More formally, the task statement asks us to divide the input data into two arithmetic sequences.

To obtain 20 points, it was enough to try all possible combinations and check whether some of them satisfy all conditions. The time complexity of this solution is ~\mathcal O(N \times 2^N)~.

For additional 30 points, it was enough to observe that the smallest number in the input must also be the smallest number in one of the arithmetic sequences. Now, let's fix the second smallest number in that sequence. Now we have also fixed the difference between two successive elements of that sequence, meaning that we can easily calculate what number could be next in the sequence. We will add new numbers to that sequence, one by one, and each time check whether the remaining numbers form another arithmetic sequence. The time complexity of this solution is ~\mathcal O(N^3)~.

As Bob the Builder would say: "Can we fix it? – Yes, we can!"

We can note that the two smallest numbers of one arithmetic sequence could only be fixed in three different ways. If we observe the three smallest numbers in the input (denoting them with ~A \le B \le C~), we can easily conclude that one arithmetic sequence must start with either ~(A, B)~, ~(A, C)~ or ~(B, C)~.

This observation leads us to an algorithm of time complexity ~\mathcal O(N^2)~.

Finally, to obtain all points, we can speed up the idea from above. More precisely, we will speed up the check whether all elements that were not included in the first sequence form another arithmetic sequence. We will keep two (multi)sets around. The first set will contain all remaining numbers, and the other will contain the differences of neighbouring elements from the first set. When we remove the number from the first set, i.e. append it to the first sequence, we need to remove the difference between it and its neighbours from the second set. We also need to add the difference between the neighbours of the removed element because they have now become neighbouring elements. The check whether the remaining number form an arithmetic sequence boils down to checking whether the smallest element from the second set is equal to the largest element in that same set. All operations on the aforementioned sets are supported by the STL collection std::set in C++. Naturally, other supported languages (with exception of C) also offer similar built-in collections.

The time complexity is ~\mathcal O(N \log N)~.