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backtracking/word_break.py
71 строка
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Aswin P Kumar
Add Word Break algorithm (#11687)
03 окт 2024, 03:03
Не верифицирован
03 окт 2024, 03:03
080e790
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""" Word Break Problem is a well-known problem in computer science. Given a string and a dictionary of words, the task is to determine if the string can be segmented into a sequence of one or more dictionary words. Wikipedia: https://en.wikipedia.org/wiki/Word_break_problem """ def backtrack(input_string: str, word_dict: set[str], start: int) -> bool: """ Helper function that uses backtracking to determine if a valid word segmentation is possible starting from index 'start'. Parameters: input_string (str): The input string to be segmented. word_dict (set[str]): A set of valid dictionary words. start (int): The starting index of the substring to be checked. Returns: bool: True if a valid segmentation is possible, otherwise False. Example: >>> backtrack("leetcode", {"leet", "code"}, 0) True >>> backtrack("applepenapple", {"apple", "pen"}, 0) True >>> backtrack("catsandog", {"cats", "dog", "sand", "and", "cat"}, 0) False """ # Base case: if the starting index has reached the end of the string if start == len(input_string): return True # Try every possible substring from 'start' to 'end' for end in range(start + 1, len(input_string) + 1): if input_string[start:end] in word_dict and backtrack( input_string, word_dict, end ): return True return False def word_break(input_string: str, word_dict: set[str]) -> bool: """ Determines if the input string can be segmented into a sequence of valid dictionary words using backtracking. Parameters: input_string (str): The input string to segment. word_dict (set[str]): The set of valid words. Returns: bool: True if the string can be segmented into valid words, otherwise False. Example: >>> word_break("leetcode", {"leet", "code"}) True >>> word_break("applepenapple", {"apple", "pen"}) True >>> word_break("catsandog", {"cats", "dog", "sand", "and", "cat"}) False """ return backtrack(input_string, word_dict, 0)