<p>The <strong>n-queens</strong> puzzle is the problem of placing <code>n</code> queens on an <code>n x n</code> chessboard such that no two queens attack each other.</p>

<p>Given an integer <code>n</code>, return <em>the number of distinct solutions to the&nbsp;<strong>n-queens puzzle</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/13/queens.jpg" style="width: 600px; height: 268px;" />
<pre>
<strong>Input:</strong> n = 4
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are two distinct solutions to the 4-queens puzzle as shown.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 9</code></li>
</ul>

class Solution {
public:
    int co=0;
    unordered_set<int> c, ld, rd;
    void h(int n, int i) {
        if (i==n) {
            ++co;
            return;
        }
        int j=0;
        for (; j<n; j++) {
            auto x=c.insert(j);
            auto y=ld.insert(j-i);
            auto z=rd.insert(j+i);
            if (x.second && y.second && z.second) {
                h(n, i+1);
                c.erase(j);
                ld.erase(j-i);
                rd.erase(j+i);
            }
            else {
                if (x.second) c.erase(j);
                if (y.second) ld.erase(j-i);
                if (z.second) rd.erase(j+i);
            }
        }
    }
    int totalNQueens(int n) {
        //vector<vector<char>> b(n, vector<char> (n, '.'));
        h(n, 0);
        return co;
    }
};
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