Add dp solution to regex_match.py

This commit is contained in:
AmirMohammad Hosseini Nasab 2022-08-12 12:52:38 +04:30
parent ba296e1e89
commit 0cd88af4ff

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@ -25,6 +25,8 @@ def recursive_match(text: str, pattern: str) -> bool:
True
>>> recursive_match('abc', 'a.c*d')
False
>>> recursive_match('aa', '.*')
True
"""
if not text and not pattern:
return True
@ -47,6 +49,51 @@ def recursive_match(text: str, pattern: str) -> bool:
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