Advent Of Code

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An unofficial home for the advent of code community on programming.dev!

Advent of Code is an annual Advent calendar of small programming puzzles for a variety of skill sets and skill levels that can be solved in any programming language you like.

AoC 2024

Solution Threads

M	T	W	T	F	S	S
						1
2	3	4	5	6	7	8
9	10	11	12	13	14	15
16	17	18	19	20	21	22
23	24	25

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🦌 - 2024 DAY 2 SOLUTIONS -🦌 (programming.dev)

submitted 3 weeks ago by Ategon@programming.dev to c/advent_of_code@programming.dev

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Day 2: Red-Nosed Reports

Megathread guidelines

Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
You can send code in code blocks by using three backticks, the code, and then three backticks or use something such as https://blocks.programming.dev if you prefer sending it through a URL

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What is this?: Here is a post with a large amount of details: https://programming.dev/post/22323136
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[–] Rin@lemm.ee 1 points 3 weeks ago (1 children)

3 nested for loops

[–] hades@lemm.ee 2 points 3 weeks ago (1 children)

It's not as simple as that. You can have 20 nested for loops with complexity of O(1) if all of them only ever finish one iteration.

Or you can have one for loop that iterates 2^N times.

[–] Rin@lemm.ee 1 points 3 weeks ago (2 children)

What do you think my complexity is?

I think it could be maybe O(n^2) because the other for loop which tries elements around the first error will only execute a constant of 5 times in the worst case? I'm unsure.

[–] hades@lemm.ee 3 points 3 weeks ago

It's O(n).

If you look at each of the levels of all reports, you will access it a constant number of times: at most twice in each call to EvaluateLineSafe, and you will call EvaluateLineSafe at most six times for each report.

[–] Gobbel2000@programming.dev 2 points 3 weeks ago

It really depends on what your parameter n is. If the only relevant size is the number of records (let's say that is n), then this solution takes time in O(n), because it loops over records only once at a time. This ignores the length of records by considering it constant.

If we also consider the maximum length of records (let's call it m), then your solution, and most others I've seen in this thread, has a time complexity in O(n * m^2) for part 2.