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O(TREE(n))
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)
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.
Get out of my office
O(fuck)
O(unnecessary zoom meeting)
We will also accept “I’ll file a ticket under the tech debt epic” (which will never get picked up unless/until it causes a prod outage/SLA violation)
It’s not like anyone cares once you have the job anyway
Usually the most straightforward solution is good enough. And when you want to improve the performance, it’s rarely about time complexity.
at job interview
“ah sure, I’ll solve it in n log(n) for you with an obscure algorithm because n² is too slow for your 1000 customers, even though there’s no perceived difference for n<10⁷.”
O(n!n!)
It works really well, until n=3, which takes a while. Don’t ask about n=4.
