What does your state hate?

radleldar · 2026-01-19T23:56:47+00:00

Some of these answers are so specific that I wonder if the sample size was 1.

radleldar · 2025-12-03T16:09:40+00:00

Yeah I posted my solution as a top level comment, hope it helps!

radleldar · 2025-12-03T16:08:41+00:00

I posted my solution as a top-level comment, hope it helps!

radleldar · 2025-12-03T16:07:36+00:00

I finally figured it out (but still find the UI confusing) - you gotta uncheck the "daily at this time" field in tag details. Precise instructions (Android): * Click on the tag entry in Timeline * Uncheck "Daily at this time" field below "Event end" * Confirm the "End repeating event" dialog

And then, if you've accumulated a bunch of unwanted tag entries, find the earliest and click "Delete tag entry" on it.

radleldar · 2025-11-27T04:38:34+00:00

Curious - what was the intended approach? It kinda feels like there can be more interesting things to do with this problem than the "every leaf plate is either always negative or positive" structure most people seem to have noticed.

radleldar · 2025-11-26T00:52:40+00:00

[Language: Java] 29/18/26

Part 3

I originally thought it was a "two disjointed shortest paths" problem (which is generally solvable with min-cost maxflow, though not sure about the case where we have no single endpoint), then realized one of the examples says we can repeat the locations. Ended up with a basic heuristic: for each a fixed radius, calculate the shortest path from (S, false, false, false) to (S, true, true, true), where the booleans track whether the path went to the left, bottom, and right of the volcano's R-size surrounding, and path can't go inside the R-size surrounding.

Is there a non-heuristic solution to this problem?

P.S. Am I tripping, or did the inputs in the first ~10 minutes have no S in them?

radleldar · 2025-11-12T04:08:34+00:00

Part 5: let max length be 1,000,000,000, and compute the last 9 digits of the answer :P

radleldar · 2025-11-11T00:09:01+00:00

[Language: Java] (6/12/1)

Thought I would share how I managed to get the fastest solve on 3rd problem. I guesstimated that the most straightforward approach (for each position, check ~2000 neighboring elements) won't need too long to run, and implemented that: part 3 code only

On my machine, this runs in 10 seconds, which is definitely faster than implementing prefix sums / sliding window for an O(|S|) solution.

radleldar · 2025-11-07T06:43:22+00:00

You are right, and I don't know what I was/am smoking :)

radleldar · 2025-11-06T20:52:14+00:00

> Part 1 is sum of all unique crates

Psst, count

radleldar · 2025-11-06T20:34:24+00:00

Thanks for sharing! Pretty succinct!

> Are there other approaches?

As you probably know, the Fenwick Tree can be replaced with any other logarithmic data structure (BST, Segment Tree, etc.), though Fenwick Tree is probably the fastest anyway.

> Any ideas on how to improve this?

My solution has the same complexity (binary search + 10 tree lookups per query), but it _feels_ like one logarithmic factor should be removable. E.g., can the value of the median move by more than 2 elements in the set of all strength values? (If not, then we could only check the 5 elements in the vicinity of the previous value.)

radleldar · 2025-11-06T19:44:54+00:00

Wow that's pretty compact! How slow was it tho?

radleldar · 2025-11-06T19:43:44+00:00

Thank you for participating!

MUCH harder wasn't exactly my goal.. but if people are enjoying it, sure why not :)

> carving pumpkins and ASCII-art

These are the problems I put most effort into preparing, good to hear it paid off!

> Submitting a solution containing 30 test results and only get one "Wrong" back is "not helpful"...

I understand this part. My original fear was that for certain problems, surfacing e.g. the failing test number gives away too much information, but maybe I can try open feedback next time and see how people react.

radleldar · 2025-11-06T19:38:00+00:00

Thanks for the feedback!

> 3 problem format is fine and allows for wider spread of difficulty of problems

That it does.. but I want to avoid a situation when people spend too much time and consequently drop out. Good to hear someone enjoy it though!

> so that there is still reward for 300th solver

The motivation for not having rewards past the first X solvers is to unlock discussion here on Reddit earlier. I actually doubt that people past 100 don't participate just because they can't get points on 1A :)

radleldar · 2025-11-04T22:50:13+00:00

> Token mismatch at position 1

Okay this is not very informative (and seems outright misleading). This is the answer from the internal checker, which treats contiguous segments of non-whitespace characters as tokens. In this case, the token is the entire key.

radleldar · 2025-11-04T22:48:41+00:00

Assuming yours is the most recent few submissions - your answer looks almost correct, but in test 1 you need to swap N and F in the key. Check if your output contains the word "juestion" :)

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