Smallest Common Multiple - Intermediate Algorithm Scripting - Free Code Camp

[nicholasrobbins]

Honestly this one totally stumped me and even looking at the solutions really made me scratch my head, but you explained everything perfectly and helped a lot. Thank you dude, you're really helping people understand some tough problems in an easy to follow way and that's no easy feat with some of these problems! 🙌

[aaronargottelopez]

I need to learn a lot from you. I spent around 80 hours in this problem and I couldn`t find the answer. It was the same thing Where art thou. Thanks for teaching.

[adamknuds]

two months ago i went through all the challenges with your videos as a guide and it was very helpful. Thanks usefulprogrammer! I'm going through them again and came up with a solution using every:

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

let incrementer = 1
while((fullArr.every(ele => (incrementer % ele) === 0)) === false){
incrementer++
}
return incrementer
}

[skatepanama]

Thank you very much for the videos they are very useful, here is my solution from another point of view:

function smallestCommons(arr) {
let higherN = Math.max(...arr); //gets the higher num in the array
let lowerN = Math.min(...arr); //same as above but lower num
let nums = []

for (let i = lowerN; i <= higherN; i++) { //create the array with all the numbers to verify
nums.push(i)
}

let divisor = higherN //create the variable to increment by the higher number

while (true) {
if (nums.every(item => divisor % item == 0)) { //if the divisor that is supposed to be always a higher number mod, verifying it with all the nums in the array
return divisor //and everyone returns true we found the solution and return it!
} else {
divisor += higherN //if not, divisor is incremented by the higher number, which always its a mod
}
}
}


smallestCommons([1, 5]);

[bassmw]

Thank you. This solution really opened my mind to a new way of solving problems with arrays. Setting the array length equal to the count, I found to be quite eye opening and ingenious.

[sarvarkhalimov]

Seems like not only I was stuck in this challenge)) when I reached the stage to find smallest Common Multiple after lining up sequence and creating an array, the rest was really strenuous. Tried many ways... and got lost))
I appreciate your work, thanks for tutorial

[pokebeaches]

You're a saint for making these videos brother. Thank you

[jovannikolic]

function smallestCommons(arr) {
if (arr[0] > arr[1]) {
[arr[0], arr[1]] = [arr[1], arr[0]];
}
console.log(arr);
}

smallestCommons([5, 1]); // logs [1, 5]

I find this way of rearranging array items very easy when we have 2 items like in this case.
Thanks for the tutorials. They're awesome!

[brianfitzsimmonsable]

Nice to see your thought process. A good example of thinking through the problem. I had got the part about proving all "true" myself but was getting "head fuzz" by then.

[ShubhamMishra-mzzw]

It's been more than 3 days, i was stuck with this problem and finally have been able to solve this.Thank you.

[quaidbergo]

Very nice explanation of the problem and your solution. Simple but with room to optimise and learn.

[julianstorm]

Thanks mate. Been struggling to follow other explanations. Seeing the steps in the log is extremely helpful. Was a bit confused about the math part. Great work man.

[koddezign]

function smallestCommons(arr) {
const min = Math.min(...arr)
const max = Math.max(...arr)
let accumulator = max
for (let i = max; i >= min; i--){
if(accumulator % i != 0){
accumulator += max;
i = max;
}}
return accumulator;
}

Is the smallest version that works that I've seen. After spending a few hours on this I was surprised to find this on the comments of the solution. Surprisingly simple

[mindrust]

That was no fckin way Intermediate Level. That was brootal

[aaaamontenegro]

yeah i always check the hints, just to try to understand them, i think is very usefull

[cristiancarrillocastillo]

It took me a few days, but I got it on my own.. Man the relief just but I'm just here looking at how other people did it.

[JService]

Thanks again for the video. For anyone wanting to use this code to pass all hypothetical tests, instead of using a large number in the while loop you can create a variable x=true and change it to x=false once the counter reaches the max value (arr.length).

[theMiaow]

instead of multiple < or whatever upperbound

just do
while (multiple)
if you found the answer set the multiple into undefined 🤯

[RusydyMuhiddin]

thank you very much! it's really helping me to understand this problem.

[zken]

Tricky lesson Mr. Ian but thanks for not only explaining it in a way I can understood but thanks for moving slowly with the lesson that I can see what I did correct and what I did not do correct. Here is a easier way to get the correct answer as well down below:
function smallestCommons(arr) {
let max = Math.max(...arr);
let min = Math.min(...arr);
let sol = max;
for (let i = max - 1; i>=min; i --) {
if (sol % i !== 0) {
sol += max;
i = max;
}
}
return sol;
}
console.log(smallestCommons([23, 18])); and your answer will then come to be 6056820.