<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>

class Trie{
public:
    int flag;
    unordered_map<char, Trie*>child;
    Trie(){
        flag = -1;
    }
};
class Solution {
public:
    void go(Trie* node, Trie* root, string& s, int spot, vector<string>& ans, int depth){
        if(node->child.count(s[spot]) == 0){
            if(depth != spot)return;
            node->child[s[spot]] = new Trie();
            if(spot == s.length() - 1){
                node->child[s[spot]]->flag = 1;
            }
            else{
                node->child[s[spot]]->flag = 0;
                go(node->child[s[spot]], root, s, spot + 1, ans, depth + 1);
            }
        }
        else{
            Trie* tempnode = node->child[s[spot]];
            if(tempnode->flag == 1){
                int pre_size = ans.size();
                if(spot == s.length() - 1)ans.push_back(s);
                else{
                    go(root, root, s, spot + 1, ans, 0);
                    if(pre_size == ans.size()){
                        go(tempnode, root, s, spot + 1, ans, depth + 1);
                    }
                }
            }
            else{
                if(s.length() - 1 == spot)return;
                go(tempnode, root, s, spot + 1, ans, depth + 1);
            }
        }
        
    }

    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        // sort(words.begin(), words.end(), vec_sort);
        Trie* root = new Trie();
        vector<string>ans;
        int len = words.size();
        for(int t=1;t<=30;t++){
            for(int s=0;s<len;s++){
                if(words[s].length() == t)go(root, root, words[s], 0, ans, 0);
            }
        }
        return ans;
    }
};
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