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102 changes: 102 additions & 0 deletions dynamic_programming/regex_match.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,102 @@
"""
Regex matching check if a text matches wildcard pattern or not.
Pattern:
'.' Matches any single character.
'*' Matches zero or more of the preceding element.
More info:
https://medium.com/trick-the-interviwer/regular-expression-matching-9972eb74c03
"""


def recursive_match(text: str, pattern: str) -> bool:
"""
Recursive matching algorithm.

Time complexity: O(2 ^ (m + n)), where m is the length of text and n is the length of pattern.
Space complexity: Recursion depth is O(m + n).

:param text: Text to match.
:param pattern: Pattern to match.
:return: True if text matches pattern, False otherwise.

>>> recursive_match('abc', 'a.c')
True
>>> recursive_match('abc', 'af*.c')
True
>>> recursive_match('abc', 'a.c*')
True
>>> recursive_match('abc', 'a.c*d')
False
>>> recursive_match('aa', '.*')
True
"""
if not text and not pattern:
return True

if text and not pattern:
return False

if not text and pattern and pattern[-1] != '*':
return False

if not text and pattern and pattern[-1] == '*':
return recursive_match(text, pattern[:-2])

if text[-1] == pattern[-1] or pattern[-1] == '.':
return recursive_match(text[:-1], pattern[:-1])

if pattern[-1] == '*':
return recursive_match(text[:-1], pattern) or recursive_match(text, pattern[:-2])

return False


def dp_match(text: str, pattern: str) -> bool:
"""
Dynamic programming matching algorithm.

Time complexity: O(m * n), where m is the length of text and n is the length of pattern.
Space complexity: O(m * n).

:param text: Text to match.
:param pattern: Pattern to match.
:return: True if text matches pattern, False otherwise.

>>> dp_match('abc', 'a.c')
True
>>> dp_match('abc', 'af*.c')
True
>>> dp_match('abc', 'a.c*')
True
>>> dp_match('abc', 'a.c*d')
False
>>> dp_match('aa', '.*')
True
"""
m = len(text)
n = len(pattern)
dp = [[False for _ in range(n + 1)] for _ in range(m + 1)]
dp[0][0] = True

for i in range(1, m + 1):
dp[i][0] = False

for j in range(1, n + 1):
dp[0][j] = pattern[j - 1] == '*' and dp[0][j - 2]

for i in range(1, m + 1):
for j in range(1, n + 1):
if pattern[j - 1] == '.' or pattern[j - 1] == text[i - 1]:
dp[i][j] = dp[i - 1][j - 1]
elif pattern[j - 1] == '*':
dp[i][j] = dp[i][j - 2] or (dp[i - 1][j] and (pattern[j - 2] == '.' or pattern[j - 2] == text[i - 1]))
else:
dp[i][j] = False

return dp[m][n]


if __name__ == "__main__":
import doctest

doctest.testmod()