Subtask 1

Let's deal with the special case ~N = 2~ first:

• If ~h_1 \ge h_2~, Ina would just need to perform the operation ~h_2 - h_1~ times to even out the two heights.
• Otherwise, it is impossible.

For every other ~N~, we can check if Ina can set all Takodachis to some height using the following process: First, we can calculate the number of operations she needs to perform. Then we iterate starting from the end, and greedily use as many +2s as possible, while fixing parity issues with the +1s. Throughout that process, it is important to make sure that at no point does the number of +1s used exceeds the number of +2s used. This process takes ~\mathcal O(N)~ time.

In this subtask, it is sufficient to continuously increment from ~max(h_i)~ until either a working height has been found or we have reached some pre-determined upper bound. Given that the maximum of ~h_i~ is ~10^3~, a generous upper bound ~H~ is ~10 \times h_i~, or ~10^4~. The total runtime complexity of the brute force is ~\mathcal O(NH)~.

Subtask 2

When ~N = 2~, the process is the same as the previous subtask.

For all other ~N~, observe the following lemma: If some height ~x~ works, then ~x+2~ also works when ~3 \mid N~; otherwise, ~x+6~ also works.

We once again set upper ~H = 10 \times h_i = 10^{13}~. Using the lemma, instead of checking all the heights from ~\max(h_i)~ to ~H~, we can check ~H~ plus each modulo of ~6~. After finding a modulo that works, we can then find the minimum height that works by decrementing in step of ~2~ if ~3 \mid N~, or in step of ~6~ otherwise. The decrementing process can be done efficiently using a binary search. This reduces the runtime to ~\mathcal O(N \log H)~, which is sufficient to get the full score.