Solving day 3 of Advent of Code 2021 in JavaScript

Advent of Code is a website that releases a new programming puzzle every day between the 1st and the 25th of December.

0:00 Part 1
10:20 Part 2

Have fun!

[danmiller]

Thanks for the video! It's nice to see a better answer. I got there but I kind of cheated and ran my for loop manually until I saw one more value for the CO2 Rating. I did it all in one for loop for each bit, and I couldn't exit the loop without stopping short since the oxygen rating took more bits to filter to one value.

[xShosTaK]

Just dropping by again to say thanks for the super helpful video. Being new to coding, a lot of these things you're doing are way over my head but I feel like I'm learning with each day by watching your videos (after failing trying myself haha).

[felipemelendez]

I'm glad I found this video; I got stuck on part two and couldn'g figure out what was wrong... that last part was tricky, thanks! Also, great explanation!

[bikefixe]

After writing some spaghetti code in part 2, I started over and realized that the recursive function could do two things: generate a gamma or epsilon rate for a set of candidate numbers (as we did in part 1), then filter the list of candidates. If candidates.length !=1, loop again, now generating a -new- gamma or epsilon based on the filtered set of numbers, then filtering again ...

[lordkim]

A slightly more short ver of part 1 (w/o counting both 0s and 1s)

const len = lines.length,
{ gammaRate, epsilonRate } = lines.reduce((res, line) => {
[...line].forEach((n, i) => res[i] += n == 1);

return res;
}, Array(lines[0].length).fill(0)).reduce((res, oneBitCounter) => {
oneBitCounter > len - oneBitCounter
? (res.gammaRate += 1, res.epsilonRate += 0)
: (res.gammaRate += 0, res.epsilonRate += 1);

return res;
}, { gammaRate: "", epsilonRate: "" });

[fdafdsaful]

Is there a more efficient way(time complexity) for the first part than doing a nested for loop? Thanks for the video