<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 Find(string tiles, int i, set<string> &s, string &temp){
        
        
//         if(i==tiles.length()){
            
//             if(temp.length()!=0){
//                 s.insert(temp);
//                 return;
//             }
//         }
        
        
//         temp.push_back(tiles[i]);
        
//         Find(tiles,i+1,s,temp);
        
//         temp.pop_back();
        
//         Find(tiles,i+1,s,temp);
        
//     }
    int numTilePossibilities(string tiles) {
        
        
//         set<string>s;
//          string temp="";
//         Find(tiles,0,s,temp);
       
//         return s.size();
        
        vector<int>count(26);
        
        for(int i=0;i<tiles.length();++i){
            count[tiles[i]-'A']++;
        }
        
        return dfs(count);
    }
    
    
    private:
    
    int dfs(vector<int> &count){
        int ans=0;
        for(int& i: count){
            
            if(i==0)
                continue;
            
            i--;
            ans+=1+dfs(count);
            i++;
        }
        return ans;
    }
};
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