Exercise 2 solution

    from collections import deque

    def last_n(lines, n):
        buf = deque(maxlen=n)
        for item in lines:
            buf.append(item)
        return list(buf)

    print(last_n(range(10), 3))
    print(last_n([1, 2], 5))
    print(last_n(range(10), 1))

Output:
    [7, 8, 9]
    [1, 2]
    [9]

The deque never holds more than n items: once full, each append drops the
oldest item from the other end. So this works on a huge or endless-looking
stream without storing it all.

list.pop(0) removes the first item, and every remaining item must then
shift one place left, so each removal costs work proportional to the length
of the list. Draining a long list this way repeats that shifting many
times over. deque.popleft() removes from the front without moving the
other items, so each removal is a constant amount of work.