alt Hacker NewsSaying that l.pop(k) has time complexity O(n-k) implies that popping something at position 5 from the end has bounded (amortized) time cost regardless of the length of the list l, ie even if we let the list grow arbitrarily.
It's a stronger claim than just saying O(n), because in the latter case you wouldn't be able to conclude that popping something 5 from the end has bounded time as the list grows.
Complexity measures the worst-case, not the amortized case. If you want to report proportional cycles for more fine-grained per-feedback, fine, report proportional-cycles, but that's not Big-O, so don't use that notation.