Add comments since I really like my day 6 solution
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2021/six.nim
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2021/six.nim
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@ -6,18 +6,33 @@ const CYCLE_DAYS = 7
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const MATURING_DAYS = 2
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const MATURING_DAYS = 2
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proc lanternFish(inputLines: seq[string], days = DEFAULT_DAYS): int =
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proc lanternFish(inputLines: seq[string], days = DEFAULT_DAYS): int =
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# a place to put newly-spawned fish that are not mature enough themselves to
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# spawn within the cycle
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var maturingFish: seq[(int, int)]
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var maturingFish: seq[(int, int)]
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# here we will keep the counts of fish that spawn on a given day in the cycle
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var fish = newSeq[int](CYCLE_DAYS)
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var fish = newSeq[int](CYCLE_DAYS)
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for i in inputLines[0].split(',').mapIt(it.parseInt()):
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inc fish[i]
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# initialize our current fish counts based on the input
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for day in 0..days+1:
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for i in inputLines[0].split(',').mapIt(it.parseInt()): inc fish[i]
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let daySeq = day mod CYCLE_DAYS
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maturingFish.add((MATURING_DAYS+1, fish[daySeq]))
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for day in 0..days:
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let cycleDay = day mod CYCLE_DAYS
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# since we're just about to decrement the days for all maturing fish, add
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# 1 to offset and spawn our maturing fish
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maturingFish.add((MATURING_DAYS+1, fish[cycleDay]))
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# age maturing fish
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for v in maturingFish.mitems(): dec v[0]
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for v in maturingFish.mitems(): dec v[0]
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# if our oldest maturing fish have now matured move them to their place in
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# the cycle
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if maturingFish[0][0] == 0:
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if maturingFish[0][0] == 0:
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fish[daySeq] += maturingFish[0][1]
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fish[cycleDay] += maturingFish[0][1]
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maturingFish.delete(0)
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maturingFish.delete(0)
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result = fish.foldl(a + b)
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# sum all fish counts plus the fish that were spawned two days ago
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fish.foldl(a + b) + (if maturingFish[0][0] <= 1: maturingFish[0][1] else: 0)
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let input = 6.loadInput()
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let input = 6.loadInput()
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time("day 6 part 1"): echo input.lanternFish()
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time("day 6 part 1"): echo input.lanternFish()
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