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tim-peters | 4 years ago
The point: for inputs like your:
a = list(range(-100, 1700)) b = list(range(1400, 1800))
it would first use a variant of binary search to find where b[0] belongs in a. "Almost all of a" is consumed by that via a relatively tiny number of compares. It would similarly do a binary-ish search to find where a[-1] belongs in b. The tail end of b, from 1700 through 1799, then also never needs to be looked at again.
In the lucky case that two input lists don't overlap at all, and a[-1] <= b[0], that's enough so that a "one-pair-at-a-time" compare _never_ needs to be done. In context (the two lists are contiguous slices of an array), no pointers at all need to be copied in that case either - it deduces that the combined array slices are already in sorted order in total time O(log(len(a)) + log(len(b))), and it's done.
adamgordonbell|4 years ago
Edit: Ok, I get it now. Don't just pop, binary search into it, so you can fast forward through. I'm not sure I'll get around to implementing that, at least not in the immediate future, but that makes total sense.
tim-peters|4 years ago