Maximum Cake Tastiness
Contest: Технокубок 2022 - Финал
Difficulty rating: 800
Tags: greedy, implementation, sortings

There are $$$n$$$ pieces of cake on a line. The $$$i$$$-th piece of cake has weight $$$a_i$$$ ($$$1 \leq i \leq n$$$).

The tastiness of the cake is the maximum total weight of two adjacent pieces of cake (i. e., $$$\max(a_1+a_2,\, a_2+a_3,\, \ldots,\, a_{n-1} + a_{n})$$$).

You want to maximize the tastiness of the cake. You are allowed to do the following operation at most once (doing more operations would ruin the cake):

- Choose a contiguous subsegment $$$a[l, r]$$$ of pieces of cake ($$$1 \leq l \leq r \leq n$$$), and reverse it.

The subsegment $$$a[l, r]$$$ of the array $$$a$$$ is the sequence $$$a_l, a_{l+1}, \dots, a_r$$$.

If you reverse it, the array will become $$$a_1, a_2, \dots, a_{l-2}, a_{l-1}, \underline{a_r}, \underline{a_{r-1}}, \underline{\dots}, \underline{a_{l+1}}, \underline{a_l}, a_{r+1}, a_{r+2}, \dots, a_{n-1}, a_n$$$.

For example, if the weights are initially $$$[5, 2, 1, 4, 7, 3]$$$, you can reverse the subsegment $$$a[2, 5]$$$, getting $$$[5, \underline{7}, \underline{4}, \underline{1}, \underline{2}, 3]$$$. The tastiness of the cake is now $$$5 + 7 = 12$$$ (while before the operation the tastiness was $$$4+7=11$$$).

Find the maximum tastiness of the cake after doing the operation at most once.

Input format:
The first line contains a single integer $$$t$$$ ($$$1 \le t \le 50$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 1000$$$) — the number of pieces of cake.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — $$$a_i$$$ is the weight of the $$$i$$$-th piece of cake.

Output format:
For each test case, print a single integer: the maximum tastiness of the cake after doing the operation at most once.

Constraints:
Time limit: 2.0 s
Memory limit: 256.0 MB

Note:
In the first test case, after reversing the subsegment $$$a[2, 5]$$$, you get a cake with weights $$$[5, \underline{7}, \underline{4}, \underline{1}, \underline{2}, 3]$$$. The tastiness of the cake is now $$$\max(5+7, 7+4, 4+1, 1+2, 2+3) = 12$$$. This is the maximum possible tastiness of the cake one can obtain by performing the operation at most once.

In the second test case, it's optimal not to do any operation. The tastiness is $$$78+78 = 156$$$.

In the third test case, after reversing the subsegment $$$a[1, 2]$$$, you get a cake with weights $$$[\underline{54}, \underline{69}, 91]$$$. The tastiness of the cake is now $$$\max(54+69, 69+91) = 160$$$. There is no way to make the tastiness of the cake greater than $$$160$$$ by performing at most one operation.

Example input:
5
6
5 2 1 4 7 3
3
32 78 78
3
69 54 91
8
999021 999021 999021 999021 999652 999021 999021 999021
2
1000000000 1000000000

Example output:
12
156
160
1998673
2000000000

Editorial:
The answer is the sum of the 2
maximum weights.
You can always pick the 2 maximum weights: if they are ai and aj (i<j), you can flip the subsegment [i,j−1]
to make them adjacent.
The result can't be larger, because the sum of the weights of any 2 pieces of cake is never greater than the sum of the 2
maximum weights.
Iterating over all pairs of pieces of cake is enough to get AC, but you can solve the problem in O(nlogn) by sorting the weights and printing the sum of the last 2 values, or even in O(n)
if you calculate the maximum and the second maximum for each prefix.
Complexity: O(t⋅n2) , O(t⋅nlogn) or O(t⋅n)