<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 TrieNode {
public:
    char c;
    vector<TrieNode*> children;
    bool end;

    // Constructor
    TrieNode(char k) : c(k), children(26, nullptr), end(false) {}
};

class Solution {
public :
    TrieNode*root;
    Solution(){
        root = new TrieNode('N');
    }
    void insert(TrieNode*r,string &word,int i){
        int n = word.size();
        if(i == n){
            r->end = 1;
            return;
        }
        int ind = word[i] - 'a';
        if(!r->children[ind]){
            r->children[ind] = new TrieNode(word[i]);
        }
        insert(r->children[ind],word,i+1);
    }
    int concatinate(string &word,vector<int>&dp,int i){
        if(dp[i] != -1)return dp[i];
        int n = word.size();
        TrieNode*now = root;
        for(int j = i;j < n;j++){
            int ind = word[j] - 'a';
            now = now->children[ind];
            if(!now)return dp[i] = 0;
            if(now -> end){
                int p = concatinate(word,dp,j+1);
                if(p >= 1)return dp[i] = 1;
            }
        }
        return dp[i] = 0;
    }
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        Solution s;
        for(string &i : words)s.insert(s.root,i,0);
        vector<string>ans;
        for(string &i : words){
            int n = i.size();
            vector<int>dp(n+1,-1);
            dp[i.size()] = 1;
            TrieNode*now = s.root;
            for(int j = 0;j < n-1;j++){
                int ind = i[j] - 'a';
                now = now->children[ind];
                if(now->end){
                    if(s.concatinate(i,dp,j+1) == 1){ans.emplace_back(i);break;}
                }
            }
        }
        return ans;
    }
};
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