<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 {
public:

    bool check(int start, int end, string &s,vector<vector<int>> &dp,unordered_set<string> &dict){
        if(end == s.size()){
            return start == end;
        }
        if(dp[start][end] != -1)
            return dp[start][end];
        
        // take end
        string temp = string(s.begin()+start, s.begin()+end+1);
        bool ans;
        if(dict.find(temp) != dict.end()){
            ans = check(end+1,end+1,s,dp,dict);
        }
        // dont take end
        dp[start][end] = ans || check(start,end+1,s,dp,dict);
        return dp[start][end];

    }

    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        vector<string> pos;
        unordered_set<string> dict;
        vector<string> curPos;

        for(auto &word:words)
            dict.insert(word);
        
        for(auto &word:words){
            vector<vector<int>> dp(word.size(),vector<int>(word.size(),-1));
            dict.erase(word);
            curPos.clear();
            if(check(0,0,word,dp,dict))
                pos.push_back(word);
            dict.insert(word);
        }
        return pos;
    }
};
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