140. Word Break II

Description

Given anon-emptystringsand a dictionarywordDictcontaining a list ofnon-emptywords, add spaces insto construct a sentence where each word is a valid dictionary word. You may assume the dictionary does not contain duplicate words.

Return all such possible sentences.

For example, given
s="catsanddog",
dict=["cat", "cats", "and", "sand", "dog"].

A solution is["cats and dog", "cat sand dog"].

UPDATE (2017/1/4):
ThewordDictparameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

Discussion

Method 1

bool tracker[from][to].

O(n^2) find words.add into tracker.

Use the tracker reconstruct result.

Method 2

dfs

class Solution {
public:
    vector<string> wordBreak(string s, vector<string>& wordDict) {
        unordered_map<string, vector<string>> tracker;
        return dfs(s, wordDict, tracker);
    }

private:
    vector<string> dfs(string s, vector<string>& wordDict,  unordered_map<string, vector<string>> &tracker) {
        vector<string> rst;

        if(tracker.find(s) != tracker.end()) {
            return tracker[s];
        }

        if(0 == s.size()) {
            rst.push_back("");
            return rst;
        }

        for(auto &word : wordDict) {
            vector<string> tmp;
            if(checkStart(s, word)) {
                cout << word << " ->" << s.substr(word.size()) << endl;
                tmp = dfs(s.substr(word.size()), wordDict, tracker);

                for(auto words : tmp) {
                    if(0 != words.size())
                        rst.push_back(word + " " + words);
                    else
                        rst.push_back(word);
                }
            }

        }
        tracker[s] = rst;

        return rst;
    }

    bool checkStart(string& s, string &word) {
        bool rst = true;
        if(s.size() < word.size()) {
            return !rst;
        }

        for(int loop = 0; loop < word.size(); ++loop) {
            if(s[loop] != word[loop]) {
                rst = false;
                break;
            }
        }

        return rst;
    }
};

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;
}