No, n³ cannot be O(n²) as otherwise that would mean that there exists a positive constant K and a positive threshold m such that for any integer n greater than m you would have n³ less than K*n², which would be the same as saying n less than K, which cannot hold for any integer n greater than m. So n³ cannot be an O(n²), which means that something that is an O(n³) is not necessarily an O(n²).
It’s the other way around, if something is an O(n²) then it is necessarily also an O(n³).
Yes. The other answer is technically correct, but yours is pragmatically correct.
If a solution is worse than O(nln(n))* then most of us are going to be looking for a pragmatic and completely alternate way to deal with it, rather than analyzing how to make it mildly less terrible.
So I’m just writing O(n^2) as a quick professional replacement for my original write in answer of “dogshit”.
isnt O(n³) usually simplified to O(n²) anyway ?
No, n³ cannot be O(n²) as otherwise that would mean that there exists a positive constant K and a positive threshold m such that for any integer n greater than m you would have n³ less than K*n², which would be the same as saying n less than K, which cannot hold for any integer n greater than m. So n³ cannot be an O(n²), which means that something that is an O(n³) is not necessarily an O(n²).
It’s the other way around, if something is an O(n²) then it is necessarily also an O(n³).
ok thanks
Yes. The other answer is technically correct, but yours is pragmatically correct.
If a solution is worse than O(nln(n))* then most of us are going to be looking for a pragmatic and completely alternate way to deal with it, rather than analyzing how to make it mildly less terrible.
So I’m just writing O(n^2) as a quick professional replacement for my original write in answer of “dogshit”.