<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 {
    int total = 0;
private:
    void dfs(vector<int>& freq) {
        total++;
        for (int& num: freq){
            if (!num)
                continue;
            num--;
            dfs(freq);
            num++;
        }
    }
public:
    int numTilePossibilities(string tiles) {
        unordered_map<char, int> tileCount;
        for (char tile: tiles) 
            tileCount[tile]++;
        vector<pair<int, int>> tileFreq(tileCount.begin(), tileCount.end());
        vector<int> freq;
        for (auto pair: tileFreq)
            freq.push_back(pair.second);
        sort(freq.begin(), freq.end());
        vector<int> factorial(8);
        for (int i = 2; i < 8; i++)
            factorial[i] = i * factorial[i-1];
        int len = tiles.length();
        dfs(freq);
        return total - 1;
    }
};
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