Discover and read the best of Twitter Threads about #adventofcode2020
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For this year's #AdventOfCode, my goal is to get each day's solution in the most readable format that will fit in a tweet. Striving for a compromise between conciseness and (subjective) elegance… 🧵
#AdventOfCode Day 1 (in Python 🐍):
# (relies on a few input assumptions)
data = [int(i) for i in open('01')]
print("Part 1:",
next(x*y for x in data for y in data if x+y == 2020))
print("Part 2:",
next(x*y*z for x in data for y in data for z in data if x+y+z == 2020))
# (relies on a few input assumptions)
data = [int(i) for i in open('01')]
print("Part 1:",
next(x*y for x in data for y in data if x+y == 2020))
print("Part 2:",
next(x*y*z for x in data for y in data for z in data if x+y+z == 2020))
#AdventOfCode Day 2 in 🐍
import re
L = [re.split('[: \-]', l) for l in open('02')]
L = [(int(p1), int(p2), [c == ch for c in pwd])
for p1, p2, ch, _, pwd in L]
print("Part 1:",
sum(n1<= sum(m)<= n2 for n1, n2, m in L),
"Part 2:",
sum(m[p1-1]^m[p2-1] for p1, p2, m in L))
import re
L = [re.split('[: \-]', l) for l in open('02')]
L = [(int(p1), int(p2), [c == ch for c in pwd])
for p1, p2, ch, _, pwd in L]
print("Part 1:",
sum(n1<= sum(m)<= n2 for n1, n2, m in L),
"Part 2:",
sum(m[p1-1]^m[p2-1] for p1, p2, m in L))
[DAY2/24]
#ChristmasIsComing
Another topic, another discovery!
Let's play with #programming in a challenging way and with no pressure! Discover the #AdventOfCode initiative!
adventofcode.com
#adventofcode2020
#ChristmasIsComing
Another topic, another discovery!
Let's play with #programming in a challenging way and with no pressure! Discover the #AdventOfCode initiative!
adventofcode.com
#adventofcode2020
"After saving Christmas five years in a row, you've decided to take a vacation at a nice resort on a tropical island. Surely, Christmas will go on without you."
"The tropical island has its own currency and is entirely cash-only. The gold coins used there have a little picture of a starfish; the locals just call them stars."
