<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]:
        n = len(words)
        trie = {}

        for i in range(n):
            word = words[i]
            curr = trie
            for char in word:
                if char not in curr:
                    curr[char] = {}
                curr = curr[char]
            curr[0] = 0

        
        @cache
        def check(i, j, pre):
            word = words[i]
            m = len(word)
            if j == m:
                return False
            
            curr = trie
            for k in range(j, m):
                char = word[k]
                if char not in curr:
                    break
                
                curr = curr[char]
                if 0 in curr:
                    if (k == m - 1 and pre) or check(i, k + 1, 1):
                        return True
            
            return False
        

        ans = []
        for i in range(n):
            if check(i, 0, 0):
                ans.append(words[i])
        
        return ans

            
            

        
        
        # def check(word, pre):
        #     m = len(word)
        #     if m == 0:
        #         return pre >= 2

        #     curr = trie
        #     for i in range(m):
        #         char = word[i]
        #         if char not in curr:
        #             break
        #         else:
        #             curr = curr[char]

        #         if 0 in curr:
        #             if check(word[i+1:], pre + 1):
        #                 return True
            
        #     return False
        

        # ans = []
        # for word in words:
        #     if check(word, 0):
        #         ans.append(word)
        
        # return ans

            



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