When (K = 0), again output YES if and only if (A = B) as in chapter 1.
When (K = 1) and (A = B), it is now possible for the answer to be YES due to duplicates, for instance, if (A = B = [1, 2, 1, 2]). We will further explain how to handle this case below.
For the special case of (N = 2), the answer is now YES if (A_1 = A_2 = B_1 = B_2). Otherwise, the same logic from chapter 1 remains.
In the general case, checking for a rotation becomes more difficult now that duplicates are allowed. We can no longer map values to their indices as in chapter 1, since there may be multiple indices for a given value. We could try rotating (A) by each (1..N) and comparing to (B), but each comparison could take (\mathcal{O}(N)), for an overall time complexity of (\mathcal{O}(N^2)). A worst-case example for typical lexicographical comparison would be (A = [1, 1, ..., 1, 2, 3]), (B = [1, 1, ..., 1, 3, 2]).
One possible solution is to concatenate (A) with a copy of itself and search for whether (B) occurs as a substring using a linear time string-matching technique such as the Knuth-Morris-Pratt, Boyer-Moore, or Z algorithm. One thing to watch out for is the (K = 1, A = B) case above, where it's especially important to rule out cutting (0) or (N) cards. This can be done by deleting the first and last characters of (A+A) before searching for (B).
Another valid approach involves hashing. We can define some hash function (h(v_1, v_2, ..., v_N)) that is sensitive to the order of its (N) inputs, but also supports computing the hash (h(v_2, ..., v_N, v_1)) of the rotated array in (\mathcal{O}(1)), assuming (h(v_1, v_2, ..., v_N)) is already given. We can then rotate just the hash of (A) one-at-a-time, comparing each to the hash of (B), for an overall (\mathcal{O}(N)) time complexity.