Add Word Break algorithm (#11687)

* Add Word Break algorithm

* Add Word Break algorithm

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Aswin P Kumar 2024-10-03 05:33:48 +05:30 committed by GitHub
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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)