<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 Solution {
public:
    unordered_map<string, bool> mp;
    vector<string> final_ans;
    int dp[10010]; 

    bool back(int pos, string& s){
        if(pos == s.length()) return true;
        if(dp[pos] != -1) return dp[pos];
        
        string currentWord = "";
        for(int i = pos; i < s.length(); ++i){
            currentWord += s[i];
            if(mp[currentWord] && currentWord != s && back(i + 1, s)){
                return dp[pos] = 1;
            }
        }
        return dp[pos] = 0;
    }

    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        for(auto& word : words) mp[word] = true;
        
        for(auto& word : words){
            memset(dp, -1, sizeof(dp)); 
            if(back(0, word)){
                final_ans.push_back(word);
            }
        }
        return final_ans;
    }
};


// class Solution {
// struct TrieNode {
//     TrieNode* arr[26];
//     bool is_end;
//     TrieNode() {
//         for (int i = 0; i < 26; i ++) arr[i] = NULL;
//         is_end = false;
//     }  
// };

// void insert(TrieNode* root, string key) {
//     TrieNode* curr = root;
//     for (int i = 0; i < key.size(); i ++) {
//         int idx = key[i] - 'a';
//         if (curr->arr[idx] == NULL)
//             curr->arr[idx] = new TrieNode();
//         curr = curr->arr[idx];
//     }
//     curr->is_end = true;
// }

// bool dfs(TrieNode* root, string key, int index, int count) {
//     if (index >= key.size())
//         return count > 1;
//     TrieNode* curr = root;
//     for (int i = index; i < key.size(); i ++) {
//         int p = key[i] - 'a';
//         if (curr->arr[p] == NULL) {
//             return false;
//         }
//         curr = curr->arr[p];
//         if (curr->is_end) {
//             if (dfs(root, key, i+1, count+1))
//                 return true;
//         }
//     }
//     return false;
// }
// public:
// vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
//     TrieNode* root = new TrieNode();
//     for (int i = 0; i < words.size(); i ++) {
//         insert(root, words[i]);
//     }
//     vector<string> ans;
//     for (int i = 0; i < words.size(); i ++) {
//         if (dfs(root, words[i], 0, 0))
//             ans.push_back(words[i]);
//     }
//     return ans;   
// }
// };
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