# 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.
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 `````` ``````class Solution: def wordBreak(self, s: str, wordDict: List[str]) -> List[str]: res = [] self.back_tracking(s, wordDict, res, []) return res def back_tracking( self, s: str, d: list[str], res: List[str], track: List[str] ) -> None: if not s: res.append(' '.join(track)) for w in d: if s.startswith(w): self.back_tracking(s[len(w):], d, res, track+[w])``````