<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:
        diag1 = {}
        diag2 = {}
        col = {}
        count = [0]
        def recur(row,n):
            if row>=n:
                count[0] += 1
                return
            for j in range(n):
                d1 = row+j
                d2 = row-j
                if d1 not in diag1 and d2 not in diag2 and j not in col:
                    # print(row,j)
                    diag1[d1] = 1
                    diag2[d2] = 1
                    col[j] = 1
                    recur(row+1,n)
                    del diag1[d1]
                    del diag2[d2]
                    del col[j]
        recur(0,n)
        return count[0]

16500