For Subtask 1, you can check every permutation of indices.

Time Complexity: ~\mathcal O(N \cdot N!)~

For Subtask 2, the intended approach is brute-forcing subsets. Another approach that probably works is randomly selecting permutations and checking them.

Time Complexity: ~\mathcal O(N^2 \cdot 2^N)~

Now to obtain ~50\%~ on any subtask, consider the smallest possible value ~s~ such that ~A_K=s~ and the largest possible value ~t~ such that ~A_K=t~. It is not hard to see that these values are obtained when the frequencies are sorted from largest to smallest and smallest to largest respectively.
What is not clear is why ~A_K~ can be equal to ~r~ for ~s \le r \le t~. However, this is a programming competition, so submitting this will obtain ~50\%~ without any need of understanding it.

Time Complexity: ~\mathcal O(N \log N)~

For Subtask 3, 4, and 5, we prove the above claim. Start with the frequencies sorted from smallest to largest and bubble-sort them until they are from largest to smallest. So two adjacent frequencies get swapped, and only if ~F_i>F_j~ and ~i>j~. It is easy to see that a swap either does not alter ~A_K~ or it increases it by exactly ~1~. Since it starts at ~s~ and ends at ~t~, then clearly, it reaches every value between ~s~ and ~t~. So it is proved.

So for Subtask 3, 4, and 5, we can simulate the bubble sort. For Subtask 3, each swap of the bubble sort is done and each time, all ~N~ values are recomputed. For Subtask 4, the ~N~ values are not recomputed. For Subtask 5, only swaps which affect the ~X~ values are done.

Time Complexity: ~\mathcal O(N^3)~, ~\mathcal O(N^2)~, and ~\mathcal O(N \log N)~

Another approach for Subtask 5 is to sort the frequencies and test every subarray of length ~X~ as ~F_1, F_2, \dots, F_X~. The proof of this is left to the reader.

Time Complexity: ~\mathcal O(N \log N)~