Let us start with the brute force solution. We simulate the calls to the something function. Using appropriate structure (for example tournament tree) we answer the sequence sum question in ~\mathcal O(1)~ time.

Now for the first improvement. If there are multiple times that number ~Y~ appears in the input (~X~ times for example): there is no need to call the something function ~X~ times. Instead, we can call a modified version of the function once, and have her increment the fields by ~X~. This eliminates all duplicates in the input sequence.

What is the complexity of this solution? The worst case for the brute force algorithm is ~K~ ones leading to ~\mathcal O(NK)~ complexity. But our new modified function solves such a case in one pass. Obviously, we can have only one number ~1~ in the worst case. The next worst thing is to enter number ~2~. Then ~3~. And so on. So the worst case is now ~\{1, 2, \dots, K\}~. Our modified function will perform ~\frac N 1 + \frac N 2 + \dots + \frac N K~ steps. But it can be easily shown that this is at most ~20N~. This is solvable in the given time limit.