<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:
    struct Node{
        bool end;
        Node *nxt[26];
        Node(){
            end = false;
            memset(nxt, 0, sizeof nxt);
        }
    }*root;
    void insert(string &s){
        Node *cur = root;
        for(int i =0; i < s.size(); i++){
            if(!cur->nxt[s[i]-'a']) cur->nxt[s[i]-'a'] = new Node();
            cur = cur->nxt[s[i]-'a'];
        }
        cur->end = true;
    }

    bool search(string &s){
        Node *cur = root;
        for(int i =0; i < s.size(); i++){
            if(!cur->nxt[s[i]-'a'])return false;
            cur = cur->nxt[s[i]-'a'];
        }
        if(cur->end)return true;
        else return false;
    }


    vector<string> ans;
    unordered_map<int, int> mp;
    int dp[10005][35];
    void solve(int idx, int cnt, string &s, int idxOfArray){
        if(idx >= s.size()){
            if(cnt > 1 && mp.find(idxOfArray) == mp.end()) {
                ans.push_back(s);
                mp[idxOfArray]++;
            }
            return;
        }
        if(dp[idxOfArray][idx]) return;
        for(int i = idx; i < s.size(); i++){
            string tmp = s.substr(idx, (i - idx) + 1);
            if(search(tmp)){
                dp[idxOfArray][i] = 1;
                solve(i+1,cnt+1, s, idxOfArray);
            }
        }
    }
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        memset(dp,0,sizeof dp);
        root = new Node();
        for(int i =0; i < words.size(); i++){
            insert(words[i]);
        }
        for(int i =0; i < words.size(); i++){
            solve(0,0,words[i], i);
        }
        return ans;
    }
};
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