Word Break II

Given anon-emptystring_s_and a dictionary_wordDict_containing a list ofnon-emptywords, 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

JAVA solution based on memorized DFS

public List<String> wordBreak(String s, Set<String> wordDict) {
    return DFS(s, wordDict, new HashMap<String, LinkedList<String>>());
}       

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

    LinkedList<String>res = new LinkedList<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;
}

Modified version:

https://leetcode.com/problems/word-break-ii/discuss/44167/My-concise-JAVA-solution-based-on-memorized-DFS

public class Solution {
    private HashMap<String,List<String>> map = new HashMap<String, List<String>>();

    public List<String> wordBreak(String s, List<String> wordDict) {
        List<String> res = new ArrayList<String>();
        if(s == null || s.length() == 0) {
            return res;
        }
        if(map.containsKey(s)) {
            return map.get(s);
        }
        if(wordDict.contains(s)) {
            res.add(s);
        }
        for(int i = 1 ; i < s.length() ; i++) {
            String t = s.substring(i);
            if(wordDict.contains(t)) {
                List<String> temp = wordBreak(s.substring(0 , i) , wordDict);
                if(temp.size() != 0) {
                    for(int j = 0 ; j < temp.size() ; j++) {
                        res.add(temp.get(j) + " " + t);
                    }
                }
            }
        }
        map.put(s , res);
        return res;
    }
}