<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:
    short co=0;
    vector<bool> c, ld, rd;
    void h(int n, short i) {
        if (i==n) {
            ++co;
            return;
        }
        short j=0;
        for (; j<n; j++) {
            if (!c[j] && !ld[i-j+(n-1)] && !rd[i+j]) {
                c[j]=true;
                ld[i-j+(n-1)]=true;
                rd[i+j]=true;
                h(n, i+1);
                c[j]=false;
                ld[i-j+(n-1)]=false;
                rd[i+j]=false;
            }
        }
    }
    int totalNQueens(int n) {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        c.assign(n, false);
        ld.assign(2*n-1, false);
        rd.assign(2*n-1, false);
        h(n, 0);
        return co;
    }
};
8902