Day 6 - Tuning Trouble

See Day 6 for a detailed description of the problem.

Continuing to solve the Advent of Code 2022 problems (see Advent of Code - Day 1).

Links:

To run the example code in this post save the code into file such as advent.jactl and take your input from the Advent of Code site (e.g. advent.txt) and run it like this:

$ cat advent.txt | java -jar jactl-2.1.0.jar advent.jactl

Part 1

Pretty easy today. We get a line of characters and have to find the position of the character after the first occurrence of 4 unique characters in a row.

We just iterate through the string looking at each 4 character substring. We sort the substring, and then use unique() to eliminate duplicates. If we still have 4 characters we know that they are unique.

def n = 4
def line = nextLine();
line.size()
    .skip(n)
    .filter{ line.substring(it - n,it).sort().unique().size() == n }
    .limit(1)[0]

Note that limit(1) returns a list. In this case the list will have a single element so we extract it using limit(1)[0].

Part 2

For part 2 the only difference is that now the number of unique characters has to be 14:

def n = 14
def line = nextLine();
line.size()
    .skip(n)
    .filter{ line.substring(it - n,it).sort().unique().size() == n }
    .limit(1)[0]