<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>

from itertools import chain
from functools import lru_cache
class Node:
    def __init__(self, c, is_end):
        self.val = c
        self.children = {}
        self.is_end = is_end

class Trie:
    def __init__(self):
        self.root = Node("", False)
    
    def insert(self, word):
        cur_node = self.find_node(word, True)
        cur_node.is_end = True
    
    def get_words(self, prefix):
        cur_node = self.find_node(prefix, False)
        if not cur_node:
            return []
        ans = self.dfs(prefix, cur_node)
        if cur_node.is_end:
            ans.append(prefix)
        return ans

    def find_node(self, prefix, is_insert):
        cur_node = self.root 
        for c in prefix:
            node = cur_node.children.get(c)
            if not node:
                if not is_insert:
                    return None
                node = Node(c, False)
                cur_node.children[c] = node
            cur_node = node
        return cur_node
    
    def dfs(self, prefix, node):
        ans = []
        for child in node.children.values():
            if child.is_end:
                ans.append(prefix + child.val)
            ans.extend(self.dfs(prefix + child.val, child))
        return ans

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

        words_set = set(words)
        trie = Trie()
        for word in words:
            trie.insert(word)
        print(trie.get_words("rat"))

        @lru_cache(None)
        def is_concat(prefix, start = 0):
            if start == len(prefix):
                return True
            last_node = trie.root
            cur_node = trie.root
            result = False
            for i in range(start, len(prefix)):
                c = prefix[i]
                last_node = cur_node
                cur_node = cur_node.children.get(c)
                if not cur_node:
                    return result
                if cur_node.is_end and not (start == 0 and i == len(prefix) - 1):
                    result |= is_concat(prefix, i + 1)
            return result

        # print(is_concat("ratcatdogcat"))
        return [w for w in words if is_concat(w)]

        

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