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.

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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])