<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:
    bool isSafe(int i, int j, int n, vector<int> row, vector<int> upperDiag, vector<int> lowerDiag){
        if(row[i]) return false;
        if(upperDiag[n - 1 + j - i]) return false;
        if(lowerDiag[i + j]) return false;
        return true;
    }
    int f(int col, int n, vector<int> row, vector<int> upperDiag, vector<int> lowerDiag){
        if(col == n) return 1;
        int count = 0;
        for(int i = 0; i < n; i++){
            if(isSafe(i, col, n, row, upperDiag, lowerDiag)){
                row[i] = 1;
                upperDiag[n - 1 + col - i] = 1;
                lowerDiag[col + i] = 1;
                count += f(col + 1, n, row, upperDiag, lowerDiag);
                row[i] = 0;
                upperDiag[n - 1 + col - i] = 0;
                lowerDiag[col + i] = 0;
            }
        }
        return count;
    }
    int totalNQueens(int n) {
        int count = 0;
        vector<int> row(n, 0), upperDiag(2 * n - 1, 0), lowerDiag(2 * n - 1, 0);
        return f(0, n, row, upperDiag, lowerDiag);
    }
};
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