<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:
    map<string,int> mp;
    vector<vector<int>> dp;
    string s; 
    int rec(int l,int r){
        if(dp[l][r] != -1 ) return dp[l][r];
        
        string temp = s.substr(l,r-l+1);
        
        if(mp.find(temp)!= mp.end() and mp[temp] == 1){
           
          dp[l][r] = 1;
          
          return dp[l][r];
        }
        if(l==r) return dp[l][r] = 0;
        int ans = 0;
        for(int mid=l; mid<r; mid++){
            ans |= (rec(l,mid)&rec(mid+1,r));
        }
        return dp[l][r] = ans;  
    }
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {     
        for(auto v:words){
            mp[v] = 1;
        }
        vector<string> ans;
        for(auto v:words){
            int n = v.length();
            dp.clear();
            dp.assign(n,vector<int>(n,-1));
            mp[v] = 0;
            s = v;
            if(rec(0,n-1)){
                ans.push_back(v);
            }
            mp[v] = 1;
        }
        return ans;
    }
};
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