<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{
        char ch;
        int isWord;
        vector<Node*> next;
        Node(char c){
            ch = c;
            isWord = 0;
            next.resize(26);
            for(int i=0;i<26;i++){
                next[i] = nullptr;
            }
        }
    };

    vector<vector<int>> dp;

    bool isConcat(int word_num,int pos,Node* &master,vector<string>& words){
        cout<<word_num<<" " << pos <<endl;
        if(pos >= words[word_num].size()){
            return true;
        }
        if(dp[word_num][pos] != -1){
            return dp[word_num][pos];
        }
        bool possible = false;
        Node* curr = master;
        for(int i=pos;i<words[word_num].size();i++){
            if(curr->next[words[word_num][i]-'a'] == nullptr){
                break;
            }
            curr = curr->next[words[word_num][i]-'a'];
            cout << curr->ch << endl;
            if(curr->isWord == 1){
                possible |= isConcat(word_num,i+1,master,words);
                if(possible && (pos != 0 || (i != words[word_num].size()-1))){
                    dp[word_num][pos] = 1;
                    return true;
                }
                possible = false;
            }
            if(i+1 != words[word_num].size() && curr->next[words[word_num][i+1] - 'a'] == nullptr){
                dp[word_num][pos] = 0;
                return false;
            }
        }

        dp[word_num][pos] = (possible == true ? 1 : 0);
        return possible;
    }

    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        int n = words.size();
        dp.assign(n+1, vector<int>(32,-1));
        Node* master = new Node('a');
        for(int i=0;i<n;i++){
            Node* curr = master;
            for(int j=0;j<words[i].size();j++){
                if(curr->next[words[i][j]-'a'] == nullptr){
                    curr->next[words[i][j]-'a'] = new Node(words[i][j]);
                }
                curr = curr->next[words[i][j]-'a'];
            }
            curr->isWord = 1;
        }
        vector<string> answer;
        for(int i=0;i<n;i++){
            if(isConcat(i,0,master,words)){
                answer.push_back(words[i]);
            }
        }
        return answer;
    }
};
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