Python/dynamic_programming/word_break.py
Christian Clauss c96241b5a5
Replace bandit, flake8, isort, and pyupgrade with ruff (#8178)
* Replace bandit, flake8, isort, and pyupgrade with ruff

* Comment on ruff rules

* updating DIRECTORY.md

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Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
2023-03-15 13:58:25 +01:00

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()