<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>

from functools import cache

class TrieNode:
    def __init__(self):
        self.next = {}
        self.is_end = False

class Solution:
    def findAllConcatenatedWordsInADict(self, words: List[str]) -> List[str]:
        root = TrieNode()

        @cache
        def traverse(p, word, i):
            nonlocal root
    
            c = word[i]
            # print(word, i, c)
            if c in p.next:
                p = p.next[c]
            else:
                return False
            # print(p.is_end)
            if i == len(word) - 1:
                return p.is_end

            if p.is_end:
                return traverse(p, word, i + 1) or traverse(root, word, i + 1)
            else:
                return traverse(p, word, i + 1)

        words = sorted(words, key=lambda x: len(x))
        res = []
        for word in words:
            # print(word)
            if traverse(root, word, 0):
                res.append(word)

            p = root
            for c in word:
                if c not in p.next:
                    p.next[c] = TrieNode()
                p = p.next[c]
            p.is_end = True

        return res

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