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* add new programme in dynamique programming wildcard_matching * add new programme in dynamique programming wildcard_matching * fix bug * fix * fix * fix * fix * fix * fix error recrusion * fix error recrusion * bug fix * add doctest * The power of enumerate() --------- Co-authored-by: Christian Clauss <cclauss@me.com>
69 lines
1.7 KiB
Python
69 lines
1.7 KiB
Python
"""
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Author : ilyas dahhou
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Date : Oct 7, 2023
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Task:
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Given an input string and a pattern, implement wildcard pattern matching with support
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for '?' and '*' where:
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'?' matches any single character.
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'*' matches any sequence of characters (including the empty sequence).
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The matching should cover the entire input string (not partial).
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Runtime complexity: O(m * n)
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The implementation was tested on the
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leetcode: https://leetcode.com/problems/wildcard-matching/
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"""
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def is_match(string: str, pattern: str) -> bool:
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"""
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>>> is_match("", "")
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True
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>>> is_match("aa", "a")
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False
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>>> is_match("abc", "abc")
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True
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>>> is_match("abc", "*c")
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True
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>>> is_match("abc", "a*")
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True
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>>> is_match("abc", "*a*")
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True
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>>> is_match("abc", "?b?")
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True
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>>> is_match("abc", "*?")
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True
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>>> is_match("abc", "a*d")
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False
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>>> is_match("abc", "a*c?")
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False
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>>> is_match('baaabab','*****ba*****ba')
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False
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>>> is_match('baaabab','*****ba*****ab')
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True
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>>> is_match('aa','*')
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True
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"""
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dp = [[False] * (len(pattern) + 1) for _ in string + "1"]
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dp[0][0] = True
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# Fill in the first row
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for j, char in enumerate(pattern, 1):
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if char == "*":
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dp[0][j] = dp[0][j - 1]
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# Fill in the rest of the DP table
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for i, s_char in enumerate(string, 1):
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for j, p_char in enumerate(pattern, 1):
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if p_char in (s_char, "?"):
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dp[i][j] = dp[i - 1][j - 1]
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elif pattern[j - 1] == "*":
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dp[i][j] = dp[i - 1][j] or dp[i][j - 1]
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return dp[len(string)][len(pattern)]
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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print(f"{is_match('baaabab','*****ba*****ab') = }")
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