Getting the time complexity of a selection sort

Question

I created a code for selection sort, but my teacher asked me what the time complexity of my code is, so I need help to get it. I'm not sure if my code is the same with the other selection sort with a worst time case of O(n^2) and a best time case of O(n).

code:

def selection(collection):
for endnum in range(len(collection)-1, 0, -1):
    print(collection)
    max_idx = endnum
    if max(collection[0:endnum]) > collection[endnum]:
        max_idx = collection.index(max(collection[0:endnum]))
    collection[endnum], collection[max_idx] = collection[max_idx], collection[endnum]

Answer 1

Selection sort doesn't have a best case. It's always O(n²), because each step needs to find the largest (or smallest) element in the unsorted portion of the array, which requires scanning the entire unsorted segment.

Your version is not different except that you rather unnecessarily compute the maximum twice and then do a third scan to find its index. However, doing three times as much work as is necessary is "just" a constant factor, so the asymptotic complexity doesn't change. The cycles you waste are real, though.