<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 {
    map<pair<string,int>,bool>mpp2;
    unordered_map<string,int>mpp;
    bool solve(string s, int i)
    {
       if(i>=s.length())return true;
       if(mpp2.find({s,i})!=mpp2.end())return mpp2[{s,i}];
        for(int j=i+1;j<s.length()+1;++j)
        {
            string s2=s.substr(i,j-i);
            if(mpp[s2]>0)
            {
               if(solve(s,j))
               return mpp2[{s,i}]= true;
            }
        }
        return mpp2[{s,i}]= false;
        
    }
public:
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        for(int i=0;i<words.size();++i)
        {
            mpp[words[i]]++;
        }
        int n=words.size();
        vector<string>ans;
        for(int i=0;i<n;++i)
        {
            mpp[words[i]]--;
            if(solve(words[i],0))
            {
                ans.push_back(words[i]);
            }
            mpp[words[i]]++;
        }
        return ans;
    }
};
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