JavaScript Challenge - Comparative Weights

This community-built FAQ covers the “Comparative Weights” code challenge in JavaScript. You can find that challenge here, or pick any challenge you like from our list.

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function scaleOfTruthN(n) {

if (n<=1){
return 0
}
if(n <= 3 && n > 1){
return 1
}

let a = (Math.cbrt(n) - Math.floor(Math.cbrt(n)))===0? 0 : 1;
let times = Math.floor(Math.cbrt(n))+a;

return times

}

scaleOfTruthN(28);

// Leave this so we can test your code:
module.exports = scaleOfTruthN;

This was the best! Couldn’t figure out why I was only passing 4 out of 5 tests until I realised (ahem, read in a website that explains the answer) that if the number of balls is 0 or 1, 0 weighs will be required. I love that this challenge blended not only programming skills but maths problem solving skills.

My answer:

function scaleOfTruthN(n) { let i = 0; //let n = number; if (n === 1 || n === 0){ return 0; } do { if (n % 3 === 2){ n = (n + 1)/3; } else if (n % 3 === 1){ n = (n + 2)/3; } else { n = n/3; }; console.log(n); i++; } while (n > 1); return i; }; scaleOfTruthN(3); // Leave this so we can test your code: module.exports = scaleOfTruthN;

Woop!!

OMG I love this. The cube root! slaps forehead

Kudos!

i get a lot fun, I was trying with a loop before, but my loop was set not properly so I was returning more then one answer the first one was good but the rest no, and there was when I saw the cubic root + 1 was the answer, except in the numbers that have a exactly cubic root

Awesome. I had a total brain freeze. I remember asking myself “what’s the mathematical function for figuring out how many times you can divide something by 2 until you get 1?” I was googling factorials, planning to google, reverse factorial or something. Can’t believe I didn’t realise I was describing square rooting.

Anyway, we’re total bosses for completing the challenge. High five!

wouldn’t that be logarithm base 2, instead of square root?

lol, yes slaps forehead again
I’m not sure if that makes me feel better or worse

I solved by doing the log base 3, rounded up. With rounding to get rid of precision errors.

function scaleOfTruthN(n) {
  return Math.ceil(Math.round(Math.log(n) / Math.log(3) * 100) / 100);
}
function scaleOfTruthN(n) {
  let times = 0;

  while(n > 1){
    n = Math.ceil(n/3);
    ++times;
  }
  return times;
}
1 Like

2.5x slow, but with recursion :wink:

function scaleOfTruthN(n) {
  if (n === 0 || n === 1) return 0;
  if (n === 2) return 1;
  const rest = n % 3;
  const part = (n - rest) / 3;
  return 1 + scaleOfTruthN1(rest === 2 ? part + 1 : part + rest);
}