<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>all distinct solutions to the <strong>n-queens puzzle</strong></em>. You may return the answer in <strong>any order</strong>.</p>

<p>Each solution contains a distinct board configuration of the n-queens&#39; placement, where <code>&#39;Q&#39;</code> and <code>&#39;.&#39;</code> both indicate a queen and an empty space, respectively.</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> [[&quot;.Q..&quot;,&quot;...Q&quot;,&quot;Q...&quot;,&quot;..Q.&quot;],[&quot;..Q.&quot;,&quot;Q...&quot;,&quot;...Q&quot;,&quot;.Q..&quot;]]
<strong>Explanation:</strong> There exist two distinct solutions to the 4-queens puzzle as shown above
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> [[&quot;Q&quot;]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 9</code></li>
</ul>

class Solution {
public:
    vector<vector<string>> solveNQueens(int n) {
        // Sets to track used columns and diagonals
        set<int> col, posDiag, negDiag;
        vector<vector<string>> res;
        vector<string> board(n, string(n, '.'));
        
        // Backtracking function
        function<void(int)> backtrack = [&](int r) {
            if (r == n) {
                res.push_back(board);
                return;
            }

            for (int c = 0; c < n; ++c) {
                if (col.count(c) || posDiag.count(r + c) || negDiag.count(r - c)) 
                    continue;
                
                // Place the queen
                col.insert(c);
                posDiag.insert(r + c);
                negDiag.insert(r - c);
                board[r][c] = 'Q';

                // Recur for the next row
                backtrack(r + 1);

                // Remove the queen (backtrack)
                col.erase(c);
                posDiag.erase(r + c);
                negDiag.erase(r - c);
                board[r][c] = '.';
            }
        };
        
        backtrack(0);
        return res;
    }
};
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