Trying a different format with problem-by-problem breakdowns, which hopefully will be more useful when I review my postmortems. I will also backfill the postmortems.

A: I first implemented the metric computation (sort sums of adjacent elements) without having any idea of how to create a different permutation to achieve the same metric. Then I realized that reversing the input preserves the adjacent sums. I won't complain about a 0:02 submission.

B: If the first nonzero element were negative, then we have to incur cost equal to its magnitude. I thought this was it, but some sample inputs didn't pass. Then I noticed that if the first nonzero element were positive, we can add the magnitude to negative terms later in the array, which existed because of the constraint that all elements sum to 0 and that the sum stays 0 after every operation. I quickly patched my solution to keep a reserve from the magnitude of positive elements and using the reserve to decrease the cost of negative elements before incurring their costs. Again, not complaining about a 0:08 submission.

C: This is where I started to struggle. Having the same number of 0s as 1s in every K subarray is the same as the subarray sum equal to K/2. I got hung up on the "system of equations" representation, where every question mark was a variable (question mark at i would be a_i), which seemed extremely hard to solve. Then I thought that I can just look at the number of equations and the number of variables, which seemed sketchy, but I YOLO-ed and incurred a WA.

At about the 0:55 mark, I realized that the system of equations implied that a_i = a_j for i = j mod K. Facepalm. I quickly coded this solution for a 1:02 submission. RIP

D: Clearly, if B is at most DA away from A, then A can reach B in the first move and win. From the problem statement, I inferred that their 10^100 limit meant that either A will reach B in finite time, or B can always move to stay DA away from A. I quickly realized that the best chance of B dodging A is on the diameter of the tree, since in no other scenario will B have the space to run away from A. This implied that if A can go from the center(s) to the ends of the diameter in DA, then A can catch B.

I made the assumption that if DA > DB (it seemed intuitive), then A will be able to reach B eventually because B will run out of space. Thinking that these three criteria were sufficient, I submitted at 1:31 for my first WA on D. Given feedback that my solution failed pretest 2, I came up with the following counterexample to my solution:

1---2---3---4---5---6 ^ ^ A B DA = 2, DB = 3 |

My solution would incorrectly say that A never reaches B, while after two moves, A will first move from 3 to 5 and B will move from 6 to 3, letting A move from 5 to 3 on the third move. This really tripped me up, leading me to think about the relative difference between DA and DB. I tried that if the relative difference is larger than the half the diameter of the tree, then A cannot catch B, for another WA at 1:44. Finally, as a last ditch attempt, I thought that if DB is twice or more than DA and if the diameter is greater than twice DA, then A cannot catch B, for my third and final WA at 1:59.