<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 check(queens: list[int]):
            m = len(queens)
            return (
                len(set(queens)) == m
                and (len(set(map(lambda t: t[0] + t[1], enumerate(queens)))) == m)
                and (len(set(map(lambda t: t[0] - t[1], enumerate(queens)))) == m)
            )

        res = []

        def backtrack(candidate: list[int]):
            if not check(candidate):
                return
            if len(candidate) == n:
                res.append(candidate)
                return
            for next_queen_col in range(n):
                candidate.append(next_queen_col)
                backtrack(candidate)
                candidate.pop()

        backtrack([])
        return len(res)
16000