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Day 2: Red-Nosed Reports

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[-] Rin@lemm.ee 1 points 2 weeks ago

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 2 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 2 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.

this post was submitted on 02 Dec 2024
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