<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):
        self.is_word = False
        self.children = {}
        self.word = None

class Tire:
    def __init__(self):
        self.root = Node()

    def add(self, word):
        cur = self.root
        for c in word:
            if c not in cur.children:
                cur.children[c] = Node()
            cur = cur.children[c]
        cur.is_word = True
        cur.word = word

class Solution:

    def __init__(self):
        self.res = []
        self.cache = {}
        self.tire = Tire()


    def findAllConcatenatedWordsInADict(self, words: List[str]) -> List[str]:
        words.sort(key = lambda x: len(x))
        min_length = len(words[0])
        max_length = len(words[-1])
        all_words = set(words)

        for w in words:
            self.tire.add(w)
        
        def cache_and_return(word, b):
            self.cache[word] = b
            return b


        def search(word, root_word):
            if word != root_word and word in all_words: return True
            if word in self.cache: return self.cache[word]

            cur_node = self.tire.root
            found = False
            for i, c in enumerate(word):
                if c not in cur_node.children:
                    break
                else:
                    cur_node = cur_node.children[c]
                    if cur_node.is_word:
                        if i != len(word)-1:
                            rest_word = word[i+1:]
                            if search(rest_word, root_word):
                                found = True

            return cache_and_return(word, found)

        for word in words:
            if len(word) < 2*min_length:
                continue
            else:
                if search(word, word):
                    self.res.append(word)

        return self.res
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