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CodeSignal sumInRange challenge in JavaScript

I'm working on this CodeSignal exercise that says:
You have an array of integers nums and an array queries, where queries[i] is a pair of indices (0-based). Find the sum of the elements in nums from the indices at queries[i][0] to queries[i][1] (inclusive) for each query, then add all of the sums for all the queries together. Return that number modulo 10^9 + 7.
This is my code:
function solution(nums, queries) {
let accumulator = 0;
let M = 1000000007;
for(let i = 0; i < queries.length; i++){
accumulator += nums.slice(queries[i][0],queries[i][1]+1).reduce((a,b) => a+b);
}
return accumulator < 0 ? ((accumulator % M) + M) % M : accumulator%M;
}
It works perfectly, BUT the penultimate hidden test throws a timeout, and I'm out of ideas on how to make this faster.
Thanks in advance for any help you may provide.
PS: if you're wondering about the difference using modulo at the end, it's because it seems JS has a bug when using negative numbers.

As tflave pointed out, using a prefix sum made the code perform faster, and it didn't timeout. Here's the code if anyone needs it:
let pre = new Array(1000,0);
function preCompute(nums)
{
let n = nums.length;
pre[0] = nums[0];
for (let i = 1; i < n; i++) {
pre[i] = nums[i] + pre[i - 1];
}
}
function solution(nums, queries)
{
preCompute(nums);
let accumulator = 0;
let M = 1000000007;
for(let i = 0; i < queries.length; i++){
if (queries[i][0] === 0) {
accumulator += pre[queries[i][1]];
} else {
accumulator += pre[queries[i][1]] - pre[queries[i][0] - 1];
}
}
return accumulator < 0 ? ((accumulator % M) + M) % M : accumulator%M;
}

Get N-sized numbers that add up to a number X

My question maybe seems like duplicated but I can't find how to do this in vanilla javascript.
(I found answer in python here.)
So basically I want to get a list of all possible combinations of number that sum up to a desired number.
For example:
function sum_to_n(n,size){}
x = sum_to_n(6,3)
// [6,0,0] [5,1,0] [4,2,0] [4,1,1] [3,3,0] [3,2,1] [2,2,2]
This is what I can do so far:
function sum_to_n(n, size, sum=[], limit) {
if (size === 1) {
sum.push(n)
return sum
}
if (!limit) {
limit = n
}
let start = Math.min(n, limit)
let stop = Math.ceil(n/size) - 1
for (let i = start; i > stop; i--) {
let tmp = [...sum];
tmp.push(i)
let combination = sum_to_n(n-i, size - 1, tmp, i)
if (combination) {
console.log(combination) // this work
}
}
}
// I don't know how to make this work
let x = sum_to_n(6, 3)
// or maybe this
for (let y in sum_to_n(6,3)) {
console.log(y)
}

This is closely related to partitions of an integer, which finds all distinct ways of breaking an integer into the sum of positive integers. For instance, the partitions of 4 are 4, 3 + 1, 2 + 2, 2 + 1 + 1, and 1 + 1 + 1 + 1. Although there is probably a more direct way of calculating your sum, it's easy to do atop a partitions function:
const partitions = (n, max = n) =>
n == 0
? [[]]
: n < 0
? []
: Array .from ({length: Math .min (n, max)}, (_, i) => i + 1) // or use a `ramge` function
.reverse()
.flatMap (x => partitions (n - x, x) .map (p => [x, ...p]))
const sumToN = (n, c) =>
partitions (n + c)
.filter (p => p .length == c)
.map (p => p .map (x => x - 1))
console .log (sumToN (6, 3))
.as-console-wrapper {max-height: 100% !important; top: 0}
(The reverse call is not strictly necessary, but the usual partition description is written in this order, as is your requested output. The Array.from (...) call would be better expressed using a range function, but is not horrible inlined like this.)
We need to do two things: we have to find only those of the correct length, and we need to allow zeros as well as the positive numbers that partition generates. We can do this by partitioning a larger number and subtracting 1 from each value in every partition, and filtering the results down to those with the proper length.
The larger number we need -- since we're going to subtract 1 for every one of our c elements in each partition -- is simply n + c.
This algorithm's performance is exponential in n. But that's necessary as there are an exponential number of such partitions, and presumably an exponential number of results for your requested output. (I don't know a proof of this, and it's possible I'm wrong. If I am, I'd love to see a sub-exponential algorithm.)

'use strict';
function _sumToN(n, size, acc = [], solutions = []) {
if (size === 0) {
const sum = acc.reduce((sum, num) => sum + num);
return sum === n ? solutions.concat([acc]) : solutions;
}
for (let i = 0; i <= n; ++i)
solutions = _sumToN(n, size - 1, acc.concat(i), solutions);
return solutions;
}
function sumToN(n, size) {
return _sumToN(n, size);
}
console.log(sumToN(6, 3));
The above uses a driver, _sumToN, which accepts an additional acc accumulator array, and a solutions array that will eventually hold all the solutions. It iterates as follows:
0 0 0
0 0 1
0 0 2
...
0 0 6
...
6 6 6
and saves off solutions when they sum to n. There are of course optimizations that can be made, but hopefully this trivial implementation helps.

let sum_to_n = (n, iteration_limit) => {
let res = [];
for (let i = 0; i < iteration_limit; i++) {
for (let j = 0; j < iteration_limit; j++) {
for (let h = 0; h < iteration_limit; h++) {
if(i+j+h == n){
res.push([i, j, h])
}
}
}
}
return res;
};
This worked for me, however it does not have a size parameter.

let getOperator = (size, arr, expect) => {
let ev = "";
for (let i = 0; i < size; i++) {
ev += `${arr[i]}+`;
}
return `${ev.slice(0, ev.length-1)} == ${expect}`;
}
let sum_to_n = (n, size, iteration_limit) => {
return new Promise(resolve => {
let identifiers = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz".split('');
let all_identifiers = [];
let res = [];
let ev = "";
for (let i = 0; i < size; i++) {
let r = identifiers[i];
all_identifiers.push(r);
ev += `for (let ${r} = 0; ${r} < iteration_limit; ${r}++){\n`;
}
let push_value_string = "";
for (let i = 0; i < all_identifiers.length; i++) {
push_value_string += `${all_identifiers[i]}, `;
}
ev += `\nif(${getOperator(size, all_identifiers, n)}){\nres.push([${push_value_string.slice(0, push_value_string.length-1)}]);\n}`;
for (let i = 0; i < size; i++) {
ev += "}";
}
console.log(ev);
eval(ev);
resolve(res);
});
};
sum_to_n(15, 3, 100).then(console.log);
Basically depending on the size it will make a JavaScript code in a text size times, save what word it used for the for loop and check if the addition of all is equal to n. Then evaluate it and resolve it.

Electronic shop hackerRank challenge in Javascript

Hi I did not pass the all the tests, only 9/16. so I want to know what is the problem in my code
problem link:https://www.hackerrank.com/challenges/electronics-shop/problem
function getMoneySpent(keyboards, drives, b) {
let arr = [];
let lenOfArr1 = keyboards.length;
let lenOfArr2 = drives.length;
let j = drives.length;
arr = keyboards.slice(0);
for (let number of drives) {
arr.push(number);
}
return (challenge(arr, lenOfArr1, b));
}
function challenge(arr, lenOfArr1, m) {
let result = [];
let j = lenOfArr1;
for (let i = 0; i < lenOfArr1; i++) {
if (arr[i] >= m || arr[j] >= m) return -1;
if (arr[i] + arr[j] < m) {
result.push(arr[i] + arr[j]);
}
i--;
j++;
if (j == arr.length) {
i++;
j = lenOfArr1;
}
}
if (result.length == 0) return -1;
return result.sort()[result.length - 1];
}

From the given constraints in the problem statement, you don't need such a complex solution.
Here is the Algorithm:
Initialize a variable maxValue to have value as -1.
Start two for loops over drives and keyboards and take all combinations and sum the value of each drive with each keyboard.
Inside the for loops, check if the sum of drive + keyboard and it should be less than or equal to the money Monica has and keep track of the maximum value you get from any combination in maxValue.
After the code computation, return maxValue.
Code:-
function getMoneySpent(keyboards, drives, b) {
let maxValue = -1;
for (let drive of drives) {
for(let keyboard of keyboards) {
let cost = drive + keyboard;
if(cost > maxValue && cost <= b) {
maxValue = cost;
}
}
}
return maxValue;
}
Overall Time complexity - O(n^2).
Hope this helps!

The following line causes your code to return -1 when there might be a solution:
if (arr[i] >= m || arr[j] >= m) return -1;
Even though one price may be above budget, you need to consider that there might be cheaper items available. You might even already have solutions in result, or they might still be found in a later iteration.
It is not clear to me why you first create a merged array of both input arrays. It does not seem to bring any benefit.
If you sort the input arrays, you can use a two-pointer system where you decide which one to move depending on whether the current sum is within limits or it is too great:
function getMoneySpent(keyboards, drives, b) {
keyboards = [...keyboards].sort((a, b) => b - a) // descending
drives = [...drives].sort((a, b) => a - b); // ascending
let k = keyboards.length - 1; // cheapest
let d = drives.length - 1; // most expensive
let maxSum = -1;
while (d >= 0 && k >= 0) {
let sum = keyboards[k] + drives[d];
if (sum === b) return b;
if (sum < b) {
if (sum > maxSum) maxSum = sum;
k--; // will increase sum
} else {
d--; // will decrease sum
}
}
return maxSum;
}

Multiplying N positive odd numbers

I'm trying to get the product of N positive odd numbers
function multOdd(n) {
var mult = 1;
var counter=[];
for (var i = 1; i <= 2*n-1; i += 2){
counter.push(i);
}
console.log(counter);
return mult=mult*counter[i];
}
console.log(multOdd(10));
I pushed the numbers into an array and attempted to get the product from them but I can't get it to work.

When you return mult=mult*counter[i] you're only returning the multipication once. It should return mult = 1 * counter[lastElement+2] which will be wrong. In your case, the last element of counter is 19, before exiting for loop i value is i= 19 + 2 = 21. You're returning mult = 1 * 21 = 21.
You can instead return the multipication value by for loop with no need for an array:
function multOdd(n) {
var mult = 1;
for (var i = 1; i <= 2*n-1; i += 2){
mult = mult * i;
}
return mult;
}

If you just want the result for n, use:
function multOdd(n) {
var result = 1;
for (var i = 1; i <= 2*n-1; i += 2){
result = result * i;
}
console.log(result);
return result;
}
console.log(multOdd(4));
If you want an array that has an array indexed by the number of odd numbers up to n you could use:
function multOdd(n) {
let result = 1;
let results = [];
for (let i = 1; i <= 2*n-1; i += 2){
result = result * i;
results[(i+1) / 2] = result;
}
console.log(results);
return results;
}
console.log(multOdd(10));

There are a few ways to get the product of an array of numbers. Here are two easy ones:
Relevant MDN
// Using `Array.prototype.reduce`
[3, 5, 7, 9].reduce((acc, val) => acc * val)
// Using a `for ... of` loop
let product = 1
for (const val of [3, 5, 7, 9]) {
product *= val
}

You could separate out the two steps of your current code into two functions:
const getFirstNOddNums = (n) => {
let oddNums = []
for (let i = 1; i <= 2*n-1; i+=2) {
oddNums.push(i)
}
return oddNums
}
const productOfArray = (arr) => {
return arr.reduce((a, b) => a * b)
}
const N = 5
const firstNOddNums = getFirstNOddNums(N)
console.log(`The first ${N} odd numbers are: ${JSON.stringify(firstNOddNums)}`)
console.log(`The product of the first ${N} odd numbers is: ${productOfArray(firstNOddNums)}`)

let multOdd = (n) => {
let total = 1;
for (let i = 1; i<= 2*n; i+=2){
total *= i;
}
return total;
}
console.log(multOdd(10));

Instead of recursion, We should use the standard mathematical formula for the product of first n positive odd integers which is
Can also be written as (in the form of pi notation)
For this latex image, I used https://codecogs.com/latex/eqneditor.php.
Factorial is
n! = n(n-1)(n-2)(n-3) ... and so on
So we can use Array(n).fill() to get an array of 10 elements and reduce them to get a factorial by
Array(n).fill().reduce((v,_,i) => (i+1) * v || 2)
Then we divide it by 2 to the power n times the n!. Which is what we want. The advantage here is that, this makes your solution, a one liner
let n = 10
let answer = Array(2*n).fill().reduce((v,_,i) => (i+1) * v || 2) / (Math.pow(2,n) * Array(n).fill().reduce((v,_,i) => (i+1) * v || 2))
console.log(answer)

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]);