Big O question

I need to find an "efficient" algorithm for a problem and the best I could figure out is something that for each element it looks at all the other elements (n^2), but as it goes through it has to look at 1 less elements. So for 5 elements it does 5 loops+4+3+2+1. Is this still considered n^2 or is it better?

Yeah, that's still O(n²). It takes 1/2 as long as n iterations that are O(n) individually, and you ignore the constant multiple.

O(bullshit) is just shit made up by jews to enforce their style of coding

>>1
That's O(Gauss(n))

https://en.wikipedia.org/wiki/Gauss#Anecdotes

>>3
Get the fuck out , self-hating Jew, you are not wanted in here!

Big O is used to SHOWTIME

The Big O was a pretty good anime, but not the best.

Looks like log to me.

>>8
n + (n−1) + (n−2) + ... + 2 + 1 =
    (n+1) + ((n−1) + 2) + ((n−2) + 3) + ... =
    (n+1) × n÷2 =
    (n² + n)÷2
It looks nothing like log.

>>3
You're a kindergarten level programmer.