<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:
    unordered_set<string> st;
    void rec(string &s, int last, string &tiles, vector<int> &was){
        if (s.size() > 0) st.insert(s);
        for (int i = 0; i < tiles.size(); i++){
            if (!was[i]){
                was[i] = 1;
                s += tiles[i];
                rec(s, last+1, tiles, was);
                s.pop_back();
                was[i] = 0;
            }
        }
    }

    int numTilePossibilities(string tiles) {
        string s = "";
        vector<int> was(tiles.size());
        rec(s, 0, tiles, was);
        return st.size();
    }
};
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