LeetCode 140. Word Break II

140. Word Break II

Difficulty:: Hard

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:
[]

Solution

Language: Java

class Solution {
    public List<String> wordBreak(String s, List<String> wordDict) {
        if (s == null || wordDict == null || s.length() == 0 || wordDict.size() == 0) {
            return new ArrayList<>();
        }
        Set<String> wordSet = new HashSet<>(wordDict);
        Map<Integer, List<String>> map = new HashMap<>();
        return dfsHelper(s, 0, wordSet, map);
    }
    
    public List<String> dfsHelper(String s, int start, Set<String> wordSet, Map<Integer, List<String>> map) {
        if (map.containsKey(start)) {
            return map.get(start);
        }
        List<String> res = new ArrayList<>();
        if (start >= s.length()) {
            res.add("");
        }
        String cur;
        for (int i = start; i < s.length(); i++) {
            if (wordSet.contains((cur = s.substring(start, i + 1)))) {
                List<String> following = dfsHelper(s, i + 1, wordSet, map);
                for (String f : following) {
                    res.add(cur + (f.length() == 0 ? "" : " ") + f);
                }
            }
        }
        map.put(start, res);
        return res;
    }
}