Add comments since I really like my day 6 solution

2021/six.nim

@@ -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]
+  # initialize our current fish counts based on the input
-  for day in 0..days+1:
+  for i in inputLines[0].split(',').mapIt(it.parseInt()): inc fish[i]
-    let daySeq = day mod CYCLE_DAYS
+
-    maturingFish.add((MATURING_DAYS+1, fish[daySeq]))
+  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()