README.md

140. Word Break II

Given a string s and a dictionary of strings wordDict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in any order.

Note that the same word in the dictionary may be reused multiple times in the segmentation.

Example 1:

Input: s = "catsanddog", wordDict = ["cat","cats","and","sand","dog"]
Output: ["cats and dog","cat sand dog"]

Example 2:

Input: s = "pineapplepenapple", wordDict = ["apple","pen","applepen","pine","pineapple"]
Output: ["pine apple pen apple","pineapple pen apple","pine applepen apple"]
Explanation: Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = "catsandog", wordDict = ["cats","dog","sand","and","cat"]
Output: []

Constraints:

• 1 <= s.length <= 20
• 1 <= wordDict.length <= 1000
• 1 <= wordDict[i].length <= 10
• s and wordDict[i] consist of only lowercase English letters.
• All the strings of wordDict are unique.
• Input is generated in a way that the length of the answer doesn't exceed 10⁵.

Solutions (Python)

1. Solution

from functools import cache


class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
        words = set(wordDict)

        @cache
        def backtracking(s: str) -> Optional[List[str]]:
            ret = []

            for i in range(1, min(len(s) + 1, 10)):
                if s[:i] in wordDict:
                    if i == len(s):
                        return ret + [s]
                    sentences = backtracking(s[i:])
                    if sentences is not None:
                        ret.extend("{} {}".format(s[:i], sentence)
                                   for sentence in sentences)

            return ret if ret != [] else None

        return backtracking(s) if backtracking(s) is not None else []