<p>Given an array of string <code>words</code>, return <em>all strings in </em><code>words</code><em> that is a <strong>substring</strong> of another word</em>. You can return the answer in <strong>any order</strong>.</p>

<p>A <strong>substring</strong> is a contiguous sequence of characters within a string</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;mass&quot;,&quot;as&quot;,&quot;hero&quot;,&quot;superhero&quot;]
<strong>Output:</strong> [&quot;as&quot;,&quot;hero&quot;]
<strong>Explanation:</strong> &quot;as&quot; is substring of &quot;mass&quot; and &quot;hero&quot; is substring of &quot;superhero&quot;.
[&quot;hero&quot;,&quot;as&quot;] is also a valid answer.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;leetcode&quot;,&quot;et&quot;,&quot;code&quot;]
<strong>Output:</strong> [&quot;et&quot;,&quot;code&quot;]
<strong>Explanation:</strong> &quot;et&quot;, &quot;code&quot; are substring of &quot;leetcode&quot;.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;blue&quot;,&quot;green&quot;,&quot;bu&quot;]
<strong>Output:</strong> []
<strong>Explanation:</strong> No string of words is substring of another string.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= words.length &lt;= 100</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 30</code></li>
	<li><code>words[i]</code> contains only lowercase English letters.</li>
	<li>All the strings of <code>words</code> are <strong>unique</strong>.</li>
</ul>

class Solution {
public:
    vector<string> stringMatching(vector<string>& words) 
    {   set <string> result;

        // Sort By Length
        sort(words.begin(),words.end(),[](string &a, string &b)
             {  return a.length() > b.length();
             });

        int size = words.size();
        for(int i=0;i<size;i++)
        {   for(int j=i+1;j<size;j++)
            {   size_t found=words[i].find(words[j]);
                if(found!=string::npos)
                {   result.insert(words[j]);
                }
            }
        }
        return vector<string>(result.begin(),result.end());
        
    }
};
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