622D - Optimal Number Permutation
Let's build the answer with the sum equal to zero. Let n be even. Let's place odd numbers in the first half of the array: the number 1 in the positions 1 and n, the number 3 in the positions 2 and n - 1 and so on. Similarly let's place even numbers in the second half: the number 2 in the position n + 1 and 2n - 1, the number 4 in the positions n + 2 and 2n - 2 and so on. We can place the number n in the leftover positions. We can build the answer for odd n in a similar way.
Easy to see that our construction will give zero sum.
Complexity: O(n).
