Given an array arr[] of N integers. Do the following operation n-1 times For every Kth operation:
Right rotate the array clockwise by 1.
Delete the (n-k+1)th last element.

eg:

A = {1, 2, 3, 4, 5, 6}. Rotate the array clockwise i.e. after rotation the array A = {6, 1, 2, 3, 4, 5} and delete the last element that is {5} so A = {6, 1, 2, 3, 4}. Again rotate the array for the second time and deletes the second last element that is {2} so A = {4, 6, 1, 3}, doing these steps when he reaches 4th time, 4th last element does not exists so he deletes 1st element ie {1} so A={3, 6}. So continuing this procedure the last element in A is {3}, so outputp will be 3

How to solve this?
It seems some kind of problem with mathematical solution (try to find the position with a closed math formula). However, here is an algorithmic approach that runs in $O(n \log^2 n)$.

Using segment tree / Fenwick tree, you can find the number of removed elements in a range (using prefix sums). Keep track of the offset (imagine having a pointer on the first element (imagine a cyclic arry and in each step we push it one element to the left instead of rotating right). Using binary search on the Fenwick tree, you can find the index of element having the distance you are looking for the offset element. Nite that after deleting half of the elements $k$ will be greater than the size and hence you can directly output the first unremoved element before the offset.