<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:
    unordered_map<string,int> mp;

    bool solve(int ind, string &s, int n, vector<int> &dp){
        if(ind == n) return true;

        if(dp[ind] != -1) return dp[ind];

        string temp = "";

        for(int k=ind;k<n;k++){
            temp += s[k];

            if(mp[temp]){
                if(solve(k+1 , s , n , dp)) return dp[ind] = true;
            }
        }

        return dp[ind] = false;
    } 

    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        for(auto x : words){
            mp[x]++;
        }

        vector<string> ans;

        for(auto x : words){
            vector<int> dp(31 , -1);
            mp.erase(x);
            if(solve(0 , x , x.size() , dp)){
                ans.push_back(x);
            }
            mp[x]++;
        }

        return ans;
    }
};
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