chore: codacy onboarding round 2

2023/aoc2023/day07/README.md

@@ -144,8 +144,8 @@ And that's our answer! On to Part 2
 
 The addition of Jokers as the `J` value has two implications:
 
-- The `hand_type` classification of a set of cards will change based on the number of Jokers present.
-- The sort key generated by `card_sort_str` needs to move `J` values *down* to sort hands correctly.
+-   The `hand_type` classification of a set of cards will change based on the number of Jokers present.
+-   The sort key generated by `card_sort_str` needs to move `J` values *down* to sort hands correctly.
 
 The latter is a very simple change, moving `J` down to the first position and shuffling others around:
 
@@ -192,9 +192,9 @@ simplified.sort();
 
 In this example, `simplified` will be `[1, 3]`. Considering how to score the 1 Joker, there are three possible outcomes:
 
-- As a distinct card, `[1, 1, 3]`, creating "Three of a Kind" (type 4);
-- Combined with the `T`, `[2, 3]`, creating "Full House" (type 5); OR
-- Combined with the `5`s, `[1, 4]`, creating "Four of a Kind" (type 6).
+-   As a distinct card, `[1, 1, 3]`, creating "Three of a Kind" (type 4);
+-   Combined with the `T`, `[2, 3]`, creating "Full House" (type 5); OR
+-   Combined with the `5`s, `[1, 4]`, creating "Four of a Kind" (type 6).
 
 Type 6 is the strongest hand possible, obtained by adding the Joker (`1`) to the highest count of cards in the set (`3`).
 

2023/aoc2023/day09/README.md

@@ -6,10 +6,10 @@ Link: <https://adventofcode.com/2023/day/9>
 
 I realized pretty quick on this one that a recursive solution would be easiest to achieve:
 
-- Take a set of numbers as an input
-- If those numbers are all 0, return 0
-- Otherwise, produce the next set of numbers as a Vector by iterating through the input over the range of indices `0..length-1` (one less than the final index), pushing `nums[i+1] - nums[i]` into the new set.
-- Finally, recurse with this new set of numbers, adding the result of the recursion to `nums[len-1]` (the final number in the input).
+-   Take a set of numbers as an input
+-   If those numbers are all 0, return 0
+-   Otherwise, produce the next set of numbers as a Vector by iterating through the input over the range of indices `0..length-1` (one less than the final index), pushing `nums[i+1] - nums[i]` into the new set.
+-   Finally, recurse with this new set of numbers, adding the result of the recursion to `nums[len-1]` (the final number in the input).
 
 Do this for every line of input, total things up, and we have an answer.