this post was submitted on 02 Dec 2023
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Advent Of Code
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An unofficial home for the advent of code community on programming.dev!
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.
AoC 2024
Solution Threads
| M | T | W | T | F | S | S |
|---|---|---|---|---|---|---|
| 1 | ||||||
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 9 | 10 | 11 | 12 | 13 | 14 | 15 |
| 16 | 17 | 18 | 19 | 20 | 21 | 22 |
| 23 | 24 | 25 |
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- Keep all content related to advent of code in some way
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console.log('Hello World')
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My solution was worse than most: replace one -> one1one You are only going to do the replace all for each number and if the "e" is also in eight it is still there for the next set of replace.
A better quick and dirty solution from Mastodon was to just add the common character first: twone -> twoone
That's an esthetic abomination, but very clever
I have a friend who says that "whatever works is elegant" and solutions like OP's is why I simultaneously love and hate that phrase.
Behold, elegance:
digits = { "one": [1,2], "two": [2,2], "three": [3,4], "four": [4,3], "five": [5,3], "six": [6,2], "seven": [7,4], "eight": [8,4], "nine": [9,3], "1": [1,1], "2": [2,1], "3": [3,1], "4": [4,1], "5": [5,1], "6": [6,1], "7": [7,1], "8": [8,1], "9": [9,1] }and what comes afterwards is even more elegant, for it works!