<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:

    void solve(vector<int> &vis, string &t, string &s, set<string> &res){

        if(s.size()>0)
        res.insert(s);

        for(int i=0;i<t.size();i++){
            if(!vis[i]){
                vis[i]=1;
                s+=t[i];
                solve(vis,t,s,res);
                s.pop_back();
                vis[i]=0;
            }
        }

    }

    int numTilePossibilities(string tiles) {
        int n = tiles.size();
        
        set<string> res;
        string s="";
        vector<int> vis(n,0);
        solve(vis,tiles,s,res);

        long long sz = res.size();

        return (int)sz;
    }
};
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