Advent of Code F# – Day 20

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Day 20

See details of the challenge here.

Today’s input looks like this:


First, let’s read the input into a list of tuples representing the lower-upper bound ranges.

Notice that I’ve also gone through the trouble of sorting the input by the lower bounds. This way, as we iterate through the ranges we can build up an overal range where:

  • the lower bound is 0
  • the upper bound is the highest upper bound value we have processed so far

and where there are gaps between the current upper bound and the lower bound in the next range, those are the allowed IPs values.

eg. for my input the first 3 ranges are:

(0, 1888888)

(1888889, 1904062)

(1900859, 2087697)

  1. after processing (0, 1888888) the overall range is 0-1888888
  2. (1888889, 1904062) forms a continuous range after 1888888, so the overall range is now 0-1904062
  3. (1900859, 2087697) overlaps with the overall range on the lower bound, that’s ok, the overall range is now 0-2087697

no gaps were found in the 3 ranges so far.

Suppose the next range is (2087700, 2100000), then a gap would appear and we will have 2 allowed IPs: 2087698 and 2087699.

To put that into code, here’s a findAllowedIPs function that will process all the sorted list of ranges and yield all the allowed IPs as a lazy sequence. ps. for part 2 where we need to find the no. of allowed IPs this is a O(n) solution where n is the no. of blacklist ranges.

To solve part 1, we need the first allowed IP.

let part1 = findAllowedIPs input |> Seq.head


Part 2

How many IPs are allowed by the blacklist?

let part2 = findAllowedIPs input |> Seq.length



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