<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:
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        set<string> word;
        for(auto i:words) word.insert(i);
        vector<string> ans;
        for(auto x:words){
            cout<<"curr word:"<<x<<"\n";
            word.erase(x);
            vector<vector<int>> dp(x.size(),vector<int>(x.size(),-1));
            int n=x.size();
            function<int(int, int)> hmm = [&](int start,int end){
                if(start>end) return 0;
                if(dp[start][end]!=-1) return dp[start][end];
                
                if(word.find(x.substr(start,end-start+1)) != word.end()) return dp[start][end]=1;
                if(start==end) return dp[start][end]=0;
                for(int k=start;k<end;k++){
                    if(hmm(start,k)==1 and hmm(k+1,end)==1) return dp[start][end]=1;
                }
                return dp[start][end]=0;
            };
            if(hmm(0,n-1)) ans.push_back(x);
            word.insert(x);
        }
        return ans;
    }
};
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