add word_break dynamic approach up -> down. (#8039)

* add word_break dynamic approach up -> down.

* updating DIRECTORY.md

* Update word_break.py

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* Update word_break.py

fix review notes

* [pre-commit.ci] auto fixes from pre-commit.com hooks

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* fix review notes

* add trie type

* Update word_break.py

add typing Any to trie.

* Update dynamic_programming/word_break.py

Co-authored-by: Caeden Perelli-Harris <caedenperelliharris@gmail.com>

* Update dynamic_programming/word_break.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* Update dynamic_programming/word_break.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* Update dynamic_programming/word_break.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* Update dynamic_programming/word_break.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* fix review notes

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@ -328,6 +328,7 @@
* [Subset Generation](dynamic_programming/subset_generation.py)
* [Sum Of Subset](dynamic_programming/sum_of_subset.py)
* [Viterbi](dynamic_programming/viterbi.py)
* [Word Break](dynamic_programming/word_break.py)
## Electronics
* [Builtin Voltage](electronics/builtin_voltage.py)

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@ -0,0 +1,111 @@
"""
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)
"""
from functools import lru_cache
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
@lru_cache(maxsize=None)
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()