1+ """
2+ Task:
3+ Given a string and a list of words, return true if the string can be
4+ segmented into a space-separated sequence of one or more words.
5+
6+ Note that the same word may be reused
7+ multiple times in the segmentation.
8+
9+ Implementation notes: Trie + Dynamic programming up -> down.
10+ The Trie will be used to store the words. It will be useful for scanning
11+ available words for the current position in the string.
12+
13+ Leetcode:
14+ https://leetcode.com/problems/word-break/description/
15+
16+ Runtime: O(n * n)
17+ Space: O(n)
18+ """
19+ 20+ import functools
21+ from typing import Any
22+ 23+ 24+ def word_break (string : str , words : list [str ]) -> bool :
25+ """
26+ Return True if numbers have opposite signs False otherwise.
27+
28+ >>> word_break("applepenapple", ["apple","pen"])
29+ True
30+ >>> word_break("catsandog", ["cats","dog","sand","and","cat"])
31+ False
32+ >>> word_break("cars", ["car","ca","rs"])
33+ True
34+ >>> word_break('abc', [])
35+ False
36+ >>> word_break(123, ['a'])
37+ Traceback (most recent call last):
38+ ...
39+ ValueError: the string should be not empty string
40+ >>> word_break('', ['a'])
41+ Traceback (most recent call last):
42+ ...
43+ ValueError: the string should be not empty string
44+ >>> word_break('abc', [123])
45+ Traceback (most recent call last):
46+ ...
47+ ValueError: the words should be a list of non-empty strings
48+ >>> word_break('abc', [''])
49+ Traceback (most recent call last):
50+ ...
51+ ValueError: the words should be a list of non-empty strings
52+ """
53+ 54+ # Validation
55+ if not isinstance (string , str ) or len (string ) == 0 :
56+ raise ValueError ("the string should be not empty string" )
57+ 58+ if not isinstance (words , list ) or not all (
59+ isinstance (item , str ) and len (item ) > 0 for item in words
60+ ):
61+ raise ValueError ("the words should be a list of non-empty strings" )
62+ 63+ 64+ # Build trie
65+ trie : dict [str , Any ] = {}
66+ word_keeper_key = "WORD_KEEPER"
67+ 68+ for word in words :
69+ trie_node = trie
70+ for c in word :
71+ if c not in trie_node :
72+ trie_node [c ] = {}
73+ 74+ trie_node = trie_node [c ]
75+ 76+ trie_node [word_keeper_key ] = True
77+ 78+ len_string = len (string )
79+ 80+ # Dynamic programming method
81+ @functools .cache
82+ def is_breakable (index : int ) -> bool :
83+ """
84+ >>> string = 'a'
85+ >>> is_breakable(1)
86+ True
87+ """
88+ if index == len_string :
89+ return True
90+ 91+ trie_node = trie
92+ for i in range (index , len_string ):
93+ trie_node = trie_node .get (string [i ], None )
94+ 95+ if trie_node is None :
96+ return False
97+ 98+ if trie_node .get (word_keeper_key , False ) and is_breakable (i + 1 ):
99+ return True
100+ 101+ return False
102+ 103+ return is_breakable (0 )
104+ 105+ 106+ if __name__ == "__main__" :
107+ print (word_break ("applepenapple" , ["apple" , "pen" ]))
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