Solution is working by i have a question

JavaScript
1

Hello guys, here is my solution to this problem.

It is working as intended, but i have a question about optimization.

In this piece of code

let res = Math.max(…arr) * 10;

Instead of starting from 1 and loop everything until variable res is divisible by all numbers in the array i started this number from math.max(…arr) * 2 and everything worked.

To make it faster i then started changing the multiplication so it starts at the highest number we can while everything still working.

Past Math.max(…arr) * 10 for example Math.max(…arr) * 11 it wont work and i understand why, but i want to understand how could i find this out this starting value without trial and error, im sure there is some mathematical formula here which I would want to understand it.

Thanks in advance.

Your code so far


function smallestCommons(arr) {

// complete the array with all numbers between the numbers in the paramenters
for (let i = Math.min(...arr) + 1; i < Math.max(...arr); i++) {
  arr.push(i);
}

// Assign a starting value to start the loop
let res = Math.max(...arr) * 10;

//Check if all numbers in the array are divisible by result
let isDivisible = () => arr.every(n => res % n === 0);

// Increment result in case all numbers in the array are not divisable by result
do {
  res++;
} while(isDivisible() !== true)

return res;
}


smallestCommons([10,2]);

Link to the challenge:

2

There is a formula for calculating least common multiple (lcm) of two numbers.

lcm(a, b) = a * b / gcd(a,b )
where gcd is the greatest common denominator (gcd) of the two numbers.

using Euclidian’s Algorithm, you can calculate the gcd

See: https://www.calculator.net/lcm-calculator.html , https://www.calculatorsoup.com/calculators/math/gcf.php ,
and https://codility.com/media/train/10-Gcd.pdf (though the last one give you more information than you need to know. Look at the Euclidian by division formula and ignore the proofs.)

1 Like
3

Thank you, exactly what i was looking for.

Will look more into this.

Best regards,