3 ms·
Python looks excruciatingly slow to me. If you want fast I believe you need to think and write in vector languages like kdb+/q. I am not a kdb+ expert by any me
by anonu 2y ago
Python looks excruciatingly slow to me. If you want fast I believe you need to think and write in vector languages like kdb+/q. I am not a kdb+ expert by any means and my code can probably use more q primitives, but here was my solution in ~2 minutes:
i1:("I I";" ")0: `:1.txt;
sum {abs last deltas x }each flip asc each i1 / answer 1
sum {x * sum x = i1[1]}each i1[0] / answer 2
- eddsolves 2y agoThat’s cool, but horribly ugly! What does production kdb+ look like? Is it similarly terse, maybe Perl-ish, or was this purely for speed? What does each flip asc do?
- anonu 2y agoi1 is Nx2. asc each i1 sorts each N in ascending order. flip transposes to 2xN.
- vjerancrnjak 2y agoEverything autocompletes in Python. from collections import * xys = list(map(int, open(0).read().split())) xs = xys[::2] ys = xys[1::2] xs.sort() ys.sort() print(sum(abs(x-y) for x,y in zip(xs,ys))) yc = Counter(ys) print(sum(((yc[x])*x for x in xs)))
- acters 2y agothis is the least amount of loc I can think of rn data = { i+1 : sorted([ x for x in list(map(int, open('input').read().split()))[i::2]]) for i in range(2) } total_distance = sum(list(map(lambda x: abs(x[0]-x[1]), zip(data[1], data[2])))) print("part 1:", total_distance) similarity_score = sum(list(map(lambda x: (x*data[2].count(x))*data[1].count(x), set(data[1]).intersection(data[2])))) print("part 2:", similarity_score)