mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-12-05 02:40:16 +00:00
c96241b5a5
* Replace bandit, flake8, isort, and pyupgrade with ruff * Comment on ruff rules * updating DIRECTORY.md --------- Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
112 lines
2.8 KiB
Python
112 lines
2.8 KiB
Python
"""
|
|
Author : Alexander Pantyukhin
|
|
Date : December 12, 2022
|
|
|
|
Task:
|
|
Given a string and a list of words, return true if the string can be
|
|
segmented into a space-separated sequence of one or more words.
|
|
|
|
Note that the same word may be reused
|
|
multiple times in the segmentation.
|
|
|
|
Implementation notes: Trie + Dynamic programming up -> down.
|
|
The Trie will be used to store the words. It will be useful for scanning
|
|
available words for the current position in the string.
|
|
|
|
Leetcode:
|
|
https://leetcode.com/problems/word-break/description/
|
|
|
|
Runtime: O(n * n)
|
|
Space: O(n)
|
|
"""
|
|
|
|
import functools
|
|
from typing import Any
|
|
|
|
|
|
def word_break(string: str, words: list[str]) -> bool:
|
|
"""
|
|
Return True if numbers have opposite signs False otherwise.
|
|
|
|
>>> word_break("applepenapple", ["apple","pen"])
|
|
True
|
|
>>> word_break("catsandog", ["cats","dog","sand","and","cat"])
|
|
False
|
|
>>> word_break("cars", ["car","ca","rs"])
|
|
True
|
|
>>> word_break('abc', [])
|
|
False
|
|
>>> word_break(123, ['a'])
|
|
Traceback (most recent call last):
|
|
...
|
|
ValueError: the string should be not empty string
|
|
>>> word_break('', ['a'])
|
|
Traceback (most recent call last):
|
|
...
|
|
ValueError: the string should be not empty string
|
|
>>> word_break('abc', [123])
|
|
Traceback (most recent call last):
|
|
...
|
|
ValueError: the words should be a list of non-empty strings
|
|
>>> word_break('abc', [''])
|
|
Traceback (most recent call last):
|
|
...
|
|
ValueError: the words should be a list of non-empty strings
|
|
"""
|
|
|
|
# Validation
|
|
if not isinstance(string, str) or len(string) == 0:
|
|
raise ValueError("the string should be not empty string")
|
|
|
|
if not isinstance(words, list) or not all(
|
|
isinstance(item, str) and len(item) > 0 for item in words
|
|
):
|
|
raise ValueError("the words should be a list of non-empty strings")
|
|
|
|
# Build trie
|
|
trie: dict[str, Any] = {}
|
|
word_keeper_key = "WORD_KEEPER"
|
|
|
|
for word in words:
|
|
trie_node = trie
|
|
for c in word:
|
|
if c not in trie_node:
|
|
trie_node[c] = {}
|
|
|
|
trie_node = trie_node[c]
|
|
|
|
trie_node[word_keeper_key] = True
|
|
|
|
len_string = len(string)
|
|
|
|
# Dynamic programming method
|
|
@functools.cache
|
|
def is_breakable(index: int) -> bool:
|
|
"""
|
|
>>> string = 'a'
|
|
>>> is_breakable(1)
|
|
True
|
|
"""
|
|
if index == len_string:
|
|
return True
|
|
|
|
trie_node = trie
|
|
for i in range(index, len_string):
|
|
trie_node = trie_node.get(string[i], None)
|
|
|
|
if trie_node is None:
|
|
return False
|
|
|
|
if trie_node.get(word_keeper_key, False) and is_breakable(i + 1):
|
|
return True
|
|
|
|
return False
|
|
|
|
return is_breakable(0)
|
|
|
|
|
|
if __name__ == "__main__":
|
|
import doctest
|
|
|
|
doctest.testmod()
|