<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:

    map<string,int> wordMap;
    
    struct TrieNode{
      TrieNode* children[26];
      bool isLeaf;

      TrieNode(){
        for(int i = 0; i < 26 ; i++){
            children[i] = NULL;
        }
        isLeaf = false;
      }
    };

    void insert(TrieNode* root, string word){
        for(int i = 0; i < word.size();i++){
            if(root->children[word[i]-'a'] == NULL){
                root->children[word[i]-'a'] = new TrieNode();
            }
            root = root->children[word[i]-'a'];
        }
        root->isLeaf = true;
    }

    bool isWord(TrieNode* root, string word){
        for(int i = 0; i < word.size();i++){
            if(root->children[word[i]-'a'] == NULL) return false;
            root = root->children[word[i]-'a'];
        }
        return root->isLeaf;
    }

    int dp[10005];

    int checkConcatenation(int l , int r, TrieNode* &root ,string word){
        if(l>=r) return 0;
        int countWords = INT_MIN;

        if(dp[l] != -1) return dp[l];
        for(int i = l ; i<=r;i++){
            string temp = word.substr(l,i-l+1);
            if(isWord(root,temp)){
            //    cout<<temp<<endl; 
               int value = checkConcatenation(i+1,r,root,word);
               if(value != INT_MIN)
               countWords = max(countWords , 1 + value);
            }
        }
        return dp[l] = countWords;
    }

    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {

        int n = words.size();
        vector<string> ans;
        TrieNode* root = new TrieNode();
        for(int i = 0; i < n ;i++){
            insert(root,words[i]);
        }
        for(int i = 0 ; i < n ; i++){
            memset(dp,-1,sizeof(dp));
            int wordCount = checkConcatenation(0, words[i].size() , root , words[i]);
            if(wordCount >=2) ans.push_back(words[i]);
        }

        return ans;
        
    }
};
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