<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 {

private:

    void solve(int col,vector<string> &board,vector<int> &leftRow,vector<int> &upperDiagonal,vector<int> lowerDiagonal,int n,int &cnt){

        if(col >= n){
            cnt++;
            return;
        }

        for(int row = 0 ;row < n;row++){

            if(leftRow[row] == 0 && upperDiagonal[row+col] == 0 && lowerDiagonal[n-1+ col - row] == 0){

                board[row][col] = 'Q';
                leftRow[row] = 1;
                upperDiagonal[row+col] = 1 ;
                lowerDiagonal[n-1+col -row] = 1;

                solve(col+1,board,leftRow,upperDiagonal,lowerDiagonal,n,cnt);

                board[row][col] = '.';
                leftRow[row] = 0;
                upperDiagonal[row+col] = 0 ;
                lowerDiagonal[n-1+col -row] = 0;

            }
        }

    }
public:
    int totalNQueens(int n) {

        vector<string> board(n);

        string s(n,'.');

        int cnt = 0 ;

        vector<int> leftRow(n,0);
        vector<int> upperDiagonal(2*n-1,0);
        vector<int> lowerDiagonal(2*n-1,0);

        solve(0,board,leftRow,upperDiagonal,lowerDiagonal,n,cnt);
        
        return cnt;
    }
};
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