It's 500! ⭐

Merry Christmas, everyone! 🎄🎅 🎁

On the last day of this fantastic adventure, we have to match keys and locks in order to access the Chief Historian's office.

Our input consists of grids (big comeback on the last day 😛) like these:

#####
.####
.####
.####
.#.#.
.#...
.....

.....
#....
#....
#...#
#.#.#
#.###
#####

The difference is that locks have their top row filled with #; keys have the bottom row filled.

Essentially, the goal is to find unique lock/key pairs that fit together without overlapping in any column.

The puzzle's description already gives us the key hint (pun unintended) to solve this: for each lock and key we can calculate the pin heights, that is the number of # in each column excluding the top (if it's a lock) or bottom (if it's a key) row.

For example, this lock has heights 0,5,3,4,3:

#####
.####
.####
.####
.#.#.
.#...
.....

And this key has heights 5,0,2,1,3:

.....
#....
#....
#...#
#.#.#
#.###
#####

This is of course the reasoning behind my main function here:

function calcHeightPerPin(key: Map<string, string>) {
    const array: number[] = [];
    for (let x = 0; x < 5; x++) {
        const height =
            [...key.entries()].filter(
                ([cell, val]) => cell.x === x && val === '#'
            ).length - 1;
        array.push(height);
    }
    return array;
}

After this, we can simply loop through all grids split between locks and keys, compare each pin height and see if the sum is ≤ 5 (as there can be gaps between a lock's and a key's pins, it doesn't have to be a perfect match):

function findFittingLocksAndKeys(keyGrids: Grid[], lockGrids: Grid[]) {
    let count = 0;
    const keys: number[][] = [];
    const locks: number[][] = [];

    for (const grid of keyGrids) {
        keys.push(calcHeightPerPin(grid));
    }

    for (const grid of lockGrids) {
        locks.push(calcHeightPerPin(grid));
    }

    for (const key of keys) {
        for (const lock of locks) {
            let fits = true;
            for (let i = 0; i < 5; i++) {
                if (key[i] + lock[i] > 5) {
                    fits = false;
                    break;
                }
            }
            if (fits) count++;
        }
    }

    return count;
}

And this is it! As per tradition, we can obtain the 50th star for free as long as we have the previous 49 in store - so clearly no Part 2 puzzle in this article! 😀

I hope you enjoyed this journey as much as I did: I'm especially pleased because, having completed all the previous years as well, I now belong to the 500-stars exclusive club! 🤩

That said, I'm off to enjoy this Christmas Day… see you next year!