<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 Node:
    def __init__(self, ch: str):
        self.ch = ch
        self.children = [None]*26
        self.is_prefix = False

class Trie:
    def __init__(self):
        self.root = Node("")
    
    def startswith(self, s: str, start: int) -> List[int]:
        res = []
        node = self.root
        
        for i in range(start, len(s)):
            offset = ord(s[i])-ord("a")
            if node.children[offset] is None:
                break
            if node.children[offset].is_prefix:
                res.append(i)
            node = node.children[offset]
        return res
    
    def add(self, s: str) -> None:
        node = self.root
        for i in range(len(s)):
            offset = ord(s[i])-ord("a")
            if node.children[offset] is None:
                node.children[offset] = Node(s[i])
            if i == len(s)-1:
                node.children[offset].is_prefix = True
            node = node.children[offset]
                

class Solution:
    def search(self, sentence: str, start: int) -> bool:
        if start == len(sentence):
            return True
        elif start > len(sentence):
            return False
        elif (sentence, start) in self.cache:
            return self.cache[(sentence, start)]
        else:
            idxs = self.trie.startswith(sentence, start)
            if len(idxs) == 0:
                self.cache[(sentence, start)] = False
                return False
            else:
                for idx in idxs:
                    res = self.search(sentence, idx+1)
                    if res:
                        self.cache[(sentence, start)] = True
                        return True
                self.cache[(sentence, start)] = False
                return False
    
    def findAllConcatenatedWordsInADict(self, words: List[str]) -> List[str]:
        words.sort(key=lambda a:len(a))
        self.trie = Trie()
        self.cache = {}
        ans = []
        for word in words:
            res = self.search(word, 0)
            if res:
                ans.append(word)
            self.trie.add(word)
        
        
        return ans
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