<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 findkey(string &s, int k){
        string key = "";
        string tmp = to_string(k);
        key+=tmp;
        key.push_back('#');
        for(char x : s){
            key.push_back(x);
        }
        return key;
    }
    bool solve(string &s, int i, string &curr, unordered_map<string,int> &umap, vector<string> &temp, unordered_map<string,bool> &dp){
        if(i==s.length()){
            if(curr.length()!=0){
                if(umap.find(curr)!=umap.end()){
                    temp.push_back(curr);
                }else{
                    return false;
                }
            }
            if(temp.size()<=1) return false;
            string tmp="";
            for(string x : temp){
                for(char y : x){
                    tmp.push_back(y);
                }
            }
            return umap.find(tmp)!=umap.end();
        }
        string key = findkey(curr,i);
        if(dp.find(key)!=dp.end()) return dp[key];
        bool ans = false;
        curr.push_back(s[i]);
        if(umap.find(curr)!=umap.end()){
            string x = curr;
            curr = "";
            temp.push_back(x);
            ans|=solve(s,i+1,curr,umap,temp,dp);
            curr=x;
            temp.pop_back();
        }
        ans|=solve(s,i+1,curr,umap,temp,dp);
        return dp[key]=ans;
    }
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        unordered_map<string,int> umap;
        for(string x : words){
            umap[x]=1;
        }
        vector<string> ans;
        for(string x : words){
            vector<string> temp;
            string curr = "";
            unordered_map<string,bool> dp;
            bool tmp = solve(x,0,curr,umap,temp,dp);
            if(tmp){
                ans.push_back(x);
            }
        }
        return ans;
        
    }
};
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