<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:
        
        self.count = 0
        
        queens = [['.' for _ in range(n)] for _ in range(n)]

        cols = set()
        posDiag = set()
        negDiag = set()

        def backtrack(i):

            if i == n:
                self.count+=1
                return
            
            for j in range(n):

                if j not in cols and i+j not in posDiag and i-j not in negDiag:

                    queens[i][j] = 'Q'
                    cols.add(j)
                    posDiag.add(i+j)
                    negDiag.add(i-j)

                    backtrack(i+1)

                    cols.remove(j)
                    posDiag.remove(i+j)
                    negDiag.remove(i-j)
                    queens[i][j] = '.'
        
        backtrack(0)
        return self.count

        

16700