<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};
    unordered_set<string>st;
    int uniq(string &s){
        vector<int>freq(26,0);
        for(auto c : s)freq[c - 'A']++;
        int res = fact[s.length()];
        for(auto f : freq) res /= fact[f];
        return res;
    }
    int dfs(string &s, string res = "", int i=0){
        if(i == s.length()){
            return !st.insert(res).second ? 0 : uniq(res);
        }
        return dfs(s, res + s[i], i+1) + dfs(s, res, i+1);
    }
    int numTilePossibilities(string tiles) {
        sort(tiles.begin(),tiles.end());
        return dfs(tiles) - 1;
    }
};
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