
                 
         return self.mem[i, j]