<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) {
        vector<vector<string>> res;
        if (n == 1) {
            res.push_back({"Q"});
            return res;
        }
        if (n == 2 || n == 3) {
            return res;
        }
        vector<int> positions;
        backtracking(n, n, positions, res);
        return res;
    }

    void backtracking(int n, int remaining, vector<int> positions, vector<vector<string>>& res) {
        int m = positions.size();
        if (remaining == 0) {
            vector<string> returnArr(positions.size(), string(positions.size(), '.'));
            for (int i = 0; i < positions.size(); i++) {
                returnArr[i][positions[i]] = 'Q';
            }
            res.push_back(returnArr);
            return;
        }
        unordered_set<int> y, diag_left, diag_right;
        for (int i = 0; i < m; ++i) {
            y.insert(positions[i]);
            diag_left.insert(positions[i] - i);
            diag_right.insert((positions[i] + i));
        }
        for (int i = 0; i < n; ++i) {
            if (y.contains(i) || diag_left.contains(i - m) || diag_right.contains((i + m))) continue;
            positions.push_back(i);
            backtracking(n, remaining - 1, positions, res);
            positions.pop_back();
        }
    }
};
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