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 $(i+\frac{N}{2})\mathrm{\%}N$ 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 $i+\frac{N}{2}$.