advent-of-code VS adventofcode

Pretty happy with this one ``` def solve() -> List[int]: seen = [] won = [] scores = [] for n in NUMS: seen.append(n) for board in BOARDS: transpose = list(zip(*board)) for i, line in enumerate(board): if (all(num in seen for num in line) or all(num in seen for num in transpose[i])) and board not in won: won.append(board) scores.append(sum(sum(num for num in line if num not in seen) for line in board) * seen[-1]) return scores

day02 of doing only one-liners and I already failed for the second part, it would have been a huge mess. The first part is pretty messy as well so if anyone can improve on it I would be very grateful ! https://github.com/masmeert/advent-of-code/blob/master/2021/day02/main.py

I actually did the same as you when I first solved it But I saw other people use the l[3] > l[n-3] trick, it didn't make much sense at first but now I get it and it's quite clever. Basically take [1 2 1 3], the windows would be w1=[1 2 1] and w2=[2 1 3], as 2 and 1 appear in both windows what really matters is w1[0] and w2[2] which are unique to each: 1+(2+1) < (2+1)+3 == 1 < 3.

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.

Easy day TypeScript, P1 - Brute Force, P2 - Formula

[LANGUAGE: TypeScript] Github - Under 40 lines, all parts running under 8ms.

[LANGUAGE: TypeScript] TypeScript - Running under 30ms (both parts)

F# This year I solved puzzles using TypeScript as well as F# - Day 18, 19, and 22 TBD

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#.