<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:

    void solve(int r, vector<string>board, unordered_set<int>cols, unordered_set<int>mdiag, unordered_set<int>adiag, vector<vector<string>> &ans, int n){
        if(r>=n){
            ans.push_back(board); 
            return;
        }
        for(int i=0; i<n; i++){
            if(cols.count(i) ||  mdiag.count(r-i) || adiag.count(r+i)) continue;
            board[r][i] = 'Q';
            cols.insert(i);
            mdiag.insert(r-i);
            adiag.insert(r+i);

            solve(r+1,board,cols,mdiag,adiag,ans,n);

            cols.erase(i);
            mdiag.erase(r-i);
            adiag.erase(r+i);
            board[r][i] = '.';
        }
    }


    int totalNQueens(int n) {
        vector<vector<string>>ans;
        vector<string>board(n, string(n,'.'));
        unordered_set<int>cols;
        unordered_set<int> mainDiag;  
        unordered_set<int> antiDiag;

        solve(0,board,cols,mainDiag,antiDiag,ans,n);
        return ans.size();
    }
};
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