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O of what now (sh.itjust.works)
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[-] xmunk@sh.itjust.works 21 points 3 months ago

Acshually, in the context of O(N^2) N can be seen to constantly be equal to N and thus, as a constant, we can ignore it in our O analysis.

Yes, my bubble sort does run in O(1)

[-] 0x0@lemmy.dbzer0.com 6 points 3 months ago

Get out of my office

[-] Alienmonkey@lemm.ee 4 points 3 months ago* (last edited 3 months ago)

Bubble sort? This wizard talk shall not pass.

My god, some of us can't read past select queries and v-lookup ruins.

On a Friday no less.

this post was submitted on 02 Aug 2024
366 points (97.4% liked)

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