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.

We can apply the backtracking1 template to solve this problem. Fill in the logic.

• is_leaf: start_index == len(s), when all the letters are used.
• get_edges: w = s[start_index:end_index+1] where start_index <= end_index < len(s), are the possible words starting at start_index.
• is_valid: is w in wordDict? w is valid if it's in the dictionary.

Implementation

1def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
2    def dfs(start_index, path):
3        if start_index == len(s):
4            ans.append(" ".join(path))
5            return
6        for end_index in range(start_index, len(s)):
7            w = s[start_index:end_index+1]
8            if w in wordDict:
9                path.append(w)
10                dfs(end_index+1, path)
11                path.pop()
12    ans = []
13    dfs(0, [])
14    return ans