<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:
    int publicN=0; 
    int ans=0;
    set<int> colSet;
    set<int> PlusSet;
    set<int> MinusSet;
    bool isValid(int i,int j)
    {
        if (colSet.find(j)!=colSet.end() || PlusSet.find(i+j)!=PlusSet.end() ||MinusSet.find(i-j)!=MinusSet.end())
        {
            return false;
        }
        else
        {
            return true;
        }
        
        

    }
    void dfs(int i)
    {
        if (i==publicN)
        {
            ans++;
        }
        else
        {
            for (int j = 0; j < publicN; j++)
            {
                if (isValid(i,j))
                {
                    colSet.insert(j);
                    PlusSet.insert(i+j);
                    MinusSet.insert(i-j);

                    dfs(i+1);
                    
                    colSet.erase(j);
                    PlusSet.erase(i+j);
                    MinusSet.erase(i-j);
                }
                
            }
            
        }
        
    }

    int totalNQueens(int n) {
        publicN = n;
        dfs(0);
        return ans;
        
    }
};
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