<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 TrieNode{
    TrieNode(bool word = false)
    {
        isWord = word;
    }

    unordered_map<char, TrieNode*> children;
    bool isWord;
};

class Trie{
    private:
        TrieNode* root;

        void addWord(string word)
        {
            TrieNode* node = root;
            for (int i = 0; i < word.length(); i ++)
            {
                if (node->children.find(word[i]) == node->children.end()) node->children[word[i]] = new TrieNode();
                node = node->children[word[i]];
            }

            node->isWord = true;
        }

    public:
        Trie(vector<string>& words)
        {
            root = new TrieNode();
            for (auto &word: words)
            {
                addWord(word);
            }
        }

        bool isConcatenated(string& curWord, int s, int count, int n, vector<int>& dp)
        {
            if (s == n) return count > 1;
            if (dp[s] != -1) return dp[s];
            TrieNode* node = root;
            for (int i = s; i < n; i ++)
            {
                if (node->children.find(curWord[i]) == node->children.end()) break;
                node = node->children[curWord[i]];
                if (node->isWord){
                    bool isConcat = isConcatenated(curWord, i+1, count+1, n, dp);
                    if (isConcat) {
                        return dp[s] = 1;
                    } 
                }
            }

            return dp[s] = 0;
        }
};

class Solution {
public:
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        Trie trie(words);
        vector<string> result;
        for (auto& word: words)
        {
            vector<int> dp(word.length(), -1);
            if (trie.isConcatenated(word, 0, 0, word.length(), dp)) result.push_back(word);
        }

        return result;
        
    }


};
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