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Calculating complexity of js alghortim

I'm trying to caluclate complexity of below method in Big O notation
function algorithm(n,m){
let result = [];
for (let i = 0; i < n.length; i++) {
const total = m.filter((x) => x === n[i]).length;
if (PrimalityTest(total)) {
result.push(n[i]);
}
}
return result;
};
function PrimalityTest(c){
if (c <= 1) {
return false;
} else if (c === 2) {
return true;
} else {
for (let i = 2; i * i <= c; i++) {
if (c % i === 0) {
return false;
}
}
return true;
}
}
So, firstly there is loop which have O(n) and then there is nested loop and primality test function so that means complexity of all is O(n * m * sqrt(c))?
Can you please confirm If my understanding is correct?

The loop for (let i = 0; i < n.length; i++) is executed n times. The function m.filter((x) => x === n[i]).length checks every element in m, so executes m-times. So we have an execution time of O(n*m).
Considering
if (PrimalityTest(total)) {
result.push(n[i]);
}
is executed n times because it is in the same loop as above. So at worst it is O(n*sqrt(c))
To sum it up: It is O(n*m)+O(n*sqrt(c)). Because O(n*m) surpasses O(n*sqrt(c)) we get as result: O(n*m).
Your solution would mean that the filter function integrates the PrimalityTest method.

Euler's Totient Function in JavaScript

I am trying to implement Euler's Totient Function (phi) in Javascript. So far this is what I have:
function phi(n) {
var result = n;
for (let i=2; i*i<=n; i++) {
if (n % i === 0) {
while (n % i === 0) {
n /= i;
result -= result / i;
}
}
}
if (n > 1) {
result -= result / n;
}
return result;
}
Unfortunately it all goes wrong when it comes up to multiples of 4. How do I improve this?

Inspired by https://www.geeksforgeeks.org/eulers-totient-function/
function phi(n) {
// return Greater Common Denominator of two given numbers
function gcd(a, b) {
if (a === 0) {
return b;
}
return gcd(b % a, a);
}
// init
var result = 1;
// walk through all integers up to n
for (let i = 2; i < n; i++) {
if (gcd(i, n) === 1) {
result++;
}
}
return result;
}

You should implement result = 1, then result ++ whenever you encounter a number coprime to the number you input. For that, you have to find the gcd function and that can be done with various methods, such as ArrayLists (like in Java) or with recursive functions.

Not the most eficient way, but rather straightforward:
function phi(n) {
let divArr = []; // this is an array for the common divisors of our n
let primeCount = 0; // this is a counter of divisors
for (let i = 0; i <= n - 1; i++) {
if (n % i === 0) {
divArr.push(i);
}
}
for (let j = n - 1; j > 0; j--) { //j is all potential coprimes
for (let k = divArr.length - 1; k >= 0; k--) { //we get the indeces of the divArr and thus we can loop through all the potentail divisors
//here we check if our potential coprimes are comprimes or not
//run possible coprimes through the list of divisors
if (j % divArr[k] === 0 && divArr[k] !== 1) { //if a potential coprime can be divided by any element of the array of n's divisors we break the arra's loop i. e. k and go the j++
break
} else if (j % divArr[k] !== 0) { //if a potential coprime j cannot be divided by any element of divArray then it's ok and we simply stick to the next k and waiting for 2 possible cases: either it will reach 1 and we primeCount++ or eventually be divided and then we break the loop
continue
} else if (divArr[k] === 1) { //if can be divided without a remainder, greatest common divisor is not zero so we should break the loop
primeCount++;
}
}
}
console.log(divArr, primeCount)
}

JavaScript (Find Minimum in Rotated Sorted Array)

I have a leetcode question that I tried to use my own method to solve it, but I got an error and I don't know what's wrong.
This is the topic:
This is my attempted solution:
var findMin = function(nums) {
if (nums.length === 0) return undefined;
var minEle = nums[0];
if (nums.length === 1) {
minEle = nums[0];
return minEle;
}
var start = 0;
var end = nums.length - 1;
if (nums[end] > nums[start]) {
end = Math.floor(end / 2);
} else {
start = Math.ceil(end / 2);
}
findMin(nums.slice(start, end + 1));
};
findMin([3,4,5,1,2]);
and the output is undefined.
I tried this too :
var findMin = function(nums) {
if (nums.length === 0) return undefined;
var minEle = nums[0];
if (nums.length === 1) {
minEle = nums[0];
return minEle;
}
var start = 0;
var end = nums.length - 1;
if (nums[end] > nums[start]) {
end = Math.floor(end / 2);
} else {
start = Math.ceil(end / 2);
}
findMin(nums.slice(start, end + 1));
return minEle;
};
findMin([3,4,5,1,2]);
And the output is 3.
I tried to debug it and I got this:
I don't understand why my recursion solution is not correct. Note that I tried to implement a solution with consideration of time complexity.

You are missing the last return statement when recursing.
return findMin(nums.slice(start, end + 1)); //added return here
var findMin = function(nums) {
if (nums.length === 0) return undefined;
var minEle = nums[0];
if (nums.length === 1) {
minEle = nums[0];
return minEle;
}
var start = 0;
var end = nums.length - 1;
if (nums[end] > nums[start]) {
end = Math.floor(end / 2);
} else {
start = Math.ceil(end / 2);
}
return findMin(nums.slice(start, end + 1));
};
console.log(findMin([3,4,5,1,2]));

If you want to find the smallest value in an array, just use Math.min.
Example:
console.log(Math.min.apply(null, [3,4,5,1,2]))

I don't think recursion is necessary for this one. At a high level we just need to go through each element and find the smallest. This could be done pretty simply with a for loop and a variable to store the smallest seen value.
var findMin = function(nums) {
if(nums.length === 0) {
return undefined;
}
if(nums.length === 1) {
return nums[0];
}
// initialize the smallest number placeholder to the first value
let smallestNum = nums[0];
for(let i = 1; i < nums.length; i++) {
if(smallestNum > nums[i]) {
smallestNum = nums[i];
}
}
return smallestNum;
}
This could be simplified with an array reducer, but the logic would be the same.
Another approach could be to assume the numbers will be ascending and once you find a number that is smaller return it. For example, in [3,4,5,1,2] it increases until we go from 5 to 1.
var findMin = function(nums) {
if(nums.length === 0) {
return undefined;
}
if(nums.length === 1) {
return nums[0];
}
for(let i = 1; i < nums.length; i++) {
if(nums[0] > nums[i]) {
return nums[i];
}
}
return nums[0];
}
In this case as soon as we find a number that is smaller than the first we break the loop. If we get all the way through the array we know the first number was the smallest.

I'm not at a rank to comment, so I have to post as an answer. It seems you're changing "end" and "start" in the if conditions, but then you're calling the function again and the function resets those variables to
var start = 0;
var end = nums.length - 1;
So whatever is changed in the second if/else statement, start will get reset to 0, and end will halve. But maybe I've missed something, so if this is irrelevant please ignore.

How to fix "Maximum call stack size exceeded" error with a function that needs to loop 1 million times

I'm doing a coding challenge where I need to first create a function that shows how many iterations of the Collatz conjecture it takes to get to 1, and then find the largest number of iterations for a number within 1 million.
This is the challenge: https://projecteuler.net/problem=14
And here's the code:
// Collatz Conjecture
function collatz(n) {
if (n <= 1) {
throw "the number needs to be greater than 1";
}
for(var i = 0; n != 1; i++) {
if (n % 2 == 0) {
n /= 2;
} else {
n = (n * 3) + 1;
}
}
return i;
}
// Greatest Collatz Within 1,000,000
function largestCollatz() {
var arr = [];
for(var i = 2; i <= 1000000; i++) {
arr.push(collatz(i));
}
return Math.max(...arr);
}
The collatz() function works as expected, but when I try to use largestCollatz(), it throws "Maximum call stack size reached.". I tried changing the 1 million to only a hundred thousand, and it works, but that doesn't satisfy the conditions of the challenge.

the problem is you use brute force algorithm that is inefficient.this is my solution to problem 14 from project Euler. it takes a few second to run. the key is you should save previous results in a dictionary so you don't have to compute those results again.:
#problem 14 project euler
import time
start=time.time()
has2={}
def collatz(x):
seq=[]
seq.append(x)
temp=x
while(temp>1):
if temp%2==0:
temp=int(temp/2)
if temp in has2:
seq+=has2[temp]
break
else:
seq.append(temp)
else:
temp=3*temp+1
if temp in has2:
seq+=has2[temp]
break
else:
seq.append(temp)
has2[x]=seq
return len(seq)
num=0
greatest=0
for i in range(1000000):
c=collatz(i)
if num<c:
num=c
greatest=i
print('{0} has {1} elements. calculation time ={2} seconds.'.format(greatest,num,time.time()-start))

Instead of spreading into Math.max, you can call Math.max on every iteration instead:
// Collatz Conjecture
function collatz(n) {
if (n <= 1) {
throw "the number needs to be greater than 1";
}
for (var i = 0; n != 1; i++) {
if (n % 2 == 0) {
n /= 2;
} else {
n = (n * 3) + 1;
}
}
return i;
}
// Greatest Collatz Within 1,000,000
function largestCollatz() {
var arr = [];
let maxSoFar = 0;
for (var i = 2; i <= 1000000; i++) {
maxSoFar = Math.max(maxSoFar, collatz(i));
}
return maxSoFar;
}
console.log(largestCollatz());
If you wanted to increase the performance of the code, you could create a lookup table for the number of iterations required for each number, and use that lookup table instead of brute force when possible.
const map = new Map();
function collatz(n) {
if (n <= 1) {
throw "the number needs to be greater than 1";
}
if (map.has(n)) {
return map.get(n);
}
const next = n % 2 === 0 ? n / 2 : (n * 3) + 1;
if (next === 1) {
return 1;
}
const result = 1 + collatz(next);
map.set(n, result);
return result;
}
// Greatest Collatz Within 1,000,000
function largestCollatz() {
var arr = [];
let maxSoFar = 0;
for (var i = 2; i <= 1000000; i++) {
maxSoFar = Math.max(maxSoFar, collatz(i));
}
return maxSoFar;
}
console.log(largestCollatz());

How to find the least common multiple of a range of numbers?

Given an array of two numbers, let them define the start and end of a range of numbers. For example, [2,6] means the range 2,3,4,5,6. I want to write javascript code to find the least common multiple for the range. My code below works for small ranges only, not something like [1,13] (which is the range 1,2,3,4,5,6,7,8,9,10,11,12,13), which causes a stack overflow. How can I efficiently find the least common multiple of a range?
function leastCommonMultiple(arr) {
var minn, max;
if ( arr[0] > arr[1] ) {
minn = arr[1];
max = arr[0];
} else {
minn = arr[0];
max = arr[1];
}
function repeatRecurse(min, max, scm) {
if ( scm % min === 0 && min < max ) {
return repeatRecurse(min+1, max, scm);
} else if ( scm % min !== 0 && min < max ) {
return repeatRecurse(minn, max, scm+max);
}
return scm;
}
return repeatRecurse(minn, max, max);
}

I think this gets the job done.
function leastCommonMultiple(min, max) {
function range(min, max) {
var arr = [];
for (var i = min; i <= max; i++) {
arr.push(i);
}
return arr;
}
function gcd(a, b) {
return !b ? a : gcd(b, a % b);
}
function lcm(a, b) {
return (a * b) / gcd(a, b);
}
var multiple = min;
range(min, max).forEach(function(n) {
multiple = lcm(multiple, n);
});
return multiple;
}
leastCommonMultiple(1, 13); // => 360360

function smallestCommons(arr) {
var max = Math.max(...arr);
var min = Math.min(...arr);
var candidate = max;
var smallestCommon = function(low, high) {
// inner function to use 'high' variable
function scm(l, h) {
if (h % l === 0) {
return h;
} else {
return scm(l, h + high);
}
}
return scm(low, high);
};
for (var i = min; i <= max; i += 1) {
candidate = smallestCommon(i, candidate);
}
return candidate;
}
smallestCommons([5, 1]); // should return 60
smallestCommons([1, 13]); // should return 360360
smallestCommons([23, 18]); //should return 6056820

LCM function for a range [a, b]
// Euclid algorithm for Greates Common Divisor
function gcd(a, b)
{
return !b ? a : gcd(b, a % b);
}
// Least Common Multiple function
function lcm(a, b)
{
return a * (b / gcd(a,b));
}
// LCM of all numbers in the range of arr=[a, b]
function range_lcm(arr)
{
// Swap [big, small] to [small, big]
if(arr[0] > arr[1]) (arr = [arr[1], arr[0]]);
for(x = result = arr[0]; x <= arr[1]; x++) {
result = lcm(x, result);
}
return result;
}
alert(range_lcm([8, 5])); // Returns 840

As this question has recently been revived, here's what I think is a simpler take on the question, writing very simple helper functions to calculate the greatest common divisor of two integers (gcd), to calculate the least common multiple of two integers (lcm), to calculate the least common multiple of an array of integers (lcmAll), to generate the range of integers between two given integers (rng), and finally, in our main function, to calculate the least common multiple of the range of integers between two given integers (lcmRng):
const gcd = (a, b) => b == 0 ? a : gcd (b, a % b)
const lcm = (a, b) => a / gcd (a, b) * b
const lcmAll = (ns) => ns .reduce (lcm, 1)
const rng = (lo, hi) => [...Array (hi - lo + 1)] .map ((_, i) => lo + i)
const lcmRng = (lo, hi) => lcmAll (rng (lo, hi))
console .log (lcmRng (1, 13))
All of these functions are simple. Although the question was tagged recursion, only gcdis recursive. If this is an attempt to play with recursion, we could rewrite lcmAll in a recursive manner with something like this:
const lcmAll = (ns) =>
ns.length == 0
? 1
: lcm(ns[0], lcmAll(ns .slice (1)))
Although I'm a big fan of recursion, I see no other reason to choose the recursive version here over the reduce one. In this case, reduce is cleaner.
And finally, if you really want the API originally requested where the range bounds are passed in an array, you could write one more wrapper:
const leastCommonMultiple = ([lo, hi]) => lcmRng (lo, hi)
leastCommonMultiple ([1, 13]) //=> 360360

I found the other answers to be somewhat confusing while I was figuring out the best way to do this with just two numbers, so I looked found the most optimal solution on Wikipedia.
https://en.wikipedia.org/wiki/Least_common_multiple#Calculation
The most efficient way to find the least common multiple of two numbers is (a * b) / greatestCommonDivisor(a, b);
To do this we need to calculate the greatest common denominator. The most efficient way to do that is using Euclid's algorithm.
https://en.wikipedia.org/wiki/Greatest_common_divisor#Euclid's_algorithm
Here is the complete solution for two numbers in case anyone else lands on this question but only needs to calculate for two numbers:
const leastCommonMultiple = (a, b) => (a * b) / greatestCommonDivisor(a, b);
const greatestCommonDivisor = (a, b) => {
const remainder = a % b;
if (remainder === 0) return b;
return greatestCommonDivisor(b, remainder);
};

Mine is not as fancy as the other answers but I think it is easy to read.
function smallestCommons(arr) {
//order our array so we know which number is smallest and which is largest
var sortedArr = arr.sort(sortNumber),
//the smallest common multiple that leaves no remainder when divided by all the numbers in the rang
smallestCommon = 0,
//smallest multiple will always be the largest number * 1;
multiple = sortedArr[1];
while(smallestCommon === 0) {
//check all numbers in our range
for(var i = sortedArr[0]; i <= sortedArr[1]; i++ ){
if(multiple % i !== 0 ){
//if we find even one value between our set that is not perfectly divisible, we can skip to the next multiple
break;
}
//if we make it all the way to the last value (sortedArr[1]) then we know that this multiple was perfectly divisible into all values in the range
if(i == sortedArr[1]){
smallestCommon = multiple;
}
}
//move to the next multiple, we can just add the highest number.
multiple += sortedArr[1];
}
console.log(smallestCommon);
return smallestCommon;
}
function sortNumber(a, b) {
return a - b;
}
smallestCommons([1, 5]); // should return 60.
smallestCommons([5, 1]); // should return 60.
smallestCommons([1, 13]); // should return 360360.
smallestCommons([23, 18]); // should return 6056820.
Edit: Turned answer into snippet.

This is a non-recursive version of your original approach.
function smallestCommons(arr) {
// Sort the array
arr = arr.sort(function (a, b) {return a - b}); // numeric comparison;
var min = arr[0];
var max = arr[1];
var numbers = [];
var count = 0;
//Here push the range of values into an array
for (var i = min; i <= max; i++) {
numbers.push(i);
}
//Here freeze a multiple candidate starting from the biggest array value - call it j
for (var j = max; j <= 1000000; j+=max) {
//I increase the denominator from min to max
for (var k = arr[0]; k <= arr[1]; k++) {
if (j % k === 0) { // every time the modulus is 0 increase a counting
count++; // variable
}
}
//If the counting variable equals the lenght of the range, this candidate is the least common value
if (count === numbers.length) {
return j;
}
else{
count = 0; // set count to 0 in order to test another candidate
}
}
}
alert(smallestCommons([1, 5]));

Hey I came across this page and wanted to share my solution :)
function smallestCommons(arr) {
var max = Math.max(arr[0], arr[1]),
min = Math.min(arr[0], arr[1]),
i = 1;
while (true) {
var count = 0;
for (j = min; j < max; j++) {
if (max * i % j !== 0) {
break;
}
count++;
}
if (count === (max - min)) {
alert(max * i);
return max * i;
}
i++;
}
}
smallestCommons([23, 18]);

function leastCommonMultiple(arr) {
/*
function range(min, max) {
var arr = [];
for (var i = min; i <= max; i++) {
arr.push(i);
}
return arr;
}
*/
var min, range;
range = arr;
if(arr[0] > arr[1]){
min = arr[1];
}
else{
min = arr[0]
}
function gcd(a, b) {
return !b ? a : gcd(b, a % b);
}
function lcm(a, b) {
return (a * b) / gcd(a, b);
}
var multiple = min;
range.forEach(function(n) {
multiple = lcm(multiple, n);
});
return multiple;
}
console.log( leastCommonMultiple([1, 13]) )

Well played on the solution. I think I got one that might be abit shorter just for future reference but ill definatly look into yours
function LCM(arrayRange) {
var newArr = [];
for (var j = arrayRange[0]; j <= arrayRange[1]; j++){
newArr.push(j);
}
var a = Math.abs(newArr[0]);
for (var i = 1; i < newArr.length; i++) {
var b = Math.abs(newArr[i]),
c = a;
while (a && b) {
a > b ? a %= b : b %= a;
}
a = Math.abs(c * newArr[i] / (a + b))
}
return console.log(a);
}
LCM([1,5]);

You may have originally had a stack overflow because of a typo: you switched between min and minn in the middle of repeatRecurse (you would have caught that if repeatRecurse hadn’t been defined in the outer function). With that fixed, repeatRecurse(1,13,13) returns 156.
The obvious answer to avoiding a stack overflow is to turn a recursive function into a non-recursive function. You can accomplish that by doing:
function repeatRecurse(min, max, scm) {
while ( min < max ) {
while ( scm % min !== 0 ) {
scm += max;
}
min++;
}
}
But perhaps you can see the mistake at this point: you’re not ensuring that scm is still divisible by the elements that came before min. For example, repeatRecurse(3,5,5)=repeatRecurse(4,5,15)=20. Instead of adding max, you want to replace scm with its least common multiple with min. You can use rgbchris’s gcd (for integers, !b is the same thing as b===0). If you want to keep the tail optimization (although I don’t think any javascript engine has tail optimization), you’d end up with:
function repeatRecurse(min, max, scm) {
if ( min < max ) {
return repeatRecurse(min+1, max, lcm(scm,min));
}
return scm;
}
Or without the recursion:
function repeatRecurse(min,max,scm) {
while ( min < max ) {
scm = lcm(scm,min);
min++;
}
return scm;
}
This is essentially equivalent to rgbchris’s solution. A more elegant method may be divide and conquer:
function repeatRecurse(min,max) {
if ( min === max ) {
return min;
}
var middle = Math.floor((min+max)/2);
return lcm(repeatRecurse(min,middle),repeatRecurse(middle+1,max));
}
I would recommend moving away from the original argument being an array of two numbers. For one thing, it ends up causing you to talk about two different arrays: [min,max] and the range array. For another thing, it would be very easy to pass a longer array and never realize you’ve done something wrong. It’s also requiring several lines of code to determine the min and max, when those should have been determined by the caller.
Finally, if you’ll be working with truly large numbers, it may be better to find the least common multiple using the prime factorization of the numbers.

function range(min, max) {
var arr = [];
for (var i = min; i <= max; i++) {
arr.push(i);
}
return arr;
}
function gcd (x, y) {
return (x % y === 0) ? y : gcd(y, x%y);
}
function lcm (x, y) {
return (x * y) / gcd(x, y);
}
function lcmForArr (min, max) {
var arr = range(min, max);
return arr.reduce(function(x, y) {
return lcm(x, y);
});
}
range(10, 15); // [10, 11, 12, 13, 14, 15]
gcd(10, 15); // 5
lcm(10, 15); // 30
lcmForArr(10, 15); //60060

How about:
// Euclid Algorithm for the Greatest Common Denominator
function gcd(a, b) {
return !b ? a : gcd(b, a % b);
}
// Euclid Algorithm for the Least Common Multiple
function lcm(a, b) {
return a * (b / gcd(a, b));
}
// LCM of all numbers in the range of arr = [a, b];
function smallestCommons(arr) {
var i, result;
// large to small - small to large
if (arr[0] > arr[1]) {
arr.reverse();
} // only happens once. Means that the order of the arr reversed.
for (i = result = arr[0]; i <= arr[1]; i++) { // all numbers up to arr[1] are arr[0].
result = lcm(i, result); // lcm() makes arr int an integer because of the arithmetic operator.
}
return result;
}
smallestCommons([5, 1]); // returns 60

function lcm(arr) {
var max = Math.max(arr[0],arr[1]),
min = Math.min(arr[0],arr[1]),
lcm = max;
var calcLcm = function(a,b){
var mult=1;
for(var j=1; j<=a; j++){
mult=b*j;
if(mult%a === 0){
return mult;
}
}
};
for(var i=max-1;i>=min;i--){
lcm=calcLcm(i,lcm);
}
return lcm;
}
lcm([1,13]); //should return 360360.

/*Function to calculate sequential numbers
in the range between the arg values, both inclusive.*/
function smallestCommons(arg1, arg2) {
if(arg1>arg2) { // Swap arg1 and arg2 if arg1 is greater than arg2
var temp = arg1;
arg1 = arg2;
arg2 =temp;
}
/*
Helper function to calculate greatest common divisor (gcd)
implementing Euclidean algorithm */
function gcd(a, b) {
return b===0 ? a : gcd(b, a % b);
}
/*
Helper function to calculate lowest common multiple (lcm)
of any two numbers using gcd function above */
function lcm(a,b){
return (a*b)/gcd(a,b);
}
var total = arg1; // copy min value
for(var i=arg1;i<arg2;i++){
total = lcm(total,i+1);
}
//return that total
return total;
}
/*Yes, there are many solutions that can get the job done.
Check this out, same approach but different view point.
*/
console.log(smallestCommons(13,1)); //360360

Here's my solution. I hope you will find it easy to follow:
function smallestCommons(arr) {
var min = Math.min(arr[0], arr[1]);
var max = Math.max(arr[0], arr[1]);
var smallestCommon = min * max;
var doneCalc = 0;
while (doneCalc === 0) {
for (var i = min; i <= max; i++) {
if (smallestCommon % i !== 0) {
smallestCommon += max;
doneCalc = 0;
break;
}
else {
doneCalc = 1;
}
}
}
return smallestCommon;
}

Here is another nonrecursive for-loop solution
function smallestCommons(arr) {
var biggestNum = arr[0];
var smallestNum = arr[1];
var thirdNum;
//make sure biggestNum is always the largest
if (biggestNum < smallestNum) {
thirdNum = biggestNum;
biggestNum = smallestNum;
smallestNum = thirdNum;
}
var arrNum = [];
var count = 0;
var y = biggestNum;
// making array with all the numbers fom smallest to biggest
for (var i = smallestNum; i <= biggestNum; i += 1) {
arrNum.push(i);
}
for (var z = 0; z <= arrNum.length; z += 1) {
//noprotect
for (y; y < 10000000; y += 1) {
if (y % arrNum[z] === 0) {
count += 1;
break;
}
else if (count === arrNum.length) {
console.log(y);
return y;
}
else {
count = 0;
z = 0;
}
}
}
}
smallestCommons([23, 18]);

function smallestCommons(arr) {
var sortedArr = arr.sort(); // sort array first
var tempArr = []; // create an empty array to store the array range
var a = sortedArr[0];
var b = sortedArr[1];
for(var i = a; i <= b; i++){
tempArr.push(i);
}
// find the lcm of 2 nums using the Euclid's algorithm
function gcd(a, b){
while (b){
var temp = b;
b = a % b;
a = temp;
}
return a;
}
function lcm(a, b){
return Math.abs((a * b) / gcd(a, b));
}
var lcmRange = tempArr.reduce(lcm);
return lcmRange;
}

function smallestCommons(arr) {
let smallest, biggest, min;
arr.reduce(function (a, b) {
biggest = Math.max(a, b);
});
const max = biggest;
arr.reduce(function (a, b) {
smallest = Math.min(a, b);
min = smallest;
});
check: while (true) {
biggest += max;
for (min = smallest; min < max; min++) {
if (biggest % min != 0) {
continue check;
}
if (min == (max - 1) && biggest % min == 0) {
console.warn('found one');
return biggest;
}
}
}
}

function smallestCommons(arr) {
let min = Math.min(arr[0], arr[1]);
let max = Math.max(arr[0], arr[1]);
let scm = max;
//calc lcm of two numbers:a,b;
const calcLcm = function(a, b) {
let minValue = Math.min(a, b);
let maxValue = Math.max(a, b);
let lcm = maxValue;
while (lcm % minValue !== 0) {
lcm += maxValue;
}
return lcm;
}
//calc scm in range of arr;
for (let i = max; i >= min; i--) {
scm = calcLcm(scm, i);
}
console.log(scm);
return scm;
}
smallestCommons([1, 13]);

this is another very simple way and have low complexity.
function smallestCommons(arr) {
let smallestNum = arr[0] < arr[1] ? arr[0] : arr[1];
let greatestNum = arr[0] > arr[1] ? arr[0] : arr[1];
let initalsArr = [];
for(let i = smallestNum; i <= greatestNum; i++){
initalsArr.push(i);
}
let notFoundFlag = true;
let gNMltpl = 0;
let filteredArrLen;
while(notFoundFlag){
gNMltpl += greatestNum;
filteredArrLen = initalsArr.filter((num)=>{
return (gNMltpl / num) === Math.floor((gNMltpl / num))
}).length;
if(initalsArr.length == filteredArrLen){
notFoundFlag = false;
}
}
return gNMltpl;
}

My solution using es6 feature is
Lcm of given numbers
const gcd = (a, b) => (!b ? a : gcd(b, a % b));
const lcm = (a, b) => a * (b / gcd(a, b));
const getLcm = (arr) => {
const numbers = arr.sort((a, b) => parseInt(a) - parseInt(b));
let result = parseInt(numbers[0]);
for (let i = 1; i < numbers.length; i++) {
result = lcm(parseInt(result), parseInt(numbers[i + 1]));
}
return result;
};
Hcf of given numbers
const getHcf = (arr) => {
const numbers = arr.sort((a, b) => parseInt(a) - parseInt(b));
let result = parseInt(numbers[0]);
for (let i = 1; i < numbers.length; i++) {
result = gcd(parseInt(numbers[i]), parseInt(result));
}
return result;
};
Call like this
console.log(getLcm([20, 15, 10, 40])). Answer 120
console.log(getHcf([2, 4, 6, 8, 16])). Answer 2

I also found myself working on this challenge on my freeCodeCamp JavaScript Certification. This is what I have been able to come up with:
function smallestCommons(arr) {
let newArr = [];
// create a new array from arr [min, min + 1,......., max - 1, max]
for (let i = Math.min(...arr); i <= Math.max(...arr); i++){
newArr.push(i);
}
// let the max of newArr be the smallestCommonMultiple initially
let largest = Math.max(...newArr);
let smallestCommonMultiple = largest;
// If the supposedly smallestCommonMultiple fail on any of elements in
//newArr add the max element until we find the smallestCommonMultiple.
while (newArr.some(element => smallestCommonMultiple % element !== 0)){
smallestCommonMultiple += largest;
}
return smallestCommonMultiple;
}
console.log(smallestCommons([23, 18]));

i think it will work.
var a = [2, 6];
function getTotalX(a) {
var num = 1e15;
var i;
var arr = [];
for (i = 1; i <=num ; i++){
arr.push(i);
}
for (i = 0; i < a.length; i++){
var filterArr = arr.filter((val, ind, arr) => (val % a[i] === 0));
}
console.log(filterArr[0]); // will return 6
}

I've made a similar function in typescript that does the same task but only without recursion...
function findLowestCommonMultipleBetween(start: number, end: number): number {
let numbers: number[] = [];
for (let i = start; i <= end; i++) {
numbers.push(i);
}
for (let i = 1; true; i++) {
let divisor = end * i;
if (numbers.every((number) => divisor % number == 0)) {
return divisor;
}
}
}

function smallestCommons(arr) {
let min = Math.min(...arr);
let max = Math.max(...arr);
let rangArr = [];
for(let i = min; i <= max; i++) rangArr.push(i);
let smallestCommon = max;
while(!rangArr.every(e => smallestCommon % e === 0)){
smallestCommon += max;
}
return smallestCommon;
}
console.log(smallestCommons([1, 13]));

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

function smallestCommons(arr) {
// Kind of a brute force method, It's not fancy but it's very simple and easy to read :P
// make an array with all the numbers in the range.
let numbersArr = [];
for (let i = Math.min(...arr); i <= Math.max(...arr); i++) {
numbersArr.push(i);
}
// keep multiplying the biggest number until it's divisible by all the numbers in the numbersArr array.
let scm = Math.max(...arr);
while (true) {
if (numbersArr.every(num => scm % num === 0)) {
return scm;
} else {
scm += Math.max(...arr);
}
}
}
smallestCommons([2, 10]); // returns 2520.
smallestCommons([1, 13]); // returns 360360.
smallestCommons([23, 18]); // returns 6056820.

function leastCommonMultiple(arr) {
// Setup
const [min, max] = arr.sort((a, b) => a - b);
// Largest possible value for LCM
let upperBound = 1;
for (let i = min; i <= max; i++) {
upperBound *= i;
}
// Test all multiples of 'max'
for (let multiple = max; multiple <= upperBound; multiple += max) {
// Check if every value in range divides 'multiple'
let divisorCount = 0;
for (let i = min; i <= max; i++) {
// Count divisors
if (multiple % i === 0) {
divisorCount += 1;
}
}
if (divisorCount === max - min + 1) {
return multiple;
}
}
}
//for a test
leastCommonMultiple([1, 5]);