<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 backtrack(int i , string tiles, vector<string>& subsets, string &curr) {
        if (i >= tiles.size()) return;
    
        curr += tiles[i]; 
        subsets.push_back(curr);  
        cout << curr << endl;
        backtrack(i + 1, tiles, subsets, curr);
        curr.pop_back();
        backtrack(i + 1, tiles, subsets, curr);


    }
    int numTilePossibilities(string tiles) {
        vector<string> subsets;
        string curr = "";
        backtrack(0, tiles, subsets, curr);

        set<string> unique;
        for (string s : subsets) {
            sort(s.begin(), s.end());
            unique.insert(s);
        }   
        int count = 0;
        for (string s : unique) {
            do {
                count++;
            } while (next_permutation(s.begin(), s.end()));
        }

        return count;
    }
};
10371