<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>

def n_queen(board,i,ans):
    if i==len(board):
        ans[0]=ans[0]+1
        return
    
    for j in range(len(board)):
        if isSafeToPlace(board,i,j):
            board[i][j]='Q'
            n_queen(board,i+1,ans)
            board[i][j]='.'
 
def isSafeToPlace(board,i,j):
    n=len(board)
    for r in range(i):
        if board[r][j]=='Q':
            return False
    
    row=i-1
    col=j-1    
    while col>=0 and row>=0:
        if board[row][col]=='Q':
            return False
        row-=1
        col-=1
    
    row=i-1
    col=j+1    
    while col<n and row>=0:
        if board[row][col]=='Q':
            return False
        row-=1
        col+=1
    return True

class Solution:
    def totalNQueens(self, n: int) -> int:
        board=[]
        ans=[0]
        for i in range(n):
            row=[]
            for j in range(n):
                row.append('.')
            board.append(row)
        i=0
        n_queen(board,i,ans)
        return ans[0]
        
16500