Subtask 1

We note that there can only be either ~2~ or ~4~ people at the table. This can be solved with conditional statements. In the case that there are only ~2~ people, they are seated across from each other, so we just compare the values of both hats. With ~4~ people, we can compare the equality of the hats at seat ~1~ and ~3~ and seat ~2~ and ~4~.

Subtask 2

All people have the same hat number ~1~. This means we can simply count the number of people at the table.

Subtask 3

Since people in even-numbered seats have hat number ~1~ and people in odd-numbered seats have hat number ~0~, they will only see matching hats is if the number of people is divisible by ~4~.

Subtask 4

There was no specific intended approach for this subtask. We observe that in a circular arrangement, the person in seat ~i~ is directly opposite to the person in seat ~i + \frac{N}{2}~.

One possible solution would be to store for each hat number, a list of the seat numbers of the people wearing that hat number. We can then iterate over the lists for each hat number and check if for each seat ~i~ in a list, if both ~i~ and ~i + \frac{N}{2}~ exist in the same list.

Subtask 5

We notice that we can do the check directly on the input as a list. For each person at seat ~i~, we check if the opposite person at seat ~i + \frac{N}{2}~ has the same hat number. Since each seat is indexed from ~1~ to ~N~, if ~i > \frac{N}{2}~ then ~i + \frac{N}{2}~ will be greater than ~N~. To ensure that we stay within bounds, there are a couple approaches we can use:

• loop from ~1~ to ~N~ and compare ~i~ and ~(i + \frac{N}{2}) \% N~ to count person at seat ~i~;
• loop from ~1~ to ~N~ and subtract ~N~ from ~i + \frac{N}{2}~ if it is greater than ~N~ to count person at seat ~i~;
• loop from ~1~ to ~N~ and count for person at seat ~i~ and seat ~i + \frac{N}{2}~.