<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:

        def place(n, row, pre_row_combs):
            # if pre_row_combs is None:
            #      return []

            combs = []

            for pre_rows in pre_row_combs:
                row_no = 0 if pre_rows is None else len(pre_rows)
                for j in range(n):
                    is_valid = True
                    if not pre_rows:
                        pass
                    else:
                        for r, c in enumerate(pre_rows):
                            if j in [c, c+(row_no-r), c-(row_no-r)]:
                                is_valid = False
                                break
                    if is_valid:
                        combs.append(pre_rows+[j] if pre_rows else [j])

            return combs

        pre_row_combs = [[i] for i in range(n)]
        for row in range(1, n):
            pre_row_combs = place(n, row, pre_row_combs)
        
        return len(pre_row_combs)

17100