<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 {
public:
vector<int> row,col,leftdia,rightdia;
    int totalNQueens(int n) {
        vector<vector<int>> grid(n,vector<int>(n,0));
        row.resize(n,0);
        col.resize(n,0);
        leftdia.resize(2*n,0);
        rightdia.resize(2*n,0);
        return solve(0,n,grid);
    }
    int solve(int i,int n,vector<vector<int>>& grid){
        if(i==n) return 1;
        int ans=0;
        for(int j=0;j<n;j++){
            if(grid[i][j]==0 && col[j]==0 && leftdia[i+j]==0 && rightdia[n-1-i+j]==0){
                grid[i][j]=1;
                col[j]=1;
                 leftdia[i+j]=1 ; 
                 rightdia[n-1-i+j]=1;
                 ans+=solve(i+1,n,grid);
                 grid[i][j]=0;
                col[j]=0;
                 leftdia[i+j]=0 ; 
                 rightdia[n-1-i+j]=0;

            }
        }
return ans;
    }
};
7872