<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 Try(int i, int n, vector<int> cot, vector<int> d1, vector<int> d2,vector<int> x, int& count) {
        for (int j = 1; j <= n; j++) {
            if (cot[j] == 1 && d1[i - j + n] == 1 && d2[i + j - 1] == 1) {
                x[i] = j;
                cot[j] = d1[i - j + n] = d2[i + j - 1] = 0;
                if (i == n)
                    count++;
                else
                    Try(i + 1, n, cot, d1, d2, x, count);
                cot[j] = d1[i - j + n] = d2[i + j - 1] = 1;
            }
        }
    }
    int totalNQueens(int n) {
        int count = 0;
        vector<int> cot(100,1); vector<int> d1(100,1); vector<int> d2(100,1); vector<int> x(100,0);
        Try(1, n, cot, d1, d2, x, count);
        return count;
    }
};
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