<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 Trie:
    def __init__(self):
        self.conts = [None] * 26
        self.end = False
    
    def val(self, c):
        return ord(c) - ord('a')
    
    def add(self, word):
        cur = self
        
        for c in word:
            idx = self.val(c)
            if cur.conts[idx] == None:
                cur.conts[idx] = Trie()
            cur = cur.conts[idx]
        
        cur.end = True
    
    def cont(self, c):
        return self.conts[self.val(c)]

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

        for w in words:
            trie.add(w)

        res = []
        for w in words:
            dp = [float('-inf')] * (len(w) + 1)
            dp[-1] = 0
            
            for i in range(len(w) - 1, -1, -1):
                node = trie
                for j in range(i, len(w)):
                    node = node.cont(w[j])
                    if not node:
                        break
                    if node.end:
                        dp[i] = max(dp[i], 1 + dp[j + 1])

            if dp[0] >= 2:
                res.append(w)
        
        return res
            
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