Function aoc::y15d08::y15d08p1

pub fn y15d08p1(input: &str) -> u32

The solution for part one of the day eight challenge.

Given the input string we initialize the total sums of each type and then loop over each string. The “code” representation we can find by simply calling the .len() function (and converting it to a u32). To calculate the “memory” representation we loop over all of the characters. If we have previously encountered the escape character then we check if we have an x which means a hexadecimal escape (and we need to skip the next two characters) or not in which case we just move on without incrementing the counter. If we have a skip value set then decrement it and move on without incrementing the counter. If we encounter the " we don’t increment the counter because it’s the start/end. Otherwise we increment the counter. If we’ve encountered the escape character then trigger the behavior described above for the next loop iteration. Finally, add the “memory” sum to the running total and then return the “code” sum minus the “memory” sum as requested by the prompt.

Example

// probably read this from the input file...
let input = "\"a\"b\"\n\"\"\n\"\x23\"";
assert_eq!(y15d08p1(input), 7);