<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{
    map <char, Trie*> next;
    bool end = false;
};
class Solution {
public:
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        Trie* root = new Trie;
        for(string word: words){
            Trie* curr = root;
            for(char ch: word){
                if(curr->next[ch]){
                    curr = curr->next[ch];
                }
                else{
                    curr->next[ch] = new Trie;
                    curr = curr->next[ch];
                }
            }
            curr->end = true;
        }
        vector <string> ans;
        for(string word: words){
            vector <pair <bool, int>> dp(word.size()+1, {false, 0});
            dp[word.size()].first = true;
            for(int i=word.size()-1; i>=0; i--){
                Trie* curr = root;
                for(int j=i; j<word.size(); j++){
                    curr = curr->next[word[j]];
                    if(!curr) break;
                    if(curr->end){
                        dp[i] = max(dp[i], {dp[j+1].first, dp[j+1].second + 1});
                    }
                }
            }
            if(dp[0].second > 1 && dp[0].first) ans.push_back(word);
        }
        return ans;
    }
};
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