<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 {
private:
    unordered_map<string, int> mp;
    void dfs(string tiles, string sub, vector<int> &visited) {
        mp[sub] = 1;
        for(int k = 0; k<tiles.size();k++) {
            if(visited[k]==1) continue;
            
            visited[k] = 1;
            sub += tiles[k];
            dfs(tiles, sub, visited);
            visited[k] = 0;
            sub.pop_back();
        }
    }
public:
    int numTilePossibilities(string tiles) {
        // return pow(2, n);
        sort(tiles.begin(), tiles.end());
        string sub = "";
        vector< int > visited(tiles.size(),false);
        dfs(tiles, sub, visited);
        int ans = 0;

        for(auto it : mp) {
            if(it.second == 1) ans++;
        }
        return ans-1;
    }
};
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