Advent of Code 2023 in JavaScript
It's been a few years since I did Advent of Code, the annual "Advent Calendar" of programming challenges. The other times I did it I had fun, but just sorta got distracted after the first week or two -- no doubt that'll happen this year, too! Until that happens, though, I'll update this post with solutions as I go.
(I had to disable copilot while solving these to prevent it from solving them for me!)
⚠️SPOILERS AHEAD⚠️ if you want to solve these yourself, don't read any further!
General Notes
The format of the puzzles largely follows the pattern:
- Take newline-separated input of a bunch of strings
- Apply some sort of conversion to each row
- Apply some sort of cumulative function to the value from each row
The base functionality to accomplish that looks like this:
/** @param {string} input */
function solvePuzzle(input) {
return input
.trim() // Ensure no trailing newlines
.split(/[\r\n]+/g) // Convert to array of strings
.reduce((sum, row) => {
const converted = someFunction(row); // Puzzle-specific
return sum + converted; // If we're doing a cumulative sum
}, 0);
So you'll see that pattern throughout the solutions, where the actual puzzle-specific logic is inside the reducer function.
Day 1: Trebuchet?!
Day 1 Part 1
The data for this puzzle is described like this:
On each line, the calibration value can be found by combining the first digit and the last digit (in that order) to form a single two-digit number.
With the examples:
1abc2 # 12
pqr3stu8vwx # 38
a1b2c3d4e5f # 15
treb7uchet # 77
The samples show that there can be more than 2 numbers, so we can't just grab the numbers and be done with it. The 4th sample shows that there can be exactly one number. So if we were to get all numbers, in order, as an array, we can just grab the first and last entry (which might be the same!).
/** @param {string} input */
function solveDay1Part1(input) {
return input
.trim() // Ensure no trailing newline
.split(/[\r\n]+/g)
.reduce((sum, row) => {
const allNums = row.match(/\d/g);
const num = +(allNums[0] + allNums.at(-1));
return sum + num;
}, 0);
}
Day 1 Part 2
For part 2, the puzzle uses the same input but adds another layer of detail:
It looks like some of the digits are actually spelled out with letters: one, two, three, four, five, six, seven, eight, and nine also count as valid "digits".
They provide the examples:
two1nine # 29
eightwothree # 83
abcone2threexyz # 13
xtwone3four # 24
4nineeightseven2 # 42
zoneight234 # 14
7pqrstsixteen # 76
Note that the spelled-out numbers can share a letter, so it's a bit more complicated than just directly using a regex like /\d|one|two|three)/, since one match can block an overlapping one. But we can do basically that, just grabbing one match at a time and keeping track of the index position.
To do that I used a "sticky" regex which, when used with .execute, checks for a match starting exactly at pattern.lastIndex.
/** @param {string} input */
function solveDay1Part2(input) {
const numNames = [
'\\d',
'one', // The index position of the number's name is also its numeric value!
'two',
'three',
'four',
'five',
'six',
'seven',
'eight',
'nine',
];
const pattern = new RegExp(numNames.join('|'), 'y');
return input
.trim() // Ensure no trailing newline
.split(/[\r\n]+/g)
.reduce((sum, row) => {
/** @type {string[]} */
const allNums = [];
for (let i = 0; i < row.length; i++) {
pattern.lastIndex = i;
const match = pattern.exec(row);
if (!match) continue;
const namedValue = numNames.findIndex((name) => name === match[0]);
allNums.push(namedValue > 0 ? `${namedValue}` : match[0]);
}
const num = +(allNums[0] + allNums.at(-1));
return sum + num;
}, 0);
}
Day 2: Cube Conundrum
The data for this puzzle is comes from the results of a "Game" that works like this:
- A bag has an unknown number of red, green, and blue cubes in it.
- Each Game is done with a different number of cubes in said bag
- For each Game, we'll get several random samples of cubes (the selected cubes are put back before the next sample)
Games are documented like this:
Game 1: 3 blue, 4 red; 1 red, 2 green, 6 blue; 2 green
Game 2: 1 blue, 2 green; 3 green, 4 blue, 1 red; 1 green, 1 blue
Game 3: 8 green, 6 blue, 20 red; 5 blue, 4 red, 13 green; 5 green, 1 red
Game 4: 1 green, 3 red, 6 blue; 3 green, 6 red; 3 green, 15 blue, 14 red
Game 5: 6 red, 1 blue, 3 green; 2 blue, 1 red, 2 green
Where the semi-colons separate each sample. Here's the function I'm using to parse this input into usable data for the two puzzles:
/**
* @param {string} input
* @returns {{id:number, samples: {red:number, green: number, blue: number}[]}[]}
*/
function parseDay2Input(input) {
const rowPattern = /Game (?<id>\d+): (?<allSamples>.*)/;
const sampleColorPattern = /(?<count>\d+) (?<color>red|green|blue)/;
return input
.trim()
.split(/[\r\n]+/g)
.map((row) => {
const { id, allSamples } = row.match(rowPattern).groups;
const samples = allSamples.split(/\s*;\s*/g).map((sample) =>
sample.split(/\s*,\s*/g).reduce((cleaned, valueString) => {
const { count, color } = valueString.match(sampleColorPattern).groups;
cleaned[color] = +count;
return cleaned;
}, {}),
);
return { id: Number(id), samples };
});
}
That parser results in data that looks like this:
[
{
"id": 1, // The Game ID
"samples": [
{ "green": 1, "blue": 1, "red": 1 },
{ "green": 3, "blue": 1, "red": 1 },
{ "green": 4, "blue": 3, "red": 1 },
{ "green": 4, "blue": 2, "red": 1 },
{ "blue": 3, "green": 3 }
]
},
// ...
]
Day 2 Part 1
For Part 1, our goal is to:
- Identify which games would have been possible if the bag contained:
- 12 red cubes
- 13 green cubes
- 14 blue cubes
- Sum the IDs of all possible games to get the puzzle solution.
To solve this, for each game we need to use the samples to get the max number of each color and make sure that maximum value isn't greater than the target values.
/** @param {string} input */
function solveDay2Part1(input) {
const maxCounts = {
red: 12,
green: 13,
blue: 14,
};
return parseDay2Input(input).reduce((sum, game) => {
const isPossible = game.samples.every((sample) =>
// Is every color <= the max allowed value?
['red', 'green', 'blue'].every(
(color) => !sample[color] || sample[color] <= maxCounts[color],
),
);
return isPossible ? sum + game.id : sum;
}, 0);
}
Day 2 Part 2
Part 2 uses the same dataset but asks a different question, which boils down to:
- Get the maximum
rgbvalues for each Game - Multiply those values together to get the "Power" for that game
- Add up the Powers for all games to get the puzzle solution
/** @param {string} input */
function solveDay2Part2(input) {
return parseDay2Input(input).reduce((sum, game) => {
const maxes = {
red: 0,
green: 0,
blue: 0,
};
// Update the max values
game.samples.forEach((sample) =>
['red', 'green', 'blue'].forEach(
(color) => (maxes[color] = Math.max(sample[color] || 0, maxes[color])),
),
);
const power = maxes.red * maxes.green * maxes.blue;
return sum + power;
}, 0);
}
Day 3: Gear Ratios
For this puzzle, we have a matrix of cells, where each cell is a numeric or symbol character. We need to be able to identify numbers, which can span multiple columns. For each cell, the puzzle parts require that we be able to ask a question about the 9 adjacent cells.
The provided example is this:
467..114..
...*......
..35..633.
......#...
617*......
.....+.58.
..592.....
......755.
...$.*....
.664.598..
Here's the parser function I used for both parts of this puzzle:
/**
* @typedef Day3Num
* @prop {number} startCol
* @prop {number} endCol
* @prop {number} value
*
* @typedef Day3Cell
* @prop {number} col
* @prop {string} value
* @prop {boolean} isSymbol
* @prop {boolean} isGear
* @prop {Day3Num} [num]
*/
/**
* @param {string} input
* @returns {Day3Cell[][]}
*/
function parseDay3Input(input) {
return input
.trim()
.split(/[\r\n]+/g)
.map((row) => {
const chars = row.split('');
/** @type {Day3Cell[]} */
const cells = [];
for (let col = 0; col < chars.length; col++) {
const value = chars[col];
const isNum = /\d/.test(value);
/** @type {Day3Num|undefined} */
let num;
if (isNum && cells[col - 1]?.num) {
// This this is part of the last number and thus
// already dealt with!
num = cells[col - 1].num;
} else if (isNum) {
// This is the start of a new number
let numStr = value;
num = {
startCol: col,
endCol: col,
value: 0,
};
for (let j = col + 1; j < chars.length; j++) {
if (/\d/.test(chars[j])) {
numStr += chars[j];
num.endCol++;
} else break;
}
num.value = Number(numStr);
}
cells.push({
col,
value,
isSymbol: /[^\d.]/.test(value),
isGear: value === '*',
num,
});
}
return cells;
});
}
This parser returns an array of rows. Each row is an array of cells. For each cell, if that cell is part of a number from the grid, it refers to a single object instance representing that number. That way we can check to make sure we aren't doing something with a number more than once!
The parsed data for a row looks like this:
[
// ... (first 38 columns)
{
"col": 39,
"value": "3", // second digit of num.value
"isSymbol": false,
"isGear": false,
"num": {
"startCol": 38,
"endCol": 40,
"value": 835
}
},
{
"col": 40,
"value": "5", // third digit of num.value
"isSymbol": false,
"isGear": false,
"num": { // This object is the SAME ONE in the prior cell!
"startCol": 38,
"endCol": 40,
"value": 835
}
},
{
"col": 41,
"value": "*",
"isSymbol": true,
"isGear": true
}
]
Day 3 Part 1
For this puzzle, we need to add up the numbers in the grid, but only those numbers that are adjacent to a "symbol" (non-numeric, non-. character).
To do that, I looped through every cell in the parsed data. For each cell, I looped through all of its neighbors. If the central cell was part of a number, and the cell I was checking was a symbol, I added it to the set of numbers to add up. This is where I took advantage of representing each parsed number with a common reference for each cell it spanned, since I could skip numbers I'd already added to the set!
/** @param {string} input */
function solveDay3Part1(input) {
const rows = parseDay3Input(input);
console.log(JSON.stringify(rows[2].slice(39, 42), null, 2));
/** @type {number[]} */
const nums = [];
/** @type {Set<Day3Num>} */
const alreadyAdded = new Set();
for (let r = 0; r < rows.length; r++) {
const row = rows[r];
for (let col = 0; col < row.length; col++) {
const cell = row[col];
if (!cell.num || alreadyAdded.has(cell.num)) continue;
// Look at all 9 spots around this cell and,
// if one of them has a symbol add its num!
outer: for (let i = -1; i <= 1; i++) {
for (let j = -1; j <= 1; j++) {
if (i === 0 && j === 0) continue; // skip self
const otherCellRow = r + i;
const otherCellCol = col + j;
const otherCell = rows[otherCellRow]?.[otherCellCol];
if (!otherCell || !otherCell.isSymbol) continue;
alreadyAdded.add(cell.num);
nums.push(cell.num.value);
break outer;
}
}
}
}
return nums.reduce((sum, num) => sum + num, 0);
}
Day 3 Part 2
For this variation of the puzzle, instead of looking at cells adjacent to numbers we're tasked to look at cells around * symbols. If that symbol is adjacent to exactly two numbers, we multiply those two numbers together to get the gear ratio. The solution is the sum of all gear ratios.
/** @param {string} input */
function solveDay3Part2(input) {
const rows = parseDay3Input(input);
/** @type {number[][]} */
const gearNums = [];
for (let r = 0; r < rows.length; r++) {
const row = rows[r];
for (let col = 0; col < row.length; col++) {
const cell = row[col];
if (!cell.isGear) continue;
// Look at all 9 spots around this cell and collect
// the unique numbers we find.
/** @type {Set<Day3Num>} */
const foundNums = new Set();
for (let i = -1; i <= 1; i++) {
for (let j = -1; j <= 1; j++) {
if (i === 0 && j === 0) continue; // skip self
const otherCellRow = r + i;
const otherCellCol = col + j;
const otherCell = rows[otherCellRow]?.[otherCellCol];
if (!otherCell || !otherCell.num) continue;
foundNums.add(otherCell.num);
}
}
if (foundNums.size !== 2) continue;
gearNums.push([...foundNums].map((num) => num.value));
}
}
return gearNums.reduce((sum, [a, b]) => {
const ratio = a * b;
return sum + ratio;
}, 0);
}