Advent of Code 2023 - December 15th
Rob van der Leek
Posted on December 15, 2023
In this series, I'll share my progress with the 2023 version of Advent of Code.
Check the first post for a short intro to this series.
You can also follow my progress on GitHub.
December 15th
The puzzle of day 15 was surprisingly easy. The first part was so trivial it made me nervous for the complexity of the second part. But also part two was really straightforward. A nice start of the weekend 😄
My pitfall for this puzzle: Not really, this day was way easier than previous days.
Solution here, do not click if you want to solve the puzzle first yourself
#!/usr/bin/env python3
import re
with open('input.txt') as infile:
lines = infile.readlines()
parts = lines[0].strip().split(',')
def calc_hash(l):
result = 0
for c in l:
result += ord(c)
result = result * 17
result = result % 256
return result
boxes = []
for _ in range(256):
boxes.append([])
for p in parts:
label = re.split('[=-]', p)[0]
box = calc_hash(label)
if p[-1] == '-':
boxes[box] = [b for b in boxes[box] if not b.startswith(label)]
else:
focal_length = p[-1]
for idx, lens in enumerate(boxes[box]):
if lens.startswith(label):
boxes[box][idx] = f'{label} {focal_length}'
break
else:
boxes[box].append(f'{label} {focal_length}')
total = 0
for idx_b, b in enumerate(boxes):
for idx_l, lens in enumerate(b):
total += (idx_b + 1) * (idx_l + 1) * int(lens[-1])
print(total)
That's it! See you again tomorrow!
Posted on December 15, 2023
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