Bitap String Match

R
```"""
Bitap exact string matching
https://en.wikipedia.org/wiki/Bitap_algorithm

Searches for a pattern inside text, and returns the index of the first occurrence
of the pattern. Both text and pattern consist of lowercase alphabetical characters only.

Complexity: O(m*n)
n = length of text
m = length of pattern

Python doctests can be run using this command:
python3 -m doctest -v bitap_string_match.py
"""

def bitap_string_match(text: str, pattern: str) -> int:
"""
Retrieves the index of the first occurrence of pattern in text.

Args:
text: A string consisting only of lowercase alphabetical characters.
pattern: A string consisting only of lowercase alphabetical characters.

Returns:
int: The index where pattern first occurs. Return -1  if not found.

>>> bitap_string_match('abdabababc', 'ababc')
5
>>> bitap_string_match('aaaaaaaaaaaaaaaaaa', 'a')
0
>>> bitap_string_match('zxywsijdfosdfnso', 'zxywsijdfosdfnso')
0
>>> bitap_string_match('abdabababc', '')
0
>>> bitap_string_match('abdabababc', 'c')
9
>>> bitap_string_match('abdabababc', 'fofosdfo')
-1
>>> bitap_string_match('abdab', 'fofosdfo')
-1
"""
if not pattern:
return 0
m = len(pattern)
if m > len(text):
return -1

# Initial state of bit string 1110
state = ~1
# Bit = 0 if character appears at index, and 1 otherwise
pattern_mask: list[int] = [~0] * 27  # 1111

for i, char in enumerate(pattern):
# For the pattern mask for this character, set the bit to 0 for each i
# the character appears.
pattern_index: int = ord(char) - ord("a")

for i, char in enumerate(text):
text_index = ord(char) - ord("a")
# If this character does not appear in pattern, it's pattern mask is 1111.
# Performing a bitwise OR between state and 1111 will reset the state to 1111
# and start searching the start of pattern again.
state <<= 1

# If the mth bit (counting right to left) of the state is 0, then we have
# found pattern in text
if (state & (1 << m)) == 0:
return i - m + 1

return -1

if __name__ == "__main__":
import doctest

doctest.testmod()
```