<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 totalNQueens(int n) {
        unordered_set<int> cols_used;
        cols_used.reserve(n);
        unordered_set<int> diagonals_used;
        diagonals_used.reserve(n * 2 - 1);
        return CheckPermutations(n, 0, cols_used, diagonals_used);
    }

private:
    int CheckPermutations(int n, int row, unordered_set<int> cols_used, unordered_set<int> diagonals_used) {
        if (row == n) {
            return 1;
        }
        int working_solutions = 0;
        for (int col = 0; col < n; ++col) {
            if (cols_used.contains(col)) continue;
            cols_used.insert(col);

            int back_diagonal = n - (row - col) - 1;
            int num_diagonals = n * 2 - 1;
            int forward_diagonal = row + col;
            if (!diagonals_used.contains(back_diagonal) && 
                !diagonals_used.contains(forward_diagonal + num_diagonals)) {
                diagonals_used.insert(back_diagonal);
                diagonals_used.insert(forward_diagonal + num_diagonals);

                working_solutions += CheckPermutations(n, row + 1, cols_used, diagonals_used);

                diagonals_used.erase(back_diagonal);
                diagonals_used.erase(forward_diagonal + num_diagonals);
            }
            
            cols_used.erase(col);
        }
        return working_solutions;
    }
};
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