<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:
      string f(int k,int n){
        string tempo="";
        for(int i=0;i<n;i++){
            if(i==k)tempo.push_back('Q');
            else tempo.push_back('.');
        }
        return tempo;
    }
    int totalNQueens(int n) {
           vector<string> temp;     queue<pair<vector<string>,pair<vector<int>,pair<vector<int>,vector<int>>>>> q;
   vector<int> sum,col,dif;
        q.push({temp,{col,{sum,dif}}});
        for(int i=0;i<n;i++){
            int t=q.size();
            for(int j=0;j<t;j++){
                vector<string> temp1=q.front().first;
                vector<int> col1=q.front().second.first;
                vector<int> sum1=q.front().second.second.first;
                 vector<int> dif1=q.front().second.second.second;
                q.pop();
                for(int k=0;k<n;k++){
         if(find(col1.begin(),col1.end(),k)!=col1.end())continue;
         if(find(sum1.begin(),sum1.end(),i+k)!=sum1.end())continue;
         if(find(dif1.begin(),dif1.end(),k-i)!=dif1.end())continue;
                    string a=f(k,n);
                    col1.push_back(k);
                    sum1.push_back(k+i);
                    dif1.push_back(k-i);
                    temp1.push_back(a);
                       q.push({temp1,{col1,{sum1,dif1}}});
                    col1.pop_back();
                     sum1.pop_back();
                     dif1.pop_back();
                    temp1.pop_back();
                }
            }
        }
       return q.size();
    }
};
30017