pub fn y15d01p2(input: &str) -> Option<u32>Expand description
The solution for part two of the day one challenge.
This solution is extremely similar to the part one solution y15d01p1
with the exception that after each character we do a check to see if we’re
on a negative floor, and if stop and return the current index.
§Example
let input = "(()))((\n"; // probably read this from the input file...
assert_eq!(y15d01p2(input), Some(5));