140. Word Break II

Problem:

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:
[]

Analysis:
Straight forward DFS is trivial will get TLE. Need to memorize the temp result. Use map<String, List<String>> to cache intermediate results.

Solution:
It's tricky to deal with end of s.

class Solution {
    public List<String> wordBreak(String s, List<String> wordDict) {
        return dfs(s, wordDict, new HashMap<String, List<String>>());
    }
    
    private List<String> dfs(String s, List<String> wordDict, Map<String, List<String>> map) {
        if (map.containsKey(s)) {
            return map.get(s);
        }
        List<String> res = new ArrayList<>();
        if (s.isEmpty()) {      
            res.add("");
            return res;
        }
        
        for (String word: wordDict) {
            if (s.startsWith(word)) {
                for(String temp: dfs(s.substring(word.length()), wordDict, map)) {
                    res.add(word + (temp.isEmpty() ? "" : " ") + temp);
                }
            }
        }
        map.put(s, res);
        return res;
    }
}