> uses fixed space I guess the "normal" way to write this is with a heap? For `min`, I guess the algorithm is something like, min(items, k=1): h = max_h

> uses fixed space I guess the "normal" way to write this is with a heap? For `min`, I guess the algorithm is something like, min(items, k=1): h = max_heap(items[:k]) for i in items[k:]: h.push(i) h.pop_max() return sorted(h) It isn't asymptotically slower than the normal algorithm for k=1, it doesn't need to make multiple passes over the data, and it doesn't mutate the original data structure, so it's a pretty sensible and natural extension to put in.