<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 solve(string tiles,string temp,int idx,int &ans,vector<int> &vis){
        ans++;
        int n=tiles.length();
        if(idx>=n){
            return;
        }
        unordered_set<char> used;
        for(int i=0;i<n;i++){
            if(!vis[i]&&used.find(tiles[i])==used.end()){
                temp+=tiles[i];
                vis[i]=1;
                solve(tiles,temp,idx+1,ans,vis);
                vis[i]=0;
                temp.pop_back();
                used.insert(tiles[i]);
            }
        }
    }
    int numTilePossibilities(string tiles) {
        sort(tiles.begin(),tiles.end());
        string temp="";
        int ans=0;
        vector<int> vis(tiles.size(),0);
        solve(tiles,temp,0,ans,vis);
        return ans-1;
    }
};
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