<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(int n,string tiles,string &ans,int &cnt,vector<int>&visit){
    if(ans.size()==n+1){
        return;
    }
    cnt++;
    unordered_set<char>st;
    for(int k=0;k<n;k++){
        if(st.find(tiles[k])!=st.end())continue;
        if(visit[k]==0){
            st.insert(tiles[k]);
            visit[k]=1;
            ans.push_back(tiles[k]);
            find(n,tiles,ans,cnt,visit);
            ans.pop_back();
            visit[k]=0;
        }
    }
}

int numTilePossibilities(string tiles) {
    int n=tiles.size(),cnt=0;
    sort(tiles.begin(),tiles.end());
    vector<int>visit(n,0);
    string ans;
    find(n,tiles,ans,cnt,visit);
    return cnt-1;
}
};
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