<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, is_word):
        self.val = ch
        self.is_word = is_word
        self.neighbors = {}

class Trie:
    def __init__(self):
        self.root = Node('#', False)

    def add_word(self, word):
        node = self.root
        for i, ch in enumerate(word):
            if ch not in node.neighbors:
                node.neighbors[ch] = Node(ch, False)
            node = node.neighbors[ch]
            if i == len(word)-1:
                node.is_word = True

    def find_breaks(self, word):
        node = self.root
        ret = []
        for i, ch in enumerate(word):
            if ch in node.neighbors:
                node = node.neighbors[ch]
                if node.is_word:
                    ret.append(i)
            else:
                break
        return ret

class Solution:
    def findAllConcatenatedWordsInADict(self, words: List[str]) -> List[str]:
        trie = Trie()
        for word in words:
            trie.add_word(word)

        ans = []
        for word in words:
            dp = [False] * len(word) + [True]
            for i in range(len(dp)-1, -1, -1):
                word_breaks = trie.find_breaks(word[i-1:])
                if i > 0:
                    for idx in word_breaks:
                        if dp[i + idx]:
                            if not(i-1 == 0 and i + idx == len(dp) - 1):
                                dp[i-1] = True
                                break
            if dp[0]:
                ans.append(word)

        return ans
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