<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:
    int fact[8] = { 1, 1, 2, 6, 24, 120, 720, 5040 }; // 0 to 7 factorial
    unordered_set<string> st;
    
    int uniquePerms(string& s)
    {
        // return the num of unique permutations
        int cnt[26] = {};
        for (auto ch : s) 
            ++cnt[ch - 'A'];
        auto res = fact[s.size()];
        // total! / n1! / n2! / n3!
        for (auto n : cnt) 
            res /= fact[n];
        return res;
    }
    
    int dfs(string& s, string seq = "", int pos = 0) {
        // unordered_set.insert  .second tells whether value is inserted, 
        // or it is already present in the set
        if (pos == s.size()) 
            return st.insert(seq).second ? uniquePerms(seq) : 0 ;
        return dfs(s, seq, pos + 1) + dfs(s, seq + s[pos], pos + 1);
    }
    
    int numTilePossibilities(string tiles) {
        sort( begin(tiles), end(tiles) );
        // remove empty option in the end
        return dfs(tiles) - 1;
    }
};

//   n
//   Σ   n! / ((m - 1)! * (n - m)!)
//          m = 1
// If we have a string of size n with i unique characters, and each character repeats m[i] times, the number of unique permutations is:
// n! / (m[1]! * m[2]! * .. * m[i]!)




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