JavaScript Challenge - Comparative Weights

Frequently Asked Questions
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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;

3

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!!

4

OMG I love this. The cube root! slaps forehead

Kudos!

5

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

6

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!

7

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

8

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

9

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);
}
10 
function scaleOfTruthN(n) {
  let times = 0;

  while(n > 1){
    n = Math.ceil(n/3);
    ++times;
  }
  return times;
}
1 Like
11

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);
}
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function scaleOfTruthN(n) {

  return Math.ceil( Math.log(n) / Math.log(3) );

}