main.py

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Part 1:

— Day 1: Not Quite Lisp —

Santa was hoping for a white Christmas, but his weather machine’s “snow” function is powered by stars, and he’s fresh out! To save Christmas, he needs you to collect fifty stars by December 25th.

Collect stars by helping Santa solve puzzles. Two puzzles will be made available on each day in the Advent calendar; the second puzzle is unlocked when you complete the first. Each puzzle grants one star. Good luck!

Here’s an easy puzzle to warm you up.

Santa is trying to deliver presents in a large apartment building, but he can’t find the right floor - the directions he got are a little confusing. He starts on the ground floor (floor 0) and then follows the instructions one character at a time.

An opening parenthesis, (, means he should go up one floor, and a closing parenthesis, ), means he should go down one floor.

The apartment building is very tall, and the basement is very deep; he will never find the top or bottom floors.

from typing import Optional

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For example:

def test_part1() -> None:

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(()) and ()() both result in floor 0.

    assert part1("(())") == 0
    assert part1("()()") == 0

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((( and (()(()( both result in floor 3.

    assert part1("(((") == 3
    assert part1("(()(()(") == 3

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))((((( also results in floor 3.

    assert part1("))(((((") == 3

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()) and ))( both result in floor -1 (the first basement level).

    assert part1("())") == -1
    assert part1("))(") == -1

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))) and )())()) both result in floor -3.

    assert part1(")))") == -3
    assert part1(")())())") == -3

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To what floor do the instructions take Santa?

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Part 1 Solution:

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For part 1, I take a very simple approach.

Take a string composed of ( and ), and return the count of (‘s minus the count of )‘s.

def part1(input: str) -> int:

#

Part 2:

Now, given the same instructions, find the position of the first character that causes him to enter the basement (floor -1). The first character in the instructions has position 1, the second character has position 2, and so on.

    return input.count("(") - input.count(")")

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For example:

def test_part2() -> None:

#

) causes him to enter the basement at character position 1.

    assert part2(")") == 1

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()()) causes him to enter the basement at character position 5.

    assert part2("()())") == 5

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What is the position of the character that causes Santa to first enter the basement?

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Part 2 Solution:

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For part 2, I have to actually iterate through the string and keep track of how many parens of each type have been encountered. I think a small match statement fits very nicely here.

Take a string composed of ( and ), and return the index of the first character at which one more ) has been encountered than (‘s. Implicitly return None if this never occurs.

def part2(input: str) -> Optional[int]:

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    floor = 0
    for index, char in enumerate(input, start=1):
        match char:
            case "(":
                floor += 1
            case ")":
                floor -= 1

        if floor == -1:
            return index

    return None


if __name__ == "__main__":
    puzzle_input = open("input.txt").read()

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Print out part 1 solution

    print("Part 1:", part1(puzzle_input))

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Print out part 2 solution

    print("Part 2:", part2(puzzle_input))