<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:
    def __init__(self):
        self.children = [None] * 26
        self.is_end = False

class Solution:
    def __init__(self):
        self.root = TrieNode()
        
    @lru_cache
    def insert(self, root, key):
        curr = root
        for i in range(len(key)):
            idx = ord(key[i]) - ord('a')
            if not curr.children[idx]:
                curr.children[idx] = TrieNode()
            curr = curr.children[idx]
        curr.is_end = True

    @lru_cache
    def dfs(self, root, key, index, count):
        if index >= len(key):
            return count > 1
        curr = root
        for i in range(index, len(key)):
            p = ord(key[i]) - ord('a')
            if not curr.children[p]:
                return False
            curr = curr.children[p]
            if curr.is_end:
                if self.dfs(root, key, i+1, count+1):
                    return True
        return False

    def findAllConcatenatedWordsInADict(self, words):
        for i in range(len(words)):
            self.insert(self.root, words[i])
        ans = []
        for i in range(len(words)):
            if self.dfs(self.root, words[i], 0, 0):
                ans.append(words[i])
        return ans
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