<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 recur(int row, int n, vector<vector<int>> vis, int &result) {
    if (row == n) { 
        result++;
        return;
    }

    for (int col = 0; col < n; col++) {
        if (!vis[row][col]) {
           vector<vector<int>> temp = vis;
           vis[row][col] = true;
           for (int i = row + 1; i < n; i++) {
                vis[i][col] = true;
           }
           int j = 1;
           for (int i = row + 1; i < n; i++) {
                if (col + j < n) vis[i][col + j] = true; 
                if (col - j >= 0) vis[i][col - j] = true;
                j++;
           }
           recur(row + 1, n, vis, result);
           vis = temp;
        }
    } 
}

int queens(int n) {
    vector<vector<int>> vis(n, vector<int>(n, false));
    int result = 0;
    recur(0, n, vis, result);
    return result;
}

    int totalNQueens(int n) {
        return queens(n);

    }
};
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