#### Wildcard Matching

P
A
R
"""
Author  : ilyas dahhou
Date    : Oct 7, 2023

Given an input string and a pattern, implement wildcard pattern matching with support
for '?' and '*' where:
'?' matches any single character.
'*' matches any sequence of characters (including the empty sequence).
The matching should cover the entire input string (not partial).

Runtime complexity: O(m * n)

The implementation was tested on the
leetcode: https://leetcode.com/problems/wildcard-matching/
"""

def is_match(string: str, pattern: str) -> bool:
"""
>>> is_match("", "")
True
>>> is_match("aa", "a")
False
>>> is_match("abc", "abc")
True
>>> is_match("abc", "*c")
True
>>> is_match("abc", "a*")
True
>>> is_match("abc", "*a*")
True
>>> is_match("abc", "?b?")
True
>>> is_match("abc", "*?")
True
>>> is_match("abc", "a*d")
False
>>> is_match("abc", "a*c?")
False
>>> is_match('baaabab','*****ba*****ba')
False
>>> is_match('baaabab','*****ba*****ab')
True
>>> is_match('aa','*')
True
"""
dp = [[False] * (len(pattern) + 1) for _ in string + "1"]
dp[0][0] = True
# Fill in the first row
for j, char in enumerate(pattern, 1):
if char == "*":
dp[0][j] = dp[0][j - 1]
# Fill in the rest of the DP table
for i, s_char in enumerate(string, 1):
for j, p_char in enumerate(pattern, 1):
if p_char in (s_char, "?"):
dp[i][j] = dp[i - 1][j - 1]
elif pattern[j - 1] == "*":
dp[i][j] = dp[i - 1][j] or dp[i][j - 1]
return dp[len(string)][len(pattern)]

if __name__ == "__main__":
import doctest

doctest.testmod()
print(f"{is_match('baaabab','*****ba*****ab') = }")