<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:
    def findAllConcatenatedWordsInADict(self, words: List[str]) -> List[str]:
        def default_trie():
            return defaultdict(default_trie)

        trie = default_trie()
        word_set = set(words)  # To quickly check if a word exists
        
        # Build Trie for the given words
        for word in words:
            if word:  # Avoid empty words
                node = trie
                for char in word:
                    node = node[char]
                node['is_end'] = True
        
        # Helper function to check if a word is a concatenation of other words
        @lru_cache(None)  # Memoization to avoid re-calculating for the same start index
        def is_concatenated(word: str, start: int) -> bool:
            node = trie
            for i in range(start, len(word)):
                char = word[i]
                if char not in node:
                    return False
                node = node[char]
                
                # If a valid word is found, recursively check the rest of the word
                if 'is_end' in node:
                    if i + 1 == len(word):
                        return start != 0  # Ensure it's a concatenation, not a single word
                    
                    # Check if the remaining part of the word can form valid concatenations
                    if is_concatenated(word, i + 1):
                        return True

            return False
        
        concatenated_words = []
        for word in words:
            if word and is_concatenated(word, 0):
                concatenated_words.append(word)
        
        return concatenated_words

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