"""
https://cp-algorithms.com/string/z-function.html
Z-function or Z algorithm
Efficient algorithm for pattern occurrence in a string
Time Complexity: O(n) - where n is the length of the string
"""
def z_function(input_str: str) -> list:
"""
For the given string this function computes value for each index,
which represents the maximal length substring starting from the index
and is the same as the prefix of the same size
e.x. for string 'abab' for second index value would be 2
For the value of the first element the algorithm always returns 0
>>> z_function("abracadabra")
[0, 0, 0, 1, 0, 1, 0, 4, 0, 0, 1]
>>> z_function("aaaa")
[0, 3, 2, 1]
>>> z_function("zxxzxxz")
[0, 0, 0, 4, 0, 0, 1]
"""
z_result = [0] * len(input_str)
left_pointer, right_pointer = 0, 0
for i in range(1, len(input_str)):
if i <= right_pointer:
min_edge = min(right_pointer - i + 1, z_result[i - left_pointer])
z_result[i] = min_edge
while go_next(i, z_result, input_str):
z_result[i] += 1
if i + z_result[i] - 1 > right_pointer:
left_pointer, right_pointer = i, i + z_result[i] - 1
return z_result
def go_next(i, z_result, s):
"""
Check if we have to move forward to the next characters or not
"""
return i + z_result[i] < len(s) and s[z_result[i]] == s[i + z_result[i]]
def find_pattern(pattern: str, input_str: str) -> int:
"""
Example of using z-function for pattern occurrence
Given function returns the number of times 'pattern'
appears in 'input_str' as a substring
>>> find_pattern("abr", "abracadabra")
2
>>> find_pattern("a", "aaaa")
4
>>> find_pattern("xz", "zxxzxxz")
2
"""
answer = 0
z_result = z_function(pattern + input_str)
for val in z_result:
if val >= len(pattern):
answer += 1
return answer
if __name__ == "__main__":
import doctest
doctest.testmod()