<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>

const int INF = 1e9;

struct TrieNode {
  TrieNode* next[26];
  bool terminal = false;

  TrieNode() {
    for (auto& v : next)
      v = nullptr;
  }
};

struct Trie {
  TrieNode* root;

  Trie() : root(new TrieNode) {}

  void insert(string s) {
    TrieNode* v = root;
   
    for (char ch : s) {
      if (!v->next[ch-'a'])
        v->next[ch-'a'] = new TrieNode;
     
      v = v->next[ch-'a'];
    }
    v->terminal = true;
  }

  unordered_map<string, int> m;
  bool search(string s) {
    TrieNode* v = root;

    auto dfs = [&](auto&& self, string t) {
      if (m.find(t) != m.end())
        return m[t];

      if (t.empty())
        return 0;

      int count = -INF;
      TrieNode* v = root;

      for (int i = 0; i < (int)t.size(); ++i) {
        if (!v->next[t[i]-'a'])
          break;

        v = v->next[t[i]-'a'];

        if (v->terminal)
          count = max(count, 1+self(self, t.substr(i+1, t.size())));
      }

      return m[t] = count;
    };

    return dfs(dfs, s) >= 2;
  }
};

class Solution {
public:
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        Trie trie;
        vector<string> ans;

        sort(words.begin(), words.end(), [](auto i, auto j) {
            return (i.length() < j.length()) || (i.length() == j.length() && i < j);
        });

        for (auto word : words) {
            trie.insert(word);
            if (trie.search(word))
                ans.push_back(word);
        }
        return ans;
    }
};
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