Function aoc::y22d15::y22d15p1

pub fn y22d15p1(input: &str, row: i32) -> u32

The solution for part one of the day fifteen challenge.

This is a somewhat naive and brute force approach to solving the problem given the puzzle input as a string and an integer of the row that we want to compute. We start by parsing the puzzle input to extract the locations of the sensors and the beacon that they detect. Because a position that already contains a beacon can’t be a position that can’t contain a beacon we need to track any beacons that are on the row that we’re interested in so that we can substract them from the final total. We then need to find the Manhattan Distance and then we check each row (both above and below) from the sensor until the end of the Manhattan distance. At each step we reduce the range along the x-axis to account for being farther from the starting point. If we match the row that we’re interested in then we save the range for later. After we’ve processed all of the sensors then we can loop through each range and insert its x-coordinate into a tracking set. Finally, to get the answer to the challenge we return the total number of x-coordinates that we have minus the number of beacons on the row.

Example

// probably read the from the input file...
let input = concat!(
    "Sensor at x=2, y=18: closest beacon is at x=-2, y=15\n",
    "Sensor at x=9, y=16: closest beacon is at x=10, y=16\n",
    "Sensor at x=13, y=2: closest beacon is at x=15, y=3",
);
assert_eq!(y22d15p1(input, 15), 9);