At first glance, this problem may seem like a messy graph problem with the barns forming some network. Luckily, the way the barns are connected means that we don't have to construct a graph explicitly.

The key observation is that because there are $K$ barns in a cycle and $K$ total pairs of connections, no two barns outside the cycle are connected. (Try drawing some cycles and counting the connections to verify this fact!) This observation means that we can split the answer into a sum of two values: the number of same-labelled barns (a) outside and (b) inside the cycle.

If a barn outside the cycle is labelled the same by Annabelle and Bessie, then its label must not appear in the lists $a_i$ or $b_i$. And because the total number of barns outside the cycle is no greater than the number of integers from $1$ to $N$ appearing in neither $a_i$ nor $b_i$, the maximum number of same-labelled barns outside the cycle is exactly equal to that number. This value can be computed in $O(N)$ time with a set data structure.

To find the maximum number of same-labelled barns inside the cycle, imagine placing the two cycles on top of each other and rotating one cycle until the maximum number of labels align. Rotating a cycle and counting the number of alignments takes $O(N)$ time, so a straightforward implementation of this algorithm would take $O(N^2)$ time. To speed it up, we can count the number of alignments for a particular rotation (say rotating $a_i$ to the right by $r$ positions) by counting the number of positions $i$ where $a_i = b_{i + r}$ (wrapping around as necessary). Specifically:

• We start with a counter array $c_0, \dots, c_{K - 1}$, initially all set to zeroes.
• We iterate over every value in $a_i$ and find the corresponding value $b_j = a_i$ if it exists.
• We then increment $c_r$ for $r = i - j$ if $i \geq j$, or $r = i - j + K$ if $j > i$.
• Finally, we take the maximum $c_r$ at the end of this process as the answer.

(Note that we also need to consider the case where the list $a_i$ or $b_i$ is reversed!)

With this improved algorithm, we no longer iterate over all barns each time we do a rotation, so it runs in $O(N)$ time.

Altogether, this solution runs in $O(N)$ time. $O(N \log N)$-time solutions using slower set data structures should also pass all test cases.

Python code implementing this solution:

n, k = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))

same_outside = set(range(1, n + 1)) - set(a) - set(b)

def max_rotation(cycle_a, cycle_b):
    position_a = {val: idx for idx, val in enumerate(cycle_a)}
    with_offset = [0] * k
    for idx, val in enumerate(cycle_b):
        if val in position_a:
            with_offset[idx - position_a[val]] += 1
    return max(with_offset)

print(len(same_outside) +
      max(max_rotation(a, b), max_rotation(a, b[::-1])))