<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 = {}
        self.end = False

class Solution:
    def findAllConcatenatedWordsInADict(self, words: List[str]) -> List[str]:
        #build trie
        root = TrieNode()
        # cur_node = root
        for w in words:
            cur_node = root
            for c in w:
                if c not in cur_node.children:
                    # print(c)
                    cur_node.children[c] = TrieNode()
                cur_node = cur_node.children[c]
            cur_node.end = True
            
        
        @cache
        def search(node, word, idx):
            # print("here", word, idx, node.end)
            if idx == len(word) and node.end == True:
                # print("here", word, idx, node.end)
                return True

            if word[idx] in node.children:
                # print(word[idx], node.children[word[idx]].end)
                return search(node.children[word[idx]], word, idx+1)
            else:
                # print("not here", word, idx, node.end)
                return False



        #search
        # for w in words:
        #     if search(root, w, 0):
        #         print(w)

        @cache
        def isConcatenated(word, idx, counts):
            if idx == len(word):
                return counts >=2
            cur_node = root
            while idx < len(word):
                c = word[idx]
                if c not in cur_node.children:
                    return False
                cur_node = cur_node.children[c]
                if cur_node.end:
                    if isConcatenated(word, idx+1, counts+1):
                        return True
                idx += 1
            
            return False
        
        res = []
        for word in words:
            if isConcatenated(word, 0, 0):
                res.append(word)
    
        return res
            
            

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