mirror of
https://github.com/TheAlgorithms/Python.git
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e3eb9daba4
* Add bitap_string_match algo * Fix types * Fix spelling and add ignore word * Add suggested changes and change return type * Resolve suggestions
80 lines
2.3 KiB
Python
80 lines
2.3 KiB
Python
"""
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Bitap exact string matching
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https://en.wikipedia.org/wiki/Bitap_algorithm
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Searches for a pattern inside text, and returns the index of the first occurrence
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of the pattern. Both text and pattern consist of lowercase alphabetical characters only.
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Complexity: O(m*n)
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n = length of text
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m = length of pattern
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Python doctests can be run using this command:
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python3 -m doctest -v bitap_string_match.py
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"""
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def bitap_string_match(text: str, pattern: str) -> int:
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"""
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Retrieves the index of the first occurrence of pattern in text.
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Args:
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text: A string consisting only of lowercase alphabetical characters.
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pattern: A string consisting only of lowercase alphabetical characters.
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Returns:
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int: The index where pattern first occurs. Return -1 if not found.
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>>> bitap_string_match('abdabababc', 'ababc')
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5
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>>> bitap_string_match('aaaaaaaaaaaaaaaaaa', 'a')
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0
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>>> bitap_string_match('zxywsijdfosdfnso', 'zxywsijdfosdfnso')
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0
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>>> bitap_string_match('abdabababc', '')
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0
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>>> bitap_string_match('abdabababc', 'c')
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9
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>>> bitap_string_match('abdabababc', 'fofosdfo')
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-1
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>>> bitap_string_match('abdab', 'fofosdfo')
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-1
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"""
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if not pattern:
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return 0
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m = len(pattern)
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if m > len(text):
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return -1
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# Initial state of bit string 1110
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state = ~1
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# Bit = 0 if character appears at index, and 1 otherwise
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pattern_mask: list[int] = [~0] * 27 # 1111
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for i, char in enumerate(pattern):
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# For the pattern mask for this character, set the bit to 0 for each i
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# the character appears.
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pattern_index: int = ord(char) - ord("a")
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pattern_mask[pattern_index] &= ~(1 << i)
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for i, char in enumerate(text):
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text_index = ord(char) - ord("a")
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# If this character does not appear in pattern, it's pattern mask is 1111.
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# Performing a bitwise OR between state and 1111 will reset the state to 1111
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# and start searching the start of pattern again.
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state |= pattern_mask[text_index]
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state <<= 1
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# If the mth bit (counting right to left) of the state is 0, then we have
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# found pattern in text
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if (state & (1 << m)) == 0:
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return i - m + 1
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return -1
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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