<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 CheckSubstring(vector<string>& dictionary, string sentence,
                        int curIndex, int lastIndex, vector<vector<int>>& dp) {
        if (curIndex == sentence.length() - 1) {
            if(lastIndex==0) return 0;
            bool check = 0;
            int subStringlen = curIndex - lastIndex + 1;
            string subString = sentence.substr(lastIndex, subStringlen);
            if (mp.count(subString)) {
                check = 1;
            }
            return dp[curIndex][lastIndex] = check;
        }
        if (dp[curIndex][lastIndex] != -1)
            return dp[curIndex][lastIndex];
        int subStringlen = curIndex - lastIndex + 1;
        string subString = sentence.substr(lastIndex, subStringlen);
        bool check =
            CheckSubstring(dictionary, sentence, curIndex + 1, lastIndex, dp);
        if (mp.count(subString)) {
            check |= CheckSubstring(dictionary, sentence, curIndex + 1,
                                    curIndex + 1, dp);
        }
        return dp[curIndex][lastIndex] = check;
    }
    bool DictionaryCheck(vector<string>& dictionary, string sentence) {
        int n = sentence.length();
        vector<vector<int>> dp(n, vector<int>(n, -1));
        bool CheckSentence = CheckSubstring(dictionary, sentence, 0, 0, dp);
        return CheckSentence;
    }
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
     for(auto i:words) mp[i]=1;
     vector<string>ans;
     for(auto i:words){
        bool check=DictionaryCheck(words,i);
        if(check) ans.push_back(i);
     }   
     return ans;
    }
};
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