<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 {
private:
    void backtrack(int i, string s, vector<bool> &used, 
                    vector<string> &result, string tiles) {
        if (!s.empty()) {
            result.push_back(s);
        }

        for (int i = 0; i < tiles.size(); i++) {
            if (used[i]) continue;
            if (i > 0 && tiles[i] == tiles[i - 1] && !used[i - 1]) continue;
            used[i] = true;
            backtrack(i + 1, s + tiles[i], used, result, tiles);
            used[i] = false;
        }
    }
public:
    int numTilePossibilities(string tiles) {
        sort(tiles.begin(), tiles.end());
        vector<bool> used(tiles.size(), false);
        vector<string> result;
        backtrack(0, "", used, result, tiles);

        return result.size();
    }
};
14178