<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:
    long long ans=0;
    vector<string> vec;
    int numTilePossibilities(string tiles) {
        sort(tiles.begin(),tiles.end());
        int size=tiles.size();
        vector<bool> visit(size,false);
        string str="";
        for(int i=0;i<tiles.size();i++){
            if(visit[i]==true ||(i>0 && tiles[i]==tiles[i-1] && !visit[i-1])) continue;
            ans++;
            str+=tiles[i];
            visit[i]=true;
            vec.push_back(str);
            dfs(tiles,visit,i,str);
            visit[i]=false;
            str.pop_back();
        }
        for(auto ss:vec) cout<<ss<<" ";
        return ans;
    }
    void dfs(string&tiles,vector<bool> & visit, int idx,string&str){

        for(int i=0;i<tiles.size();i++){
            if(visit[i]==true ||(i>0 && tiles[i]==tiles[i-1] && !visit[i-1])) continue;
            ans++;
            
            str+=tiles[i];
            vec.push_back(str);
            visit[i]=true;
            dfs(tiles,visit,i,str);
            visit[i]=false;
            str.pop_back();
        }

    }
};
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