algorithm-test/src/bm.py

from typing import Dict


def make_km_table(pattern: str) -> Dict[str, int]:
    """
    Build a lookup table from a pattern string.

    Iterates over the string and creates a dictionary where the key is a
    search character and the value is the distance from the end of the string
    to the *last* occurrence in the pattern. We ignore the final character.

    Example:
      pattern = "ABCDAB"
                 543210 <- distance from end of pattern
      table = {
        "A": 1,   # The second 'A'
        "B": 4,   # The first 'B', because the other 'B' is the end of the string
        "C": 3,
        "D": 2,
      }
    """
    PATTERN_LENGTH = len(pattern)
    table = {}

    # Note: Because we always start each loop in the search by comparing the
    # last character in the pattern, we can skip it in the table. This is
    # especially helpful if the last character also appears elsewhere in the
    # pattern because when using this table we will have already tried matching
    # the last character, so we want to try the *next* occurrence of that
    # character in the pattern, if it exists.
    for index_from_left, char in enumerate(pattern[:-1]):
        table[char] = PATTERN_LENGTH - index_from_left - 1
    return table


class Bm(object):
    def __init__(self, text: str, pattern: str):
        self.text = text
        self.pattern = pattern
        self.table = make_km_table(pattern)

    def decide_slide_width(self, c: str) -> int:
        """
        If a character 'c' doesn't match in the search, this decides how far
        down to "slide" the pattern for the next search.
        """

        # Note: The lookup table only has characters from the pattern in it.
        # If the test character 'c' is not in the table, then we should skip
        # down the entire length of the pattern
        try:
            return self.table[c]
        except KeyError:
            return len(self.pattern)

    def search(self) -> int:
        """
        Return the index of the *first* occurrence of the pattern in the
        text, or '-1' if the pattern is not in the text.
        """
        PATTERN_LENGTH = len(self.pattern)
        if not self.pattern or not self.text or len(self.text) < PATTERN_LENGTH:
            return -1

        # The search works on one character at a time in the text. The current
        # location is the "head" (like a read-head on a disk)
        # It starts at the last character in the pattern, and gets moved in the
        # loop until we reach the end of the text
        head = PATTERN_LENGTH - 1
        while head < len(self.text):
            # When the last character of the pattern matches the current head
            # position, walk backwards through the text and match each
            # character one by one until either they all match (return) or one
            # doesn't match (slide the head down)
            if self.text[head] == self.pattern[-1]:
                for index_from_right, char in enumerate(reversed(self.pattern)):
                    if self.text[head - index_from_right] != char:
                        head += self.decide_slide_width(self.text[head])
                        break
                # If the for loop finished without breaking we found a full
                # match!
                else:
                    # Note: head is always at the end of the pattern, but the
                    # search should return the index of the start of the pattern
                    return head - PATTERN_LENGTH + 1
            # When the last character of the pattern *doesn't* match the current
            # head, always slide down the head
            else:
                head += self.decide_slide_width(self.text[head])

        # If the main loop doesn't return, there was no match
        return -1