advent_of_code
adventofcode
advent_of_code | adventofcode | |
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4 | 86 | |
1 | 16 | |
- | - | |
8.4 | 7.3 | |
4 months ago | 4 months ago | |
Rust | TypeScript | |
MIT License | - |
Stars - the number of stars that a project has on GitHub. Growth - month over month growth in stars.
Activity is a relative number indicating how actively a project is being developed. Recent commits have higher weight than older ones.
For example, an activity of 9.0 indicates that a project is amongst the top 10% of the most actively developed projects that we are tracking.
advent_of_code
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[2022][Friendly Reminder] Don't commit your input files to Git
Caution: the method I used is not recommended. Use at your own risk, know what you're doing, and have plenty of backups, etc. That said, here are my notes.
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-🎄- 2022 Day 9 Solutions -🎄-
Once I figured out how to elegantly move the tail using signum() in part 2 it became so much cleaner. Full code is on GitHub, but here's the core of it:
- -🎄- 2022 Day 6 Solutions -🎄-
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-🎄- 2021 Day 6 Solutions -🎄-
Group the fish by their timer, which can be anything from 0 to 8. The neat part is the rotate_left to decrease the timers of each group. Fish with timer 0 will automatically end up at timer 8, and so the only thing left to do is adding another copy with timer 6. Core function: rust fn multiply(timers: Vec, generations: usize) -> usize { // Index equals timer value, so index 0 contains the count of fish with timer 0 let mut counts_by_timer = vec![0usize; 9]; timers.into_iter().for_each(|f| { counts_by_timer[f] += 1; }); for _ in 0..generations { // A left rotation represents the timer (=index) decreasing by 1. // The fish with timer 0 will not only produce new fish with timer 8, // but also reset their timer to 6 let count_of_fish_with_timer_0 = counts_by_timer[0]; counts_by_timer.rotate_left(1); counts_by_timer[6] += count_of_fish_with_timer_0; // == counts_by_timer[8] } counts_by_timer.into_iter().sum() } Full code at Github
adventofcode
- -❄️- 2023 Day 10 Solutions -❄️-
- -❄️- 2023 Day 9 Solutions -❄️-
- -❄️- 2023 Day 8 Solutions -❄️-
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-❄️- 2023 Day 7 Solutions -❄️-
The only code differs for two parts is as follows - https://github.com/bhosale-ajay/adventofcode/blob/master/2023/ts/D07.test.ts - under 80 lines.
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-❄️- 2023 Day 6 Solutions -❄️-
Easy day TypeScript, P1 - Brute Force, P2 - Formula
- -❄️- 2023 Day 5 Solutions -❄️-
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-❄️- 2023 Day 4 Solutions -❄️-
[LANGUAGE: TypeScript] Github - Under 40 lines, all parts running under 8ms.
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-❄️- 2023 Day 1 Solutions -❄️-
[LANGUAGE: TypeScript] TypeScript - Running under 30ms (both parts)
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-🎄- 2022 Day 25 Solutions -🎄-
F# This year I solved puzzles using TypeScript as well as F# - Day 18, 19, and 22 TBD
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-🎄- 2022 Day 24 Solutions -🎄-
F# - After looking at some Python solution calculated the position of blizzard for nth time instead of maintaining the grid, which makes it easier with F#.
What are some alternatives?
adventofcode - adventofcode.com solutions
AoC - my personal repo for the advent of code yearly challenge
adventofcode - Advent of Code 2022 as part of getting back into Python https://adventofcode.com/
adventofcode - Advent of Code solutions of 2015, 2016, 2017, 2018, 2019, 2020, 2021, 2022 and 2023 in Scala
adventofcode - Advent of Code challenge solutions
advent-of-code-go - All 8 years of adventofcode.com solutions in Go/Golang; 2015 2016 2017 2018 2019 2020 2021 2022
Advent-of-Code - Advent of Code
advent-of-code
aoc2021 - Advent of Code 2021 on my homemade 16-bit CPU SCAMP
rockstar - The Rockstar programming language specification
nom - Rust parser combinator framework
advent_of_code - Solutions to programming puzzles on Advent of Code