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O of what now (sh.itjust.works)
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[-] xmunk@sh.itjust.works 21 points 2 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)

[-] Alienmonkey@lemm.ee 4 points 2 months ago* (last edited 2 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.

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this post was submitted on 02 Aug 2024
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