<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:
    bool isEnd;
    unordered_map<char, TrieNode*> children;
    
    TrieNode() : isEnd(false) {}
};

class Trie {
public:
    TrieNode* root;
    
    Trie() {
        root = new TrieNode();
    }
    
    void insert(const string& word) {
        TrieNode* node = root;
        for (char c : word) {
            if (!node->children.count(c)) {
                node->children[c] = new TrieNode();
            }
            node = node->children[c];
        }
        node->isEnd = true;
    }
    
    bool search(const string& word, int start, int count, unordered_map<int, bool>& memo) {
        if (start == word.size()) {
            return count > 1;
        }
        if (memo.count(start)) return memo[start];
        
        TrieNode* node = root;
        for (int i = start; i < word.size(); ++i) {
            if (!node->children.count(word[i])) {
                break;
            }
            node = node->children[word[i]];
            if (node->isEnd && search(word, i + 1, count + 1, memo)) {
                return memo[start] = true;
            }
        }
        return memo[start] = false;
    }
};

class Solution {
public:
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        Trie trie;
        vector<string> result;
        unordered_map<int, bool> memo; // Memoization map
        
        for (const string& word : words) {
            if (!word.empty()) {
                trie.insert(word);
            }
        }
        
        for (const string& word : words) {
            if (!word.empty()) {
                memo.clear(); // Clear memoization for each new word
                if (trie.search(word, 0, 0, memo)) {
                    result.push_back(word);
                }
            }
        }
        
        return result;
    }
};

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