**ps. look out for all my other solutions for Advent of Code challenges here.**

## Day 20

See details of the challenge here.

Today’s input looks like this:

1873924193–1879728099

2042754084–2076891856

4112318198–4113899685

1039794493–1057170966

3791841434–3797494664

1518668516–1518748952

1946127596–1953926346

4058215215–4086224696

3429681642–3455096313

2599576643–2604275147

1800210010–1801990849

1761160149–1766904471

2774395403–2774748831

1520470679–1542287000

2343327790–2346083217…

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)

- after processing (0, 1888888) the overall range is 0–1888888
- (1888889, 1904062) forms a continuous range after 1888888, so the overall range is now 0–1904062
- (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

**solution where**

*O(n)**is the no. of blacklist ranges.*

**n**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**