Based on the problem in GeeksForGeeks here. I came across a solution here.

The question is as follows

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.

Now, find the element which is left at last.

If (n-k+1)th last element doesnt exist, delete the first.

Test Cases -

Input

2

4

1 2 3 4

6

1 2 3 4 5 6

Output

2

3

Explanation - Testcase 2: 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.

Can someone please help me understand the solution. Primarily I need help in the following block:

if(n==1) cout<<arr[0]<<endl;
else if(n%2) {
    ll ind = n-3;
    ind = floor(ind/4);
    ind = 3+ind;
    cout<<arr[ind-1]<<endl;
} else {
    ll ind = n-2;
    ind = floor(ind/4);
    ind = 2+ind;
    cout<<arr[ind-1]<<endl;
}