<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:
    string substring(string s,int i,int j){
        string temp;
        for(int k=i;k<j;k++)
        temp.push_back(s[k]);
        return temp;
    }
    bool solve(int i,int j,string &s,vector<vector<int>>&dp,map<string,int> &m){
        if(i==s.size()){
        string tt = substring(s,j,i);
        if(m[tt])
        return 1;
        return 0;
        }
         if(dp[i][j]!=-1)
        return dp[i][j];

        string tt = substring(s,j,i+1);
        int ans = 0;
        if(m[tt])
        ans|=solve(i+1,i+1,s,dp,m);
        
        ans|=solve(i+1,j,s,dp,m);
        return dp[i][j]=ans;
    }
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        map<string,int> m;
        for(auto &b:words){
            m[b]++;
        }
        vector<string> ans;
        for(auto b:words){
            m[b]--;
            vector<vector<int>>dp(b.size(),vector<int>(b.size(),-1));
            if(solve(0,0,b,dp,m))
            ans.push_back(b);
            m[b]++;
        }
        return ans;
    }
};
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