10. Regular Expression Matching.py

'''
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
'''

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        m = len(s)
        n = len(p)
        dp = [[False for i in range(n+1)] for j in range(m+1)]
        
        dp[0][0] = True
        
        for i in range(2,n+1):
            if p[i-1]=='*':
                dp[0][i] = dp[0][i-2]
                
        for i in range(1,m+1):
            for j in range(1,n+1):
                if s[i-1]==p[j-1] or p[j-1]=='.':
                    dp[i][j] = dp[i-1][j-1]
                elif p[j-1] == '*':
                    op1 = dp[i][j-2]
                    if p[j-2]==s[i-1] or p[j-2]=='.':
                        op2 = dp[i-1][j]
                    else:
                        op2 = False
                    dp[i][j] = op1 or op2
                else:
                    dp[i][j] = False
         
        return dp[m][n]