<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 Solution:
    def findAllConcatenatedWordsInADict(self, words: List[str]) -> List[str]:
        h={}
        for w in words:
            l=len(w)
            if l in h:
                h[l].append(w)
            else:
                h[l]=[w]
        lengths=[l for l in h]
        lengths.sort()

        def aux(word):
            #checks if word can be constructed by concatenating stuff
            nonlocal h
            m=len(word)
            dp=[False for i in range(m)]
            #dp[i]: True if word[:i+1] can be constructed by concatenating at least 1 word
            for i in range(m-1):
                if (i+1) in h and word[:i+1] in h[i+1]:
                    dp[i]=True
            for i in range(m):
                if not dp[i]:
                    for j in range(i-1,-1,-1):
                        if dp[j] and (i-j) in h and word[j+1:i+1] in h[i-j]:
                            dp[i]=True
            return dp[-1]
        ans=[]
        for w in words:
            if aux(w):
                ans.append(w)
        return ans
        
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