<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 dfs(string& cur, unordered_set<string>& us, vector<int>& cnt, int& result) {
        for (int i = 0 ; i < 26 ; ++i) {
                if (cnt[i] == 0) continue;
                --cnt[i];
                cur.push_back(i + 'A');
                if (us.count(cur) == 0) {
                        us.emplace(cur);
                        ++result;
                        dfs(cur, us, cnt, result);
                }
                cur.pop_back();
                ++cnt[i];
        }
}
    int numTilePossibilities(string tiles) {
        int result = 0;
        unordered_set<string> us;

        vector<int> cnt(26, 0);
        for (const auto& c : tiles) ++cnt[c-'A'];

        for (int i = 0 ; i < 26 ; ++i) {
                if (cnt[i] == 0) continue;
                --cnt[i];
                string cur(1, i + 'A');
                ++result;
                us.emplace(cur);
                dfs(cur, us, cnt, result);
                ++cnt[i];
        }
        return result;        
    }
};
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