Day 4: Giant Squid
Challenge
You're already almost 1.5km (almost a mile) below the surface of the ocean, already so deep that you can't see any sunlight. What you can see, however, is a giant squid that has attached itself to the outside of your submarine.
Maybe it wants to play bingo?
Bingo is played on a set of boards each consisting of a 5x5 grid of numbers. Numbers are chosen at random, and the chosen number is marked on all boards on which it appears. (Numbers may not appear on all boards.) If all numbers in any row or any column of a board are marked, that board wins. (Diagonals don't count.)
The submarine has a bingo subsystem to help passengers (currently, you and the giant squid) pass the time. It automatically generates a random order in which to draw numbers and a random set of boards (your puzzle input). For example:
7,4,9,5,11,17,23,2,0,14,21,24,10,16,13,6,15,25,12,22,18,20,8,19,3,26,1
22 13 17 11 0
8 2 23 4 24
21 9 14 16 7
6 10 3 18 5
1 12 20 15 19
3 15 0 2 22
9 18 13 17 5
19 8 7 25 23
20 11 10 24 4
14 21 16 12 6
14 21 17 24 4
10 16 15 9 19
18 8 23 26 20
22 11 13 6 5
2 0 12 3 7
After the first five numbers are drawn (
7
, 4
, 9
, 5
, and 11
), there are no winners, but the boards are marked as follows (shown here adjacent to each other to save space):
22 13 17 11 0 3 15 0 2 22 14 21 17 24 4
8 2 23 4 24 9 18 13 17 5 10 16 15 9 19
21 9 14 16 7 19 8 7 25 23 18 8 23 26 20
6 10 3 18 5 20 11 10 24 4 22 11 13 6 5
1 12 20 15 19 14 21 16 12 6 2 0 12 3 7
After the next six numbers are drawn (
17
, 23
, 2
, 0
, 14
, and 21
), there are still no winners:
22 13 17 11 0 3 15 0 2 22 14 21 17 24 4
8 2 23 4 24 9 18 13 17 5 10 16 15 9 19
21 9 14 16 7 19 8 7 25 23 18 8 23 26 20
6 10 3 18 5 20 11 10 24 4 22 11 13 6 5
1 12 20 15 19 14 21 16 12 6 2 0 12 3 7
Finally,
24
is drawn:
22 13 17 11 0 3 15 0 2 22 14 21 17 24 4
8 2 23 4 24 9 18 13 17 5 10 16 15 9 19
21 9 14 16 7 19 8 7 25 23 18 8 23 26 20
6 10 3 18 5 20 11 10 24 4 22 11 13 6 5
1 12 20 15 19 14 21 16 12 6 2 0 12 3 7
At this point, the third board wins because it has at least one complete row or column of marked numbers (in this case, the entire top row is marked:
14 21 17 24 4
).
The score of the winning board can now be calculated. Start by finding the sum of all unmarked numbers on that board; in this case, the sum is
188
. Then, multiply that sum by the number that was just called when the board won, 24
, to get the final score, 188 * 24 =
4512
.
To guarantee victory against the giant squid, figure out which board will win first. What will your final score be if you choose that board?
🔗 Part Two
On the other hand, it might be wise to try a different strategy: let the giant squid win.
You aren't sure how many bingo boards a giant squid could play at once, so rather than waste time counting its arms, the safe thing to do is to figure out which board will win last and choose that one. That way, no matter which boards it picks, it will win for sure.
In the above example, the second board is the last to win, which happens after
13
is eventually called and its middle column is completely marked. If you were to keep playing until this point, the second board would have a sum of unmarked numbers equal to 148
for a final score of 148 * 13 =
1924
.
Figure out which board will win last. Once it wins, what would its final score be?
🔗 Part A
We'll start off by reading in the numbers and boards, using the
paragraphs
function from the
Utilities module.
import { paragraphs } from "./util";
const input = paragraphs();
const numbers = input[0]
.split(",")
.map((s) => parseInt(s));
const boards = input.slice(1).map((str) => {
return str.split("\n").map((line) =>
line
.trim()
.split(/\s+/)
.map((s) => {
return parseInt(s);
})
);
});
Our logic will be as follows:

Get the next number in the list

Loop through the boards

Enter the number in the board

If the board is now winning, return the result

outer: for (
let i = 0;
i < numbers.length;
i++
) {
const called = numbers[i];
for (let j = 0; j < boards.length; j++) {
const board = boards[j];
if (enterNumber(board, called)) {
console.log(
called *
board
.reduce(
(flattened, row) =>
flattened.concat(row),
[]
)
.reduce((sum, i) => sum + i, 0)
);
break outer;
}
}
}
This requires a new
enterNumber
function which, given a board and a number:

Checks if the board contains the number

If it does

Check if the placed number leads to a completed row or column

Return
true


Otherwise, continue looping

Return
false
at the end
function enterNumber(
board: number[][],
number: number
) {
for (let i = 0; i < board.length; i++) {
for (
let j = 0;
j < board[i].length;
j++
) {
if (board[i][j] === number) {
board[i][j] = 0;
return (
board[i].every(
(square) => square === 0
) 
board.every((row) => row[j] === 0)
);
}
}
}
return false;
}
🔗 Part B
On the other hand, it might be wise to try a different strategy: let the giant squid win.
You aren't sure how many bingo boards a giant squid could play at once, so rather than waste time counting its arms, the safe thing to do is to figure out which board will win last and choose that one. That way, no matter which boards it picks, it will win for sure.In the above example, the second board is the last to win, which happens after13
is eventually called and its middle column is completely marked. If you were to keep playing until this point, the second board would have a sum of unmarked numbers equal to148
for a final score of148 * 13 =
1924
.Figure out which board will win last. Once it wins, what would its final score be?
We'll reuse much of our code, since entering and checking bingo card wins is the same.
Since we only want the last winning card, we'll keep track of a list of winning cards.
const winningBoards: Set<number> = new Set(
[]
);
Now, instead of logging and breaking as soon as we have a winning board, we'll add it to
winningBoards
. When that set is full (the size of the boards array) we've finished our last board and can print the result.
for (let j = 0; j < boards.length; j++) {
if (winningBoards.has(j)) {
continue;
}
const board = boards[j];
if (enterNumber(board, called)) {
winningBoards.add(j);
if (
winningBoards.size === boards.length
) {
// print result
}
}
}