<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:
    set<string> s1;
    int fact[8]={1,1,2,6,24,120,720,5040};
    int unique(string s){
        vector<int> cnt(26);
        for(char c:s){
            cnt[c-'A']++;
        }
        int res=fact[s.size()];
        for(int n:cnt){
            res/=fact[n];
        }
        return res;
    }
    int dfs(string& s,string seq="",int pos=0){
        if(pos>=s.size()){
           if(s1.count(seq)==0){
              s1.insert(seq);
              return unique(seq); 
           }
           else
           return 0;
        }
        return dfs(s,seq,pos+1)+dfs(s,seq+s[pos],pos+1);
    }
    int numTilePossibilities(string tiles) {
        sort(tiles.begin(),tiles.end());
        return dfs(tiles)-1;
    }
};
12148