Word Break II

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.

Note:

• The same word in the dictionary may be reused multiple times in the segmentation.

• You may assume the dictionary does not contain duplicate words.

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:
[]

class Solution {
    // Using DFS directly will lead to TLE, so I just used HashMap to save the
    // previous results to remove duplicated branches, as the following:

    public List<String> wordBreak(String s, List<String> wordDict) {
        Set<String> dict = new HashSet<>();
        for (String x : wordDict)
            dict.add(x);
        return DFS(s, dict, new HashMap<String, List<String>>());
    }

    // DFS function returns an array including all substrings derived from s.
    public List<String> DFS(String s, Set<String> wordDict, HashMap<String, List<String>> map) {
        if (map.containsKey(s))
            return map.get(s);

        List<String> res = new ArrayList<String>();
        if (s.length() == 0) {
            res.add("");
            return res;
        }
        for (String word : wordDict) {
            if (s.startsWith(word)) {
                List<String> sublist = DFS(s.substring(word.length()), wordDict, map);
                for (String sub : sublist)
                    res.add(word + (sub.isEmpty() ? "" : " ") + sub);
            }
        }
        map.put(s, res);
        return res;
    }
}