Simple and elegant

Continuing the discussion from freeCodeCamp Challenge Guide: Smallest Common Multiple:

Found this from the discussion section submitted by a user.

function smallestCommons(arr) {

  var max = Math.max(arr[0], arr[1]);
  var min = Math.min(arr[0], arr[1]);
  var mltple = max;

  for(var i = max; i >= min; i--){
    if(mltple % i !== 0){
      mltple += max; 
      i = max;
    } 
  }

  return mltple;  
}

Need help to explain this please…

Thanks~

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what do you need explained?
what do you understand, what does confuse you?

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sorry but someone can explain me the procces, i understand the min and max metodh but the if inside the for loop dont understand it,if someone can explain me thanks;
i know that this solution is basic and readable but well i want know why/how give the result…

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I had a similar thought process for this problem, but couldn’t figure how to write it up until I saw his code (I was trying to use a while loop in the for loop, etc what a mess).

I will break it down how I understood it using smallestCommons ([1,3])
So the min = 1 & max = 3
He has set his variable mltple equal to max (or 3 in this case)

The for loop is created to go 3 --> 2 --> 1 then break

The if statement condition essentially says that the remainder of mltple divided by i does not equal 0 and if that is true then execute the below code which is max is added to mltple and i is reset to max (or 3). If the if statement is false, where the remainder does = 0, then you will decrease i by 1 and run through the loop again.

So starting with i = 3
3 % 3 !== 0 is false --> move to i =2
3 % 2 !== 0 is true --> 3 + 3, i = 3
6 % 3 !==0 is false --> move to i =2
6 % 2 !==0 is false --> move to i = 1
6 % 1 !==0 is false --> for loop breaks because of initial condition

then the code says to return mltple, which is 6 at the time of the for loop breaking.

As you can see, 6 is divisible by all the numbers in the range of 1 to 3. This is why you have to work incrementally from greatest to smallest.

Hope that helps!

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it goes through the multiples of max (represented by mltple). if the multiple is not evenly divisible by max to min (represented by i), it goes to the next multiple (mltple += max) and resets i (i = max). when mltple is evenly divisible by max to min, the program ends (return mltple).

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