<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:
    def totalNQueens(self, n: int) -> int:
        stack = []
        y_visited = set()
        diagonal_left, diagonal_right = set(), set()
        for i in range(n):
            stack.append((0, i, 1, False))
        ans = 0
        
        while stack:
            x, y, count, flag = stack.pop()
            if not flag:
                stack.append((x, y, count, True))
                y_visited.add(y)
                diagonal_left.add(x + y)
                diagonal_right.add(x - y)
                if count == n:
                    ans += 1
                    continue
                nx = x + 1
                for ny in range(n):
                    if ny in y_visited:
                        continue
                    if nx + ny in diagonal_left:
                        continue
                    if nx - ny in diagonal_right:
                        continue
                    stack.append((nx, ny, count + 1, False))
            else:
                y_visited.remove(y)
                diagonal_left.remove(x + y)
                diagonal_right.remove(x - y)
        return ans
        
        
16400