<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.res = 0
        for j in range(n):
            col, pos, neg = [False]*n, [False]*(n*2-1), [False]*(n*2-1)
            self.add(n, col, pos, neg, 0, j)
            self.backtrack(n, col, pos, neg, 0, j)
        return self.res

    def backtrack(self, n, col, pos, neg, i, j):
        if i == n-1:
            self.res += 1
            return

        nextI = i + 1
        for nextJ in range(n):
            if col[nextJ] or pos[self.pos(n, nextI, nextJ)] or neg[self.neg(nextI, nextJ)]:
                continue
            self.add(n, col, pos, neg, nextI, nextJ)
            self.backtrack(n, col, pos, neg, nextI, nextJ)
            self.remove(n, col, pos, neg, nextI, nextJ)

    def add(self, n, col, pos, neg, i, j):
        col[j] = True
        pos[self.pos(n,i,j)] = True
        neg[self.neg(i,j)] = True

    def remove(self, n, col, pos, neg, i, j):
        col[j] = False
        pos[self.pos(n, i,j)] = False
        neg[self.neg(i,j)] = False

    def pos(self, n, i, j):
        return n - 1 + j - i
    
    def neg(self, i, j):
        return i + j
16600