Passed div2 d at the last minute, hooray!

What, how!?

n=3,k=12

I played you program with my program. It was fun. Your testcase is (n,k) = (5,10). You pick Bob and I take away 1 stone making n == 4. You cannot pick all 4 now.

Since you can swap ai and aj if |i-j| >= k, that means you can swap a[0] with a[k], a[k+1], ..., a[n-1], a[1] with a[k+1], a[k+2], ..., a[n-1]. Which means all indices i < n-k can be swapped with all indices j >= k. Now there are three cases. First case : If n-k >= k i.e k <= n/2 then all numbers can be swapped around which means just sort the array and return it. Second case : If k > n then no number can be swapped, so just return the original array. Third case : Put the first n-k numbers and n-1-k+1 numbers in another array, sort them. Print the first n-k numbers then print the numbers from the original array from n-k+1 to k-1 and then print the remaining arrays from the new sorted array.

I used a multiset ,so I think my approach was similar to yours and I got AC. My approach--> The indexes where swap operation will NOT be allowed will follow the following condition-> (abs(i-0) <k && abs(i-(n-1) ) <k).

I stored all these indices in a set ns (non swappable indexes) and I stored rest of the indices in another set s (swappable indexes) and their respective values in a multiset and I again traversed theses swappable indices from left to right and filled the values of the multiset on these indices in an ascending order.

Values at non-swappable indexes will remain the same as they were in the initial array. Finally I printed the answer array. This is my submission

I think the tests are weak too, I literally just simulated the $$$k$$$ constraints $$$2 \cdot n$$$ times and checked if the upper bound at least the lower bound, and it passes all test cases. If this is actually correct, I don't have a proof for this, it's just random intuition lmao. Otherwise, the tests are just bad.

No they are different. Consider the linear system:

$$$b_0 + b_{n-1} = a_0$$$.

$$$b_0 + b_1 = a_1$$$.

$$$b_1 + b_2 = a_2$$$.

...

$$$b_{n-2} + b_{n-1} = a_{n-1}$$$.

Intuitively, $$$b_1 = a_1 - b_0$$$, $$$b_2 = a_2 - a_1 + b_0$$$, $$$b_3 = a_3 - a_2 + a_1 - b_0$$$, therefore the parity of $$$n$$$ matters, because it decides the sign of $$$b_0$$$ in the expression.