Smallest Common Multiple - can somebody explain why my code doesn't work?

JavaScript
1 
function smallestCommons(arr) {
  let big = 1;
  let num=1;
  let small=0;
  let flag=0;
  let result=0;
  if (arr[0]>arr[1]) { 
    big=arr[0];
   small=arr[1];
  }
  else big=arr[1]; small=arr[0];
  console.log(small);
  for(let i =small; i<=big; i++) {
    num*=i;
  }
  console.log(num);
  for (let j=num; j>=small; j--){
    for(let i =small; i<=big; i++) {
      if(j%i==0) flag++;
    }
    if(flag===big) result=j;
    flag=0;
    console.log(result);
  }
  return result;
}

console.log(smallestCommons([2,4]));
2

I haven’t had time to fully review your code. BUT see this error in response to your second for loop

Screen Shot 2020-07-13 at 11.25.29 AM
Screen Shot 2020-07-13 at 11.25.29 AM754×84 13.8 KB

3

j is decreasing; i don’t understand why there would be an infinite loop

let j=num; j>=small; j--
4

I added some comments to your code.

function smallestCommons(arr) {
  let big=1;
  let num=1;
  let small=0;
  let flag=0;
  let result=0;

  // Let's add some comments here
  // Find biggest and smallest in range
  if (arr[0]>arr[1]) { 
    big=arr[0];
    small=arr[1];
  } else {
    // If you don't put {} around this, only the first will be guarded by the else.
    big=arr[1];
    small=arr[0];
  }
  // Log smallest value
  console.log(small);

  // Find product of all numbers in the range
  for (let i=small; i<=big; i++) {
    num*=i;
  }
  // Log product
  console.log(num);

  // Find smallest common multiple
  // Start with product and decrement
  // Note: why not start small and stop when you find the first common multiple? This way will take a *very* long time, especially when the range or numbers are big.
  // Also Note: Why decrement by 1? If j % big === 0, then you won't have (j - 1) % big === 0 unless big === 1. I'd decrement by a different (bigger) number than 1.
  for (let j=num; j>=small; j--) {
    // Flag if the number is divisible by a value in the range 
    for (let i=small; i<=big; i++) {
      if (j%i==0) {
        flag++;
      }
    }
    // Check if 'big' flags have been set
    // Note: Can you have 'big' flags set? In the example you could have a flag for 2, 3, and 4. But that is 3 flags and big === 4.
    if (flag===big) {
      result=j;
    }
    // Reset flag and log
    flag=0;
    console.log(result);
  }
  return result;
}

// Test the function
console.log(smallestCommons([2,4]));

Your code has some serious efficiency problems, which will trigger the infinite loop protections.

5

it works for the first three tests in the task; thank you though

6

I just fixed the issues I highlighted and with those fixes your code will pass for all tests. (Provided that you comment out the console.log()s.)