N-Queens II
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0052.N-Queens%20II/images/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>
Backtracking
class Solution {
public:
    int totalNQueens(int n) {
        bitset<10> cols;
        bitset<20> dg;
        bitset<20> udg;
        int ans = 0;
        function<void(int)> dfs = [&](int i) {
            if (i == n) {
                ++ans;
                return;
            }
            for (int j = 0; j < n; ++j) {
                int a = i + j, b = i - j + n;
                if (cols[j] || dg[a] || udg[b]) continue;
                cols[j] = dg[a] = udg[b] = 1;
                dfs(i + 1);
                cols[j] = dg[a] = udg[b] = 0;
            }
        };
        dfs(0);
        return ans;
    }
};