Word Break II

Given a string s and a dictionary of words dict, add spaces in s to
construct a sentence where each word is a valid dictionary word.
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"].

Solution: check before constructing the sentences.

class Solution {
public:
    vector<string> wordBreak(string s, unordered_set<string> &dict) {
        vector<string> res;
        if (!wordBreakPossible(s, dict)) return res; //! not pass if ignoring here
        wordBreakRe(s, dict, 0, "", res);
        return res;
    }
    
    void wordBreakRe(const string &s, const unordered_set<string> &dict, 
                     int start, string sentence, vector<string> &res) {
        if (start == s.size()) {
            res.push_back(sentence);
            return;
        }
        if (start != 0) sentence.push_back(' ');
        for (int i = start; i < s.size(); ++i) {
            string word = s.substr(start, i-start+1);
            if (dict.find(word) == dict.end())
                continue;
            wordBreakRe(s, dict, i+1, sentence + word, res);
        }
    }
    
    bool wordBreakPossible(const string &s, const unordered_set<string> &dict) {
        int N = s.size();
        bool canBreak[N+1];
        memset(canBreak, false, sizeof(canBreak));
        canBreak[0] = true;
        for (int i = 1; i <= N; ++i) {
            for (int j = i-1; j >= 0; --j) {
                if (canBreak[j] && dict.find(s.substr(j, i-j)) != dict.end()) {
                    canBreak[i] = true;
                    break;
                }
            }
        }
        return canBreak[N];
    }
};