<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 rec(int n, vector<vector<int>>& ans, unordered_set<int>& cols, unordered_set<int>& diag1, unordered_set<int>& diag2, vector<int>& output, int index) {
        if (index == n) {
            if (output.size() == n) {
                ans.push_back(output);
            }
            return;
        }
        
        for (int i = 0; i < n; i++) {
            if (cols.find(i) == cols.end() && diag1.find(index - i) == diag1.end() && diag2.find(index + i) == diag2.end()) {
                output.push_back(i+1);
                cols.insert(i);
                diag1.insert(index - i);
                diag2.insert(index + i);
                
                rec(n, ans, cols, diag1, diag2, output, index + 1);
                
                output.pop_back();
                cols.erase(i);
                diag1.erase(index - i);
                diag2.erase(index + i);
            }
        }
    }
    int totalNQueens(int n) {
        vector<vector<int>> ans;
        unordered_set<int> cols, diag1, diag2;
        vector<int> output;
        rec(n, ans, cols, diag1, diag2, output, 0);
        return ans.size();
    }
};
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