<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 res=0;
    bool ischeck( vector<pair<int,int>> track, int r, int c){
        for(int i=0; i<track.size(); i++){
            if(track[i].first ==r || track[i].second == c) return false;
            if(abs(track[i].first-r) == abs(track[i].second-c)) return false;
        }return true;
    }
    void backtrack(int n, int r, int c,int count, vector<pair<int,int>>& track){
        if(count == n){
            res++;
            return;
        }if(r == n){
            return;
        }if(c == n){
            if(count < r+1){
                return;
            }backtrack(n,r+1,0,count,track);
        }else if(ischeck(track,r,c)){
            track.push_back(make_pair(r,c));
            backtrack(n,r+1,0,count+1,track);
            track.pop_back();
        }backtrack(n,r,c+1,count,track);
    }
    int totalNQueens(int n) {
        vector<pair<int,int>> track;
        backtrack(n,0,0,0,track);
        return res;
    }
};
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