<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]]:
        arr = []
        available = [[True for _ in range(n)] for _ in range(n)]
        curstate = [['.' for _ in range(n)] for _ in range(n)]
        def mask_available(available, i, j):
            for k in range(n):
                available[i][k] = False
                available[k][j] = False
            for k in range(n):
                if i+k < n and j+k < n:
                    available[i+k][j+k] = False
                if i-k >= 0 and j-k >= 0:
                    available[i-k][j-k] = False
                if i-k >= 0 and j+k < n:
                    available[i-k][j+k] = False
                if i+k < n and j-k >= 0:
                    available[i+k][j-k] = False

        def backtrack(available, curstate):
            cursum = len([1 for s in curstate for c in s if c == "Q"])
            # print(cursum)
            # print(curstate)
            # print(available)
            if cursum == n:
                curset = [''.join(s) for s in curstate]
                arr.append(curset)
                return
            elif all(available) == False:
                return 
            
            for i in range(n):
                for j in range(n):
                    if available[i][j]:
                        available_bk = [[x for x in a] for a in available]
                        curstate[i][j] = 'Q'
                        # print(available_bk)
                        mask_available(available, i, j)
                        # print(available_bk)
                        backtrack(available, curstate)
                        available = available_bk
                        available[i][j] = False
                        curstate[i][j] = '.'
        backtrack(available, curstate)
        # print(available, curstate)
        return arr
16700