<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:# using DP + Trie
    def findAllConcatenatedWordsInADict(self, words: List[str]) -> List[str]:
        def addToTrie(word):
            temp_trie = trie
            for char in word:
                if char not in temp_trie:
                    temp_trie[char] = {}
                temp_trie = temp_trie[char]
            temp_trie['end'] = 1

        def makeTrie(words):
            for word in words:
                addToTrie(word)
        trie = {}
        makeTrie(words)
        # print(trie)

        def wordBreaker(s, words):
            dp = [0] * (len(s) + 1)
            dp[-1] = 1
            for i in range(len(s)):
                if dp[i-1]:
                    temp_trie = trie
                    for j in range(i, len(s)):
                        char = s[j]
                        if char not in temp_trie:
                            break
                        temp_trie = temp_trie[char]
                        if 'end' in temp_trie:
                            dp[j] = max(dp[j], dp[i-1] + 1)
            # print(dp)
            if dp[-2] > 2:
                return 1
            return 0
        result = []
        for word in words:
            if wordBreaker(word, words):
                result.append( word )
        return result


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