freeCodeCamp Intermediate Algorithm Scripting: Smallest Common Multiple

smallestCommons.js

// Find the smallest common multiple of the provided parameters 
// that can be evenly divided by both, as well as by all sequential 
// numbers in the range between these parameters.

// The range will be an array of two numbers that will not necessarily 
// be in numerical order.

// For example, if given 1 and 3, find the smallest common multiple of 
// both 1 and 3 that is also evenly divisible by all numbers between 1 
// and 3. The answer here would be 6.

function smallestCommons(arr) {

  // function to calculate gcd (greatest common divisor)
  function gcd(a, b) {
    let d=0;
    while ((a % 2 == 0) && (b % 2 == 0)) {
      a /= 2;
      b /= 2;
      d++;
    }
    while (a != b) {
      if (a % 2 == 0) { a /= 2; }
      else if (b % 2 == 0) { b /= 2; }
      else if (a > b) { a = (a-b)/2; }
      else { b = (b-a)/2; }
    }
    let result=a*Math.pow(2,d);
    return result;
  }

  // we will get min value and max, in case array is inverse order
  let max=Math.max(...arr);
  let min=Math.min(...arr);

  // create array with all the number from min to max
  let newArr=[];
  for (let i=min; i<=max; i++) {
    newArr.push(i);
  }

  // now, we can calculate the smallest common multiple (scm) using
  // formula scm = (a * b)/gcm(a,b);
  // we use reduce, since we can calculte the scm in any order

  return newArr.reduce((acum, item) => { 
    acum=(acum*item)/gcd(acum, item);
    return acum;
  });
}

smallestCommons([1, 5]);
smallestCommons([5, 1]);
smallestCommons([2, 10]);
smallestCommons([1, 13]);
smallestCommons([23, 18]);