I solved the Smallest Common Multiple challenge, but it was too long?

JavaScript
1

I’m grateful that I solved the challenge, but I feel like I did it the long way. How can I improve my code?

function smallestCommons(arr) {
  let ran = []
  let sorted = arr.sort((a,b) => a - b)
  for (let i = sorted[0]; i <= sorted[1]; i++) {
    ran.push(i);
  }

  function primeFactors(n) {
    const factors = [];
    let divisor = 2;
  
    while (n >= 2) {
      if (n % divisor == 0) {
        factors.push(divisor);
        n = n / divisor;
      } else {
        divisor++;
      }
    }
    return factors;
  }

  let factors = []
  for (let i = 0; i < ran.length; i++) {
    if (ran[i] == 1) {
      factors.push([1])
    } else {
      factors.push(primeFactors(ran[i]))
    }
  }

  const countOccurrences = (arr, val) => arr.reduce((a, v) => (v === val ? a + 1 : a), 0)

  let final = []
  let count = [];
  for (let i = 1; i <= ran[ran.length - 1]; i++) {
    for (let j = 0; j < factors.length; j++) {
      if (countOccurrences(factors[j], i) > count) {
        count = countOccurrences(factors[j], i)
      }
    }
    final.push([i, count]);
    count = 0;
  }

  let realFinal = []
  for (let i = 0; i < final.length; i++) {
    for (let j = 0; j < final[i][1]; j++) {
      realFinal.push(final[i][0])
    }
  }

  let superFinal = realFinal.reduce((a,b) => a * b, 1)
  return superFinal;
}
2

The divisors approach is an interesting one. I’d first compare it to the one here:

The biggest difference between solution 4 and yours is that they keep less intermediate values than you do.

3

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