<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> finalAns;

    bool check(string s, set<string>& st, map<string,bool>& dp)
    {
        if(s.size() == 0)return true;
        if(dp.find(s) != dp.end())return dp[s];

        bool ans = false;
        for(int i = 0; i < s.size(); i++)
        {
            string temp = s.substr(0,i+1);
            if(st.count(temp))
            {
                bool val = check(s.substr(i+1),st,dp);
                ans = ans || val;
            }
        }

        return dp[s] = ans;
    }

    void solve(set<string>& st, vector<string>& words)
    {
        int n = words.size();
        for(int i = 0; i < n; i++)
        {
            string s = words[i];
            st.erase(s);
            map<string,bool> dp;

            if(check(s,st,dp))
            {
                finalAns.push_back(s);
            }

            st.insert(s);
        }
    }

    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        set<string> st(words.begin(),words.end());
        solve(st,words);
        return finalAns;
    }
};
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