<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 {
    int dfs(int n, vector<string>& board, unordered_set<int> columns, unordered_set<int> diagonals, unordered_set<int>antidiagonals, int row){
        if(row>=n){
            // if(columns.size()==n){
            //     return 1;
            // }
            // else{
            //     return 0;
            // }
            return 1;
        }
        int res=0;
        for(int c=0;c<n;c++){
            int d = row-c;
            int ad = row+c;
            if(columns.count(c) || diagonals.count(d) || antidiagonals.count(ad) || board[row][c]!='.'){
                continue;
            }
            board[row][c] = 'Q';
            columns.insert(c);
            diagonals.insert(d);
            antidiagonals.insert(ad);
            res += dfs(n, board, columns, diagonals, antidiagonals, row+1);
            board[row][c] = '.';
            columns.erase(c);
            diagonals.erase(d);
            antidiagonals.erase(ad);
            

        }
        return res;
    }
public:
    int totalNQueens(int n) {
        unordered_set<int> columns;
        unordered_set<int> diagonals;
        unordered_set<int> antidiagonals;
        string emptyRow = "";
        for(int i=0;i<n;i++){
            emptyRow += '.';
        }
        vector<string> emptyBoard;
        for(int i=0;i<n;i++){
            emptyBoard.push_back(emptyRow);
        }
        return dfs(n, emptyBoard, columns, diagonals, antidiagonals, 0);

    }
};
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