<p>Given an array of strings <code>words</code> (<strong>without duplicates</strong>), return <em>all the <strong>concatenated words</strong> in the given list of</em> <code>words</code>.</p>

<p>A <strong>concatenated word</strong> is defined as a string that is comprised entirely of at least two shorter words (not necessarily distinct)&nbsp;in the given array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;cat&quot;,&quot;cats&quot;,&quot;catsdogcats&quot;,&quot;dog&quot;,&quot;dogcatsdog&quot;,&quot;hippopotamuses&quot;,&quot;rat&quot;,&quot;ratcatdogcat&quot;]
<strong>Output:</strong> [&quot;catsdogcats&quot;,&quot;dogcatsdog&quot;,&quot;ratcatdogcat&quot;]
<strong>Explanation:</strong> &quot;catsdogcats&quot; can be concatenated by &quot;cats&quot;, &quot;dog&quot; and &quot;cats&quot;; 
&quot;dogcatsdog&quot; can be concatenated by &quot;dog&quot;, &quot;cats&quot; and &quot;dog&quot;; 
&quot;ratcatdogcat&quot; can be concatenated by &quot;rat&quot;, &quot;cat&quot;, &quot;dog&quot; and &quot;cat&quot;.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;cat&quot;,&quot;dog&quot;,&quot;catdog&quot;]
<strong>Output:</strong> [&quot;catdog&quot;]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= words.length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= words[i].length &lt;= 30</code></li>
	<li><code>words[i]</code> consists of only lowercase English letters.</li>
	<li>All the strings of <code>words</code> are <strong>unique</strong>.</li>
	<li><code>1 &lt;= sum(words[i].length) &lt;= 10<sup>5</sup></code></li>
</ul>

struct Trie
{
    std::array<Trie*, 26> _childs;
    bool isWord = false;

    Trie * emplace(string & w, int i)
    {
        if(i == w.size())
        {
            isWord = true;
            return this;
        }
        else
        {
            int j = w[i]-'a';
            if(!_childs[j])
            {
                _childs[j] = new Trie();
            }

            return _childs[j]->emplace(w, i+1);
        }
    }

    bool has(string & w, int i)
    {
        if(i == w.size())
        {
            return isWord;
        }
        else
        {
            int j = w[i]-'a';
            return _childs[j] ? _childs[j]->has(w, i+1) : false;
        }
    }
};


class Solution {
public:
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        //std::unordered_set<string> set;
        Trie set;
        for(auto & w : words)
        {
            set.emplace(w, 0);
        }

        vector<string> ans;
        for(auto & w : words)
        {
            std::vector<bool> visted(w.size(), false);
            if(isConcatenated(w, 0, set, visted))
            {
                ans.emplace_back(w);
            }
        }
        return ans;
    }

    bool isConcatenated(string & w, int i, Trie & set, std::vector<bool> & visted)
    {
        bool ret = (i == w.size());
        string key;
        for(int j=i; j<w.size(); ++j)
        {
            key += w[j];
            if(!visted[j] && set.has(key,0) && key.size() < w.size())
            {   
                visted[j] = true;
                ret |= isConcatenated(w, j+1, set, visted);
            }

            if(ret)
            {
                break;
            }
        }

        return ret;
    }
};

// class Solution {
//     bool dfs(const string& word, int length, vector<bool>& visited,
//              const unordered_set<string>& dictionary) {
//         if (length == word.length()) 
//         {
//             return true;
//         }
        
//         if (visited[length]) 
//         {
//             return false;
//         }

//         visited[length] = true;
//         for (int i = word.length() - (length == 0 ? 1 : 0); i > length; --i) 
//         {
//             if (dictionary.count(word.substr(length, i - length)) &&
//                 dfs(word, i, visited, dictionary)) 
//             {
//                 return true;
//             }
//         }
//         return false;
//     }

// public:
//     vector<string> findAllConcatenatedWordsInADict(vector<string>& words) 
//     {
//         unordered_set<string> dictionary(words.begin(), words.end());
//         vector<string> answer;
//         for (const string& word : words) 
//         {
//             vector<bool> visited(word.length());        // ?
//             if (dfs(word, 0, visited, dictionary)) 
//             {
//                 answer.push_back(word);
//             }
//         }
//         return answer;
//     }
// };
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