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* Type annotations for `strings/autocomplete_using_trie.py` * Update autocomplete_using_trie.py * Update detecting_english_programmatically.py * Update detecting_english_programmatically.py * Update frequency_finder.py * Update frequency_finder.py * Update frequency_finder.py * Update word_occurrence.py * Update frequency_finder.py * Update z_function.py * Update z_function.py * Update frequency_finder.py
91 lines
2.5 KiB
Python
91 lines
2.5 KiB
Python
"""
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https://cp-algorithms.com/string/z-function.html
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Z-function or Z algorithm
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Efficient algorithm for pattern occurrence in a string
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Time Complexity: O(n) - where n is the length of the string
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"""
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def z_function(input_str: str) -> list[int]:
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"""
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For the given string this function computes value for each index,
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which represents the maximal length substring starting from the index
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and is the same as the prefix of the same size
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e.x. for string 'abab' for second index value would be 2
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For the value of the first element the algorithm always returns 0
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>>> z_function("abracadabra")
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[0, 0, 0, 1, 0, 1, 0, 4, 0, 0, 1]
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>>> z_function("aaaa")
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[0, 3, 2, 1]
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>>> z_function("zxxzxxz")
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[0, 0, 0, 4, 0, 0, 1]
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"""
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z_result = [0 for i in range(len(input_str))]
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# initialize interval's left pointer and right pointer
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left_pointer, right_pointer = 0, 0
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for i in range(1, len(input_str)):
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# case when current index is inside the interval
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if i <= right_pointer:
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min_edge = min(right_pointer - i + 1, z_result[i - left_pointer])
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z_result[i] = min_edge
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while go_next(i, z_result, input_str):
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z_result[i] += 1
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# if new index's result gives us more right interval,
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# we've to update left_pointer and right_pointer
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if i + z_result[i] - 1 > right_pointer:
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left_pointer, right_pointer = i, i + z_result[i] - 1
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return z_result
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def go_next(i: int, z_result: list[int], s: str) -> bool:
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"""
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Check if we have to move forward to the next characters or not
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"""
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return i + z_result[i] < len(s) and s[z_result[i]] == s[i + z_result[i]]
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def find_pattern(pattern: str, input_str: str) -> int:
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"""
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Example of using z-function for pattern occurrence
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Given function returns the number of times 'pattern'
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appears in 'input_str' as a substring
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>>> find_pattern("abr", "abracadabra")
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2
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>>> find_pattern("a", "aaaa")
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4
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>>> find_pattern("xz", "zxxzxxz")
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2
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"""
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answer = 0
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# concatenate 'pattern' and 'input_str' and call z_function
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# with concatenated string
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z_result = z_function(pattern + input_str)
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for val in z_result:
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# if value is greater then length of the pattern string
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# that means this index is starting position of substring
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# which is equal to pattern string
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if val >= len(pattern):
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answer += 1
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return answer
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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