Add comments since I really like my day 6 solution

This commit is contained in:
Daniel Flanagan 2021-12-06 09:46:49 -06:00
parent 5339b86631
commit 6761e4725d
Signed by: lytedev
GPG key ID: 5B2020A0F9921EF4

View file

@ -6,18 +6,33 @@ const CYCLE_DAYS = 7
const MATURING_DAYS = 2
proc lanternFish(inputLines: seq[string], days = DEFAULT_DAYS): int =
# a place to put newly-spawned fish that are not mature enough themselves to
# spawn within the cycle
var maturingFish: seq[(int, int)]
# here we will keep the counts of fish that spawn on a given day in the cycle
var fish = newSeq[int](CYCLE_DAYS)
for i in inputLines[0].split(',').mapIt(it.parseInt()):
inc fish[i]
for day in 0..days+1:
let daySeq = day mod CYCLE_DAYS
maturingFish.add((MATURING_DAYS+1, fish[daySeq]))
# initialize our current fish counts based on the input
for i in inputLines[0].split(',').mapIt(it.parseInt()): inc fish[i]
for day in 0..days:
let cycleDay = day mod CYCLE_DAYS
# since we're just about to decrement the days for all maturing fish, add
# 1 to offset and spawn our maturing fish
maturingFish.add((MATURING_DAYS+1, fish[cycleDay]))
# age maturing fish
for v in maturingFish.mitems(): dec v[0]
# if our oldest maturing fish have now matured move them to their place in
# the cycle
if maturingFish[0][0] == 0:
fish[daySeq] += maturingFish[0][1]
fish[cycleDay] += maturingFish[0][1]
maturingFish.delete(0)
result = fish.foldl(a + b)
# sum all fish counts plus the fish that were spawned two days ago
fish.foldl(a + b) + (if maturingFish[0][0] <= 1: maturingFish[0][1] else: 0)
let input = 6.loadInput()
time("day 6 part 1"): echo input.lanternFish()