Day 01 - Not Quite Lisp

Language: Rust

Problem https://adventofcode.com/2015/day/1

Part 1: The input is a string of parentheses. ( means go up one floor, ) means go down one. Start at floor 0 and find the final floor after processing everything.

A fold over the characters works perfectly here:

let sum = input.chars().fold(0, |acc, c| {
    acc + match c {
        '(' => 1,
        ')' => -1,
        _ => 0,
    }
});

Part 2: Find the position of the first character that sends Santa into the basement, meaning the floor hits -1. I walk through the characters, update the floor, and return the index as soon as it drops:

for (i, c) in input.chars().enumerate() {
    match c {
        '(' => floor += 1,
        ')' => floor -= 1,
        _ => {}
    }
    if floor == -1 {
        return (i + 1).to_string();
    }
}

Position is 1-indexed so i + 1.

Solution: https://github.com/Elyrial/AdventOfCode/blob/main/src/solutions/year2015/day01.rs

Language: C

Problem https://adventofcode.com/2015/day/1

Part 1: The input is a string of parentheses. ( means go up one floor, ) means go down one. Start at floor 0 and find the final floor after processing everything.

A single pass with a pointer through the string does it:

while (*p) {
    if (*p == '(')
        floor++;
    else if (*p == ')')
        floor--;
    p++;
}

Part 2: Find the position of the first character that sends Santa into the basement, meaning the floor hits -1. Same loop, but check after each step and return the 1-indexed position:

while (*p) {
    if (*p == '(')
        floor++;
    else if (*p == ')')
        floor--;
    if (floor == -1)
        return (long long)(p - input) + 1;
    p++;
}

The pointer difference p - input gives the 0-based index, so adding 1 gives the 1-indexed position.

Solution: https://github.com/Elyrial/AdventOfCode_C/blob/main/src/solutions/year2015/day01.c