Circular Array Rotation | Editorial by naveen161

**sort explanation:-** Given array arr[] of N elements and given r (clockwise rotation of r times of array ) 
                                        and q (queries is nothing just like location of values after rotation)
                                         example :- given arr[] = {3,2,1,5,6} , r = 2, (no of q = 2)  q1 = 2, q2 = 3;
                                        after first rotation  --> {6,3,2,1,5}
                                        after second rotation --> {5, 6,3,2,1}
                                        q1 = 2 (so index of 2) = 1
                                        q2 = 3 (so index of 3) = 5
                                        answer is 1, 5
                               solution :- if rotation is > array size    simple rotate the value using rotate function in c++ :- std::rotate(a.begin(), a.begin()+a.size()-r, a.end());  
                                                  if rotation < array size     take the mod m =  (r%size of array) because it given the rotation of value 
                                                  then use :-  std::rotate(a.begin(), a.begin()+a.size()-m, a.end());
                                                  
 solution of function below that:-
                                                
      vector<int> circularArrayRotation(vector<int> a, int k, vector<int> queries) {
     vector <int> q;  
     int s = a.size();
     if(s >= k)
     {
     std::rotate(a.begin(), a.begin()+a.size()-k, a.end()); 
     for(auto i = queries.begin();i != queries.end();i++)
     {
         q.push_back(a.at(*i));
     }
     return q;
     }else{
         int m = k%s;
          std::rotate(a.begin(), a.begin()+a.size()-m, a.end()); 
          for(auto i = queries.begin();i != queries.end();i++)
         {
            q.push_back(a.at(*i));
         }
       return q;
     }
}

Circular Array Rotation | Editorial by naveen161

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