<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:
    trie *children[26];
    bool end;

    
    trie(){
        this->end=false;
        for(int i=0;i<26;i++) this->children[i]=NULL;
    }
    void insert(string &key){
        trie *root=this;
        for(auto i:key){
            if(root->children[i-'a']==NULL){
                root->children[i-'a']=new trie();
            }
            root=root->children[i-'a'];
        }
        root->end=true;
    }

    
};

class Solution {
public:
    map<string,pair<int,bool>> res;
    pair<int,bool> search(trie *root,string &key){
        trie *currNode=root;
        for(int i=0;i<key.size();i++){
            if(currNode->end==true){
                string x=key.substr(i);
                if(res.count(x)){
                 if(res[x].second) return res[key]={2,true};
                }else{
                    pair<int,bool> check=search(root,x);
                    res[x]=check;
                    if(check.second) return res[key]={2,true};
                }
            }
            if(currNode->children[key[i]-'a']==NULL) return {0,false};
            currNode=currNode->children[key[i]-'a'];
        }
        if(currNode->end==true) return res[key]={1,true};
        return res[key]={0,false};
    }
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        trie *root=new trie();
        for(auto i:words){
            root->insert(i);
        }
        vector<string> v;
        for(auto i:words){
            pair<int,bool> check=search(root,i);
            if(check.second && check.first>1) v.push_back(i);
        }
        return v;
    }
};
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