<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(object):

    def __init__(self):
        super().__init__()
        self.next = {}
    
    def insert(self, w):
        cur = self
        for ch in w+'#':
            if ch not in cur.next:
                cur.next[ch] = TrieNode()
            cur = cur.next[ch]
    
    def prefixlen(self, w):
        cur = self
        i = 0
        while True:
            if '#' in cur.next:
                yield i
            if i >= len(w):
                break
            if w[i] not in cur.next:
                break
            cur = cur.next[w[i]]
            i += 1

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

        def segmented(w):
            q = deque([0])
            inserted = [False]*(len(w)+1)
            inserted[0] = True
            while q:
                i = q.popleft()
                if i == len(w):
                    return True
                for l in trie.prefixlen(w[i:]):
                    if inserted[i+l]:
                        continue
                    inserted[i+l] = True
                    q.append(i+l)
            return False

        orig = set(words)
        cache = {}
        def _bt(w):
            if w in cache:
                return cache[w]
            cache[w] = False
            for i in range(1, len(w)):
                pref, suff = w[:i], w[i:]
                if pref in orig and (suff in orig or _bt(suff)):
                    cache[w] = True
                    break
            return cache[w]

        return [w for w in words if _bt(w)]

        for w in sorted(words, key=len):
            #print(w)
            if segmented(w):
                res.append(w)
            trie.insert(w)
        return res

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