freeCodeCamp: Intermediate Algorithm Scripting: Smallest Common Multiple

solution.js

function smallestCommons(arr) {
  let firstNum = arr[0],
      secondNum = arr[1],
      primeFactors = [[],[]], //save the prime factors here
      divider = 2, // start diving by two
      counts = [{},{}]; //count how many times each prime factor appears

  //get the firsNum prime factors
  while (firstNum > 1) {
    if (firstNum % divider === 0) {
      primeFactors[0].push(divider);
      firstNum = firstNum / divider;
      divider = 2;
    } else {
      divider++
    }
  }

  //get the secondNum prime factors
  while (secondNum > 1) {
    if (secondNum % divider === 0) {
      primeFactors[1].push(divider);
      secondNum = secondNum / divider;
      divider = 2;
    } else {
      divider++
    }
  }

  //get the counts of each prime factor
  for (let i = 0; i < arr.length; i++) {
    for (let j = 0; j < primeFactors[i].length; j++) {
      let num = primeFactors[i][j];
      counts[i][num] = counts[i][num] ? counts[i][num] + 1 : 1;
    } 
  }

  //merge both counts. when there are common numbers, drop the smaller one
  function extend(obj, src) {
    //iterate between each src key
    Object.keys(src).forEach(function(key) {
      // if obj doesn't have this property
      if (!obj.hasOwnProperty(key)) {
        // add the key: value
        obj[key] = src[key];
      } else {
        // if obj property value is bigger than src property value, saves the former, else saves the latter
        obj[key] = obj[key] > src[key] ? obj[key] : src[key];
      }
	  });
    return obj;
  }
  let multiples = extend(counts[0], counts[1]);

  //calculate the smallest common multiple
  let scm = 1;
  Object.keys(multiples).forEach(function(key) {
    scm *= Math.pow(key, multiples[key]);	
  });
  console.log(scm);

  // iterate between the arr range
  let arrRange = [];
  for (let i = arr[0]; i <= arr[1]; i++) {
    arrRange.push(i);
  }

  //divide the scm by each value in the range, if it cannot be evently divided, multiply by two, and keep until it gets the right result
  let value = arr[0],
      stop = false;
  while (value <= arr[1]) {
    if (scm % value === 0) {
      if (value === arr[1]) {
        console.log(scm);
      }
      console.log(value);
      value++;
    } else {
      value = arr[0];
      scm = scm * 2;
    }
  }

  return arr;
}


smallestCommons([2, 10]);