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dada una cuerda s
y un diccionario de cuerdas palabraDict
, add spaces in s
to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in cualquier orden.
Nota que la misma palabra en el diccionario pueda reutilizarse varias veces en la segmentación.
Ejemplo 1:
Aporte: s = "catsanddog", wordDict = ["cat","cats","and","sand","dog"] Producción: ["cats and dog","cat sand dog"]
Ejemplo 2:
Aporte: s = "pineapplepenapple", wordDict = ["apple","pen","applepen","pine","pineapple"] Producción: ["pine apple pen apple","pineapple pen apple","pine applepen apple"] Explicación: Note that you are allowed to reuse a dictionary word.
Ejemplo 3:
Aporte: s = "gatos y perros", wordDict = ["gatos","perros","arena","y","gato"] Producción: []
Restricciones:
1 <= s.length <= 20
1 <= palabraDict.longitud <= 1000
1 <= wordDict[i].length <= 10
s
ypalabraDict[i]
constan únicamente de letras inglesas minúsculas.- Todas las cuerdas de
palabraDict
son único. - Input is generated in a way that the length of the answer doesn’t exceed 105.
Pitón
# time complexity: O(2^n)
# space complexity: O(2^n)
from typing import List
class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
word_set = set(wordDict)
memoization = {}
return self._dfs(s, word_set, memoization)
def _dfs(self, remainingStr: str, wordSet: set, memoization: dict) -> List[str]:
if remainingStr in memoization:
return memoization[remainingStr]
if not remainingStr:
return [""]
results = []
for i in range(1, len(remainingStr) + 1):
currentWord = remainingStr[:i]
if currentWord in wordSet:
for nextWord in self._dfs(remainingStr[i:], wordSet, memoization):
results.append(
currentWord + (" " if nextWord else "") + nextWord)
memoization[remainingStr] = results
return results
s = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
print(Solution().wordBreak(s, wordDict))