<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 {
private:

    // normal word break - 1
    // this helper will run for average O(N^2)
    // iterative
    bool wordBreak(string s, unordered_set<string> &st,
                   vector<int> &dp) {
        int n = s.size();
        dp[n]=1;
        for(int i=n-1;i>=0;i--){
            string temp;
            for(int j=i;j<n;j++){
                temp.push_back(s[j]);
                if(st.find(temp)!=st.end() && dp[j+1]){
                    dp[i]=true;
                    break;
                }
            }
        }
        return dp[0];
    }
public:
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        unordered_set<string> st;
        for(auto x : words){
            st.insert(x);
        }
        vector<string>ans;
        for(auto x : words){
            // now problem has become similar to word break - I
            int n = x.size();
            vector<int> dp(n+1,0);
            // as elements are distinct and we are going to check for each word
            // set contains the same word. so remove it and check
            st.erase(x);
            if(wordBreak(x,st,dp)){
                ans.push_back(x);
            }
            // after checking put that word again
            st.insert(x);
        }
        return ans;
    }
    //total T.C. will be O(N^2)*M. M is length of words. that is words.length
};
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