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🤶 - 2025 DAY 8 SOLUTIONS - 🤶
(programming.dev)
An unofficial home for the advent of code community on programming.dev! Other challenges are also welcome!
Advent of Code is an annual Advent calendar of small programming puzzles for a variety of skill sets and skill levels that can be solved in any programming language you like.
Everybody Codes is another collection of programming puzzles with seasonal events.
Solution Threads
| M | T | W | T | F | S | S |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 8 | 9 | 10 | 11 | 12 |
Icon base by Lorc under CC BY 3.0 with modifications to add a gradient
console.log('Hello World')
(Browser-based) Javascript
My initial approach was to consider each junction as a circuit, and then merge the closest circuits. Took me way to long to realize the problem statement for part 1 made this unworkable (as you need to count redundant connections as "attempts"). Thankfully part 2 doesn't care about how many connections you make, so I was able to reuse that code to solve part 2.
To solve part 1 "the right way", I first thought I had to store a circuit as a collection of pairs of junctions (literally, the collection of connections). Oh god was that messy code; 3 layers of nested for-loops and it still wasn't getting close to the answer. I eventually realized I could reference the junctions' indices in the input to simplify things, allowing me to manipulate simple sets of ints. Combine with pre-calculating the distances once before starting the connecting/merging and you end up with a surprisingly simple and snappy algorithm.
Given I realized these optimizations after restarting part 1, my solution for part 2 doesn't take advantage of any of them and takes an eye-watering 22 seconds to terminate on average. It probably re-computes more distances than it computes new ones, for each iteration...
Code