<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 TrieNode:
    def __init__(self):
        self.children = {}
        self.end = False

class Trie:
    def __init__(self):
        self.root = TrieNode()
    
    def add_word(self, word):
        cur = self.root
        for c in word:
            if c not in cur.children:
                cur.children[c] = TrieNode()
            cur = cur.children[c]
        cur.end = True
    
class Solution:
    def findAllConcatenatedWordsInADict(self, words: List[str]) -> List[str]:
        trie = Trie()
        root = trie.root
        for word in words:
            trie.add_word(word)
        res = []
        # todo: Create a function to check is concatenated
        def is_concatenated(word) -> bool:
            cur = root
            cache = {}
            # return maximum number of words in dictionary if it is comprised, if not return 0 
            def dfs(cur, i) -> int:
                if (cur, i) in cache: return cache[(cur, i)]
                if i == len(word):
                    return 0 if cur == root else -inf
                if word[i] not in cur.children: return -inf
                local_res = -inf
                if cur.children[word[i]].end:
                    local_res = 1 + dfs(root, i + 1)
                local_res = max(local_res, dfs(cur.children[word[i]], i + 1))
                cache[(cur, i)] = local_res
                return cache[(cur, i)]
            count = dfs(root, 0)
            # print(word, count)
            return count >= 2
        for word in words:
            if is_concatenated(word):
                res.append(word)
        return res

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