<http://cantrip.org/sortfast.html>
Anyway I thought it was interesting.
vector.push(x)
Most of the time, it's O(1). Sometimes it's O(n). If you double the size of the backing array when it runs out of space, it's amortized O(1).Often such a grid will be quite well abstracted from its data source - it might be executing a remote query to return data in the new order every time - but I bet there are some examples out there that are backed by a local dataset and carry out an actual sort operation when you hit the header... and fall into a quicksort worst case if the user clicks the same header twice in a row.
It can be n*n for properly-sorted arrays too, depending on pivot selection (for randomized pivot selection, it's n log n).
But I'm pretty sure data sorted in either direction (i.e., 'reversed' or not, ascending or descending), and taking a pivot from either end, is n^2. It doesn't have to be reversed; you always end up with everything unsorted ending up on one side or the other of the pivot, with each recursive step being just one less comparison than the step prior, meaning it has N-1 + N-2 + ... + 1 comparisons regardless of which way the array is sorted, or N(N-1)/2 comparisons total (Gauss' formula but starting at one less than the total number of items N, since that's the number of comparisons each step), which is O(N^2). There is no cause where it's linear time, unless you are first iterating across the array to select the first pivot that is out of place (which may be a reasonable optimization, but can also be made to apply regardless of what direction the array is sorted).
From: https://stackoverflow.com/a/185576/368409
/* This is O(scary), but seems quick enough in practice. */
But, yes. (naive) Quicksort's amortized complexity being O(nlogn), but its O(n^2) on already sorted data, is all I ever needed to learn to take away that lesson. When sorting already sorted data is worse than sorting randomized data, it's a quick realization that "amortized cost" = "read the fine print".