<p>You have <code>n</code>&nbsp;&nbsp;<code>tiles</code>, where each tile has one letter <code>tiles[i]</code> printed on it.</p>

<p>Return <em>the number of possible non-empty sequences of letters</em> you can make using the letters printed on those <code>tiles</code>.</p>

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

<pre>
<strong>Input:</strong> tiles = &quot;AAB&quot;
<strong>Output:</strong> 8
<strong>Explanation: </strong>The possible sequences are &quot;A&quot;, &quot;B&quot;, &quot;AA&quot;, &quot;AB&quot;, &quot;BA&quot;, &quot;AAB&quot;, &quot;ABA&quot;, &quot;BAA&quot;.
</pre>

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

<pre>
<strong>Input:</strong> tiles = &quot;AAABBC&quot;
<strong>Output:</strong> 188
</pre>

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

<pre>
<strong>Input:</strong> tiles = &quot;V&quot;
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= tiles.length &lt;= 7</code></li>
	<li><code>tiles</code> consists of uppercase English letters.</li>
</ul>

class Solution {
public:
    
    vector<string> subset(string tiles)
    {
        int n=tiles.size();
        vector<string> all;
        for(int i=0;i<(1<<n);i++)
        {
            string sub;
            for(int j=0;j<n;j++)
            {
                if(i & (1<<j))
                {
                    sub.push_back(tiles[j]);
                }
            }
            if(!sub.empty())
            {
                if(sub.size()==1)
                all.push_back(sub);
                else
                {
                    sort(sub.begin(),sub.end());
                    do
                    {
                        all.push_back(sub);
                    }while(next_permutation(sub.begin(),sub.end()));
                }
            }
        }
        return all;
    }

    int numTilePossibilities(string tiles)
    {
        vector<string> all;
        all=subset(tiles);
        set<string> al=set<string>(all.begin(),all.end());
        
        return al.size();    
    }
};
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