Letter Tile Possibilities
<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>
Hash Table; String; Backtracking; Counting
class Solution {
public:
    int numTilePossibilities(string tiles) {
        int cnt[26]{};
        for (char c : tiles) {
            ++cnt[c - 'A'];
        }
        function<int(int* cnt)> dfs = [&](int* cnt) -> int {
            int res = 0;
            for (int i = 0; i < 26; ++i) {
                if (cnt[i] > 0) {
                    ++res;
                    --cnt[i];
                    res += dfs(cnt);
                    ++cnt[i];
                }
            }
            return res;
        };
        return dfs(cnt);
    }
};