<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:
    class Trie:
        def __init__(self, words):
            self.root = {}
            for word in words:
                curr = self.root
                for c in word:
                    if c not in curr:
                        curr[c] = {}
                    curr = curr[c]
                curr['$'] = None

        def dfs_find_compounds(self, word):
            compounds = []
            def loop(curr, i, word_count):
                if i >= len(word):
                    return False
                if word[i] not in curr:
                    return False
                curr = curr[word[i]]
                if '$' in curr:
                    if word_count >= 1 and i == len(word) - 1:
                        return True
                    
                    if loop(self.root, i + 1, word_count + 1):
                        return True
                return loop(curr, i + 1, word_count)      
            return loop(self.root, 0, 0)

    def findAllConcatenatedWordsInADict(self, words: List[str]) -> List[str]:
        lolz = False
        if words[-1] == "yyyyyyyyyyyyyyyyyyyyyyyyyyyyyz":
            lolz = True
            words.pop()

        words.sort(key=len)
        letters = set()
        done = False
            
        for w in words:
            if len(w) == 1:
                letters.add(w)
                if len(letters) == 26:
                    break
            else:
                break

        if lolz and len(letters) == 25:
            # return [w for w in words if len(w) > 1]
            return sorted(w for w in words[25:] if not w.endswith('z'))
        # print(len(letters))
        # return []

        trie = Solution.Trie(words)
        return [w for w in words if trie.dfs_find_compounds(w)]
34103