<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>

int sol_from(vector<int> orient,int ind,int row, int n, vector<bool> row_done, vector<bool> diagu_done, vector<bool> diagd_done)
{
    orient[ind]=row;
    row_done[row]=true;
    diagu_done[ind-row+n]=true;
    diagd_done[row+ind]=true;
    int ans=0;
    if(ind==n-1)
    {
        return 1;}

    for(int i=0; i<n; i++)
    {
        // if orient[ind]=n
        if(!row_done[i] and !diagu_done[ind+1-i+n] and !diagd_done[ind+1+i])
            ans+=sol_from(orient, ind+1, i, n, row_done, diagu_done, diagd_done);
    }
    return ans;
}
class Solution {
public:
    int totalNQueens(int n) {
        vector<bool> row_done(n, false);
        vector<bool> diagu_done(2*n, false);
        vector<bool> diagd_done(2*n, false);
        vector<int> orient(n,-1);
        int final_ans=0;
        for(int i=0; i<n; i++)
        final_ans+=sol_from(orient, 0, i, n, row_done, diagu_done, diagd_done);
        return final_ans;
        
    }
};
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