<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:
        TrieNode* node[26];
        bool ends;
        TrieNode(){
            ends = 0;
            for(int i = 0; i < 26; i++){
                node[i] = nullptr;
            }
        }

};
class Trie{
    public:
    TrieNode* root;
    Trie(){
        root = new TrieNode();
    }
    void insert(string& s){
        int n = s.size();
        TrieNode* ptr = root;
        for(int i = 0; i < n; i++){
            int ch = s[i] - 'a';
            if(ptr->node[ch] == nullptr){
                ptr->node[ch] = new TrieNode();
            }
            ptr = ptr->node[ch];
        }
        ptr->ends = 1;
    }
};

class Solution {
public:
    bool rec(int i, int j, vector<string>& words, TrieNode* root, map<string, int>& mp) {
        bool ans = 0;
        if (j >= words[i].size()) {
            return 1;
        }
        string s = words[i].substr(j);
        // cout << i << " " << j << " " << s << endl;
        int n = words[i].size();
        if (mp.find(s) != mp.end()) {
            // cout << "memo " << s << endl;
            return mp[s];
        }

        TrieNode* ptr = root;  // Don't modify the original root reference
        for (int k = j; k < n; k++) {
            int ch = words[i][k] - 'a';
            if (ptr->node[ch] == nullptr) {
                break;
            }
            ptr = ptr->node[ch];
            if (ptr->ends) {
                ans |= rec(i, k + 1, words, root, mp);  // Pass the original root
            }
        }

        return mp[s] = ans;
    }

    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        int n = words.size();
        if (n <= 2) {
            return {};
        }

        // Sort words by length
        sort(words.begin(), words.end(), [&](const string& a, const string& b) {
            return a.size() < b.size();
        });
        // for(auto x : words){
        //     cout << x << " ";
        // }
        // cout << endl;
        // Remove duplicates
        words.erase(unique(words.begin(), words.end()), words.end());

        map<string, int> mp;
        Trie tree;

        // Insert first two words into Trie
        tree.insert(words[0]);
        tree.insert(words[1]);
        mp[words[0]] = 1;
        mp[words[1]] = 1;
        vector<string> ans;
        for (int i = 2; i < n; i++) {
            if (rec(i, 0, words, tree.root, mp)) {
                ans.push_back(words[i]);
                tree.insert(words[i]);
            } else {
                mp[words[i]] = 1;
                tree.insert(words[i]);  // Insert current word into Trie
            }
        }
        return ans;
    }
};

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