<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 numTilePossibilities(string tiles) {
        vector<int> counts(26);
        
        for(auto l: tiles)
        {
            counts[l - 'A']++;
        }
        
        queue<char> usable_chars;
        
        for(int i = 0; i < 26; i++)
            if(counts[i])
                usable_chars.push(i + 'A');
        
        return recursive_count(counts, usable_chars) - 1;
    }
    
    int recursive_count(vector<int>& counts, queue<char>& chars)
    {
        if(chars.size() == 0)
            return 1;
         
        int res = 1;
        
        int n = chars.size();
        
        for(int i = 0; i < n; i++)
        {
            char c = chars.front();
            chars.pop();
                
            counts[c - 'A']--;
            
            if(counts[c - 'A'])
                chars.push(c);
            
            res += recursive_count(counts, chars);
            
            if(!counts[c - 'A'])
                chars.push(c);
            
            counts[c - 'A']++;
        }
        
        return res;
    }
};
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