<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>all distinct solutions to the <strong>n-queens puzzle</strong></em>. You may return the answer in <strong>any order</strong>.</p>

<p>Each solution contains a distinct board configuration of the n-queens&#39; placement, where <code>&#39;Q&#39;</code> and <code>&#39;.&#39;</code> both indicate a queen and an empty space, respectively.</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> [[&quot;.Q..&quot;,&quot;...Q&quot;,&quot;Q...&quot;,&quot;..Q.&quot;],[&quot;..Q.&quot;,&quot;Q...&quot;,&quot;...Q&quot;,&quot;.Q..&quot;]]
<strong>Explanation:</strong> There exist two distinct solutions to the 4-queens puzzle as shown above
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> [[&quot;Q&quot;]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 9</code></li>
</ul>

class Solution:

    def solveNQueens(self, n: int) -> List[List[str]]:
        self.vis = [[0] * (2 * n) for i in range(3)]
        self.ans = []
        self.mapping = ["." * i + "Q" + "." * (n - 1 - i) for i in range(n)]

        def cal(x, y):
            return y, x - y + n, y + x

        def dfs(cnt=0, stack=[]):
            if cnt == n:
                self.ans.append(stack[:])
                return

            for i in range(n):
                tags = cal(cnt, i)

                _f = False
                for k in range(3):
                    if self.vis[k][tags[k]]:
                        _f = True
                if _f:
                    continue

                for k in range(3):
                    self.vis[k][tags[k]] = 1

                stack.append(self.mapping[i])
                dfs(cnt + 1)
                stack.pop()
                for k in range(3):
                    self.vis[k][tags[k]] = 0

        dfs()
        return self.ans

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