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Largest Triple Products without using sort?

I implemented the Largest Triple Products algorithm, but I use sort which makes my time complexity O(nlogn). Is there a way to implement it without a temporary sorted array?
The problem:
You're given a list of n integers arr[0..(n-1)]. You must compute a list output[0..(n-1)] such that, for each index i (between 0 and n-1, inclusive), output[i] is equal to the product of the three largest elements out of arr[0..i] (or equal to -1 if i < 2, as arr[0..i] then includes fewer than three elements).
Note that the three largest elements used to form any product may have the same values as one another, but they must be at different indices in arr.
Example:
var arr_2 = [2, 4, 7, 1, 5, 3];
var expected_2 = [-1, -1, 56, 56, 140, 140];
My solution:
function findMaxProduct(arr) {
// Write your code here
if(!arr || arr.length === 0) return [];
let helper = arr.slice();
helper.sort((a,b)=>a-b); // THIS IS THE SORT
let ans = [];
let prod = 1;
for(let i=0; i<arr.length; i++) {
if(i < 2) {
prod *= arr[i];
ans.push(-1);
}
else {
if(i === 3) {
prod *= arr[i];
ans.push(prod);
} else if(arr[i] < helper[0]) {
ans.push(prod);
} else {
const min = helper.shift();
prod /= min;
prod *= arr[i];
ans.push(prod);
}
}
}
return ans;
}
Thanks

You don't need to sort it. You just maintain an array of the largest three elements at each index.
For the first three elements it is simple you just assign the product of them to the third element in the result.
For the next elements, you add the current element to the three-largest-element-array and sort it and take the elements from 1 to 3 ( the largest three ) and assign the product of those at that index in result array. Then update the three-element-array with largest three.
Complexity :
This sort and slice of three-element-array should be O(1) because each time atmost 4 elements are there in the array.
Overall complexity is O(n).
You can do it as follows :
function findMaxProduct(arr) {
if(!arr) return [];
if (arr.length < 3) return arr.slice().fill(-1)
let t = arr.slice(0,3)
let ans = arr.slice().fill(-1,0,2) //fill first two with -1
ans[2] = t[0]*t[1]*t[2];
for(let i=3; i<arr.length; i++) {
t.push(arr[i]);
t = t.sort().slice(1,4);
ans[i] = t[0]*t[1]*t[2];
}
return ans;
}

I am keeping the array ordered (manually). Then just get the first 3 elements.
function findMaxProduct(arr) {
let results = [];
let heap = [];
for (let i = 0; i < arr.length; i++) {
// Insert the new element in the correct position
for (let j = 0; j < heap.length; j++) {
if (arr[i] >= heap[j]) {
heap.splice(j, 0, arr[i]);
break;
}
}
// No position found, insert at the end
if (heap.length != i + 1) {
heap.push(arr[i]);
}
if (i < 2) {
results.push(-1);
} else {
results.push(heap[0] * heap[1] * heap[2]);
}
}
return results;
}

You can make an array that holds three currently largest integers, and update that array as you passing through original array. That's how you will always have three currently largest numbers and you will be able to solve this with O(n) time complexity.

I think there's a faster and more efficient way to go about this. This is a similar thought process as #Q2Learn, using Python; just faster:
def findMaxProduct(arr):
#create a copy of arr
solution = arr.copy()
# make first 2 elements -1
for i in range(0,2):
solution[i] = -1
#for each item in copy starting from index 2, multiply item from 2 indices b'4 (notice how each index of arr being multiplied is reduced by 2, 1 and then 0, to accommodate each move)
for i in range(2, len(arr)):
solution[i] = arr[i-2] * arr[i-1] * arr[i]
return solution
check = findMaxProduct(arr)
print(check)

Single Scan Algorithm O(n)
We don't need to necessarily sort the given array to find the maximum product. Instead, we can only find the three largest values (x, y, z) in the given stage of iteration:
JavaScript:
function findMaxProduct(arr) {
let reults = []
let x = 0
let y = 0
let z = 0
for(let i=0; i<arr.length; i++) {
n = arr[i]
if (n > x) {
z = y
y = x
x = n
}
if (n < x && n > y) {
z = y
y = n
}
if (n < y && n > z) {
z = n
}
ans = x*y*z
if (ans === 0) {
results.push(-1)
} else {
results.push(ans)
}
return ans;
}
Python:
def findMaxProduct(arr):
results = []
if not arr:
return []
x = 0
y = 0
z = 0
for i, n in enumerate(arr):
if n > x:
z = y
y = x
x = n
if n < x and n > y:
z = y
y = n
if n < y and n > z:
z = n
ans = x*y*z
if ans == 0:
results.append(-1)
else:
results.append(ans)
print(results)

public int[] LargestTripleProducts(int[] input)
{
var ansArr = new int[input.Length];
var firstLargetst = input[0];
var secondLargetst = input[1];
ansArr[0] = ansArr[1] = -1;
for (int i = 2; i < input.Length; i++)
{
ansArr[i] = firstLargetst * secondLargetst * input[i];
if (firstLargetst < input[i] && firstLargetst < secondLargetst)
{
firstLargetst= input[i];
continue;
}
if (secondLargetst < input[i] && secondLargetst < firstLargetst)
{
secondLargetst= input[i];
}
}
return ansArr;
}

Python solution based on #SomeDude answer above. See explanation there.
def findMaxProduct(arr):
if not arr:
return None
if len(arr) < 3:
for i in range(len(arr)):
arr[i] = -1
return arr
three_largest_elem = arr[0:3]
answer = arr.copy()
for i in range(0, 2):
answer[i] = -1
answer[2] = three_largest_elem[0] * three_largest_elem[1] * three_largest_elem[2]
for i in range(3, len(arr)):
three_largest_elem.append(arr[i])
three_largest_elem = sorted(three_largest_elem)
three_largest_elem = three_largest_elem[1:4]
answer[i] = three_largest_elem[0] * three_largest_elem[1] * three_largest_elem[2]
return answer #Time: O(1) n <= 4, to Overall O(n) | Space: O(1)

Python has it's in-built package heapq, look at it for it.
Credit: Martin
> Helper function for any type of calculations
import math
> Heap algorithm
import heapq
> Create empty list to append output values
output = []
def findMaxProduct(arr):
out = []
h = []
for e in arr:
heapq.heappush(h, e)
if len(h) < 3:
out.append(-1)
else:
if len(h) > 3:
heapq.heappop(h)
out.append(h[0] * h[1] * h[2])
return out
Hope this helps!

Javascript Program for find common elements in two array

Recently I had an interview question as follows:
Let us consider we have two sorted arrays of different length. Need to find the common elements in two arrays.
var a=[1,2,3,4,5,6,7,8,9,10];
var b = [2,4,5,7,11,15];
for(var i=0;i<a.length;i++){
for(var j=0;j<b.length;j++){
if(a[i]==b[j]){
console.log(a[i],b[j])
}
}
}
I wrote like above. The interviewer said let now assume a have 2000 elements and b have 3000 elements. Then how you wrote in a more efficient way?
Please explain your answers with sample code. So I can understand more clearly.

The easiest way!!
var a = [1,2,3,4,5,6,7,8,9,10];
var b = [2,4,5,7,11,15];
for(let i of a){
if(b.includes(i)){
console.log(i)
}
}
--------- OR --------------
var c = a.filter(value => b.includes(value))
console.log(c)

Since the arrays are sorted, binary search is the key.
Basically, you're searching an item in an array.
You compare the item against the middle index of the array (length / 2)
If both are equal, you found it.
If item is inferior than the one at the middle index of the array, compare item against the index being at index length / 4 -> ((0 + length / 2) / 2), if it's inferior, at index ((length / 2) + length) / 2 (the middle of upper part) and so on.
That way, if in example you have to search item in a 40 000 length array, at worse, you find out that item isn't in the array with 16 comparisons :
I'm searching for "something" in an array with 40 000 indexes, minimum index where I can find it is 0, the maximum is 39999.
"something" > arr[20000]. Let's assume that. I know that now the minimum index to search is 20001 and the maximum is 39999. I'm now searching for the middle one, (20000 + 39999) / 2.
Now, "something" < arr[30000], it limits the search from indexes 20001 to 29999. (20000 + 30000) / 2 = 25000.
"something" > arr[25000], I have to search from 25001 to 29999. (25000 + 30000) / 2 = 27500
"something" < arr[27500], I have to search from 25001 to 27499. (25000 + 27500) / 2 = 26250
"something" > arr[26250], I have to search from 26251 to 27499. (26250 + 27500) / 2 = 26875
"something" < arr[26875], I have to search from 26251 to 26874. (26250 + 26875) / 2 = 26563
And so on... Of course, you have to round and stuff to avoid floating indexes
var iteration = 1;
function bSearch(item, arr)
{
var minimumIndex = 0;
var maximumIndex = arr.length - 1;
var index = Math.round((minimumIndex + maximumIndex) / 2);
while (true)
{
++iteration;
if (item == arr[index])
{
arr.splice(0, minimumIndex);
return (true);
}
if (minimumIndex == maximumIndex)
{
arr.splice(0, minimumIndex);
return (false);
}
if (item < arr[index])
{
maximumIndex = index - 1;
index = Math.ceil((minimumIndex + maximumIndex) / 2);
}
else
{
minimumIndex = index + 1;
index = Math.floor((minimumIndex + maximumIndex) / 2);
}
}
}
var arrA;
var arrB;
for (var i = 0; i < arrA.length; ++i)
{
if (bSearch(arrA[i], arrB))
console.log(arrA[i]);
}
console.log("number of iterations : " + iteration);

You could use a nested approach by checking the index of each array and find the values by incrementing the indices. If equal values are found, increment both indices.
Time complexity: max. O(n+m), where n is the length of array a and m is the length of array b.
var a = [1, 2, 3, 4, 5, 6, 8, 10, 11, 15], // left side
b = [3, 7, 8, 11, 12, 13, 15, 17], // right side
i = 0, // index for a
j = 0; // index for b
while (i < a.length && j < b.length) { // prevent running forever
while (a[i] < b[j]) { // check left side
++i; // increment index
}
while (b[j] < a[i]) { // check right side
++j; // increment
}
if (a[i] === b[j]) { // check equalness
console.log(a[i], b[j]); // output or collect
++i; // increment indices
++j;
}
}

since both arrays are sorted just save the lastest match index . then start your inner loop from this index .
var lastMatchedIndex = 0;
for(var i=0;i<a.length;i++){
for(var j=lastMatchIndex ;j<b.length;j++){
if(a[i]==b[j]){
console.log(a[i],b[j]);
lastMatchedIndex = j;
break;
}
}
}
=================
UPDATE :
As Xufox mentioned in comments if a[i] is lower than b[i] then u have break loop since it has no point to continue the loop .
var lastMatchedIndex = 0;
for(var i=0;i<a.length;i++){
if(a[i]<b[i]){
break;
}
for(var j=lastMatchIndex ;j<b.length;j++){
if(a[i]==b[j]){
console.log(a[i],b[j]);
lastMatchedIndex = j;
break;
}
if(a[i]<b[j]){
lastMatchedIndex = j;
break;
}
}
}

An optimal strategy would be one where you minimize the amount of comparisons and array readings.
Theoretically what you want is to alternate which list you are progressing through so as to avoid unnecessary comparisons. Giving that the lists are sorted we know that no number to the left of any index in a list can ever be smaller than the current index.
Assuming the following list A = [1,5], list B = [1,1,3,4,5,6] and indexes a and b both starting at 0, you would want your code to go like this:
A[a] == 1, B[b] == 1
A[a] == B[b] --> add indexes to results and increase b (B[b] == 1)
A[a] == B[b] --> add indexes to results and increase b (B[b] == 3)
A[a] < B[b] --> don't add indexes to results and increase a (A[a] == 5)
A[a] > B[b] --> don't add indexes to results and increase b (B[b] == 4)
A[a] > B[b] --> don't add indexes to results and increase b (B[b] == 5)
A[a] == B[b] --> add indexes to results and increase b (B[b] == 6)
A[a] < B[b] --> don't add indexes to results and increase a (A is at the end, so we terminate and return results)
Below is my JavaScript performing the above described algorithm:
//Parameters
var listA = [];
var listB = [];
//Parameter initialization
(function populateListA() {
var value = 0;
while (listA.length < 200) {
listA.push(value);
value += Math.round(Math.random());
}
})();
(function populateListB() {
var value = 0;
while (listB.length < 300) {
listB.push(value);
value += Math.round(Math.random());
}
})();
//Searcher function
function findCommon(listA, listB) {
//List of results to return
var results = [];
//Initialize indexes
var indexA = 0;
var indexB = 0;
//Loop through list a
while (indexA < listA.length) {
//Get value of A
var valueA = listA[indexA];
var result_1 = void 0;
//Get last result or make a first result
if (results.length < 1) {
result_1 = {
value: valueA,
indexesInA: [],
indexesInB: []
};
results.push(result_1);
}
else {
result_1 = results[results.length - 1];
}
//If higher than last result, make new result
//Push index to result
if (result_1.value < valueA) {
//Make new object
result_1 = {
value: valueA,
indexesInA: [indexA],
indexesInB: []
};
//Push to list
results.push(result_1);
}
else {
//Add indexA to list
result_1.indexesInA.push(indexA);
}
//Loop through list b
while (indexB < listB.length) {
//Get value of B
var valueB = listB[indexB];
//If b is less than a, move up list b
if (valueB < valueA) {
indexB++;
continue;
}
//If b is greather than a, break and move up list a
if (valueB > valueA) {
break;
}
//If b matches a, append index to result
result_1.indexesInB.push(indexB);
//Move up list B
indexB++;
}
//Move up list A
indexA++;
}
//Return all results with values in both lines
return results.filter(function (result) { return result.indexesInB.length > 0; });
}
//Run
var result = findCommon(listA, listB);
//Output
console.log(result);

We could iterate one array and find the duplicate in the other, but each time we find a match, we move to the matched element + 1 for the next iteration in the nested loop. It works because both arrays are sorted. So each match the array to compare is shorter (from left to right).
We could also break the nested loop when the element of the second array is greater than the first (it's shorter from right to left), because we will never find a match (since the array is ordered, there are only greater values remaining), here and example finding duplicates in two arrays of 10k elements, takes roughly 15 miliseconds:
var arr = [];
var arr2 = [];
for(let i = 0; i<9999; i++){
arr.push(i);
arr2.push(i+4999)
}
var k = 0;//<-- the index we start to compare
var res = [];
for (let i = 0; i < arr2.length; i++) {
for (let j = k; j < arr.length; j++) {
if (arr2[i] === arr[j]) {
res.push(arr2[i]);
k = j + 1;//<-- updates the index
break;
} else if (arr[j] > arr2[i]) {//<-- there is no need to keep going
break;
}
}
}
console.log(res.length)
I did not print res, because it has 5000 elements.

You can build a hash with first array (irrespective of they are sorted or not) and iterate the second array and check for existence in the hash!
let arr1 = [10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150],
arr2 = [15,30,45,60,75,90,105,120,135,150,165]
hash = arr1.reduce((h,e)=> (h[e]=1, h), {}), //iterate first array once
common = arr2.filter(v=>hash[v]); //iterate secod array once
console.log('Cpmmon elements: ', common);

Not sure but this may help
let num1 = [2, 3, 6, 6, 5];
let num2 = [1, 3, 6, 4];
var array3 = num1.filter((x) => {
return num2.indexOf(x) != -1
})
console.log(array3);

I sometimes find it convenient to turn one list into a hashset.
var hashA = {};
for(var i=0; i<a.length; i++) {hashA[a[i]] = true;}
then you can search the hashset.
for(var i=0; i<b.length; i++) {if(hashA[b[i]]) {console.log(b[i]);}}
This isnt as fast as the binary search of course because you have to take time to build the hashset, but its not bad, and if you need to keep the list and do a lot of future searching it might be the best option. Also, I know javascript objects arent really just hashsets, its complicated, but it mostly works pretty well.
Honestly though, for 3000 items I wouldnt change the code. Thats still not big enough to be an issue. That will run in like 30ms. So it also depends on how often its going to run. Once an hour? Forget about it. Once per millisecond? Definitely gotta optimize that.

if we are talking about the algorithm to find common elements between two array, then here is my opinion.
function common(arr1, arr2) {
var newArr = [];
newArr = arr1.filter(function(v){ return arr2.indexOf(v) >= 0;})
newArr.concat(arr2.filter(function(v){ return newArr.indexOf(v) >= 0;}));
return newArr;
}
but if you are going to think on performance also, then you should try another ways also.
first check the performance for javascript loop here, it will help you to figure out best way
https://dzone.com/articles/performance-check-on-different-type-of-for-loops-a
https://hackernoon.com/javascript-performance-test-for-vs-for-each-vs-map-reduce-filter-find-32c1113f19d7

Find the largest subarray length with delimiter

I'm trying to solve this problem:
Given two parameters: an array a and integer i, find the largest array length where the sum of all elements is <= i.
For example, having the following array: [3, 1, 2, 1], and i = 4, all the combinations that are <= i are: [3], [1], [2], [3, 1], [1, 2], [1, 2, 1]. The largest subarray is [1, 2, 1], so the return of the function should be 3 (the array length).
What would be a an efficient approach to solve this problem?
This is my algorithm until now, but I know that I'm missing something:
function sumArray(a) {
return a.reduce((a, b) => a + b, 0)
}
function maxLength(a, i) {
let max = 0
let array = [a[0]]
a.splice(1, a.length).forEach(number => {
array.push(number)
if (sumArray(array) <= k) max = array.length
else array.splice(array.indexOf(Math.max.apply(null, array)), 1)
})
return max
}

Here's how I would do it.
First, we'll get the longest sub-array by taking only the smallest elements out of the original, because that way the sum will be smallest possible as we go along. So:
const original = [3, 1, 2, 1];
const maxsum = 4;
// To make sure we take only the smallest, let's just
// sort the array straight away. That way the smallest
// element will always be the first.
// The .slice call is to make a copy, so we don't change
// the original:
const sorted = original.slice().sort();
// Here's your sum function. Looks legit.
function sumArray(a) {
return a.reduce((a, b) => a + b, 0)
}
// Now take items from the front of the sorted array and
// put them in the new array until either the original is
// empty or the max is reached.
let subarray = [];
while (sorted.length) {
// Get the next smallest element. It's always the
// first one because of the sort.
const smallest = sorted.shift();
// If the sum of what we have plus this new element
// is bigger than the max, our work is done:
if (sumArray(subarray) + smallest > maxsum) break;
// Otherwise, add it to our sub array and continue.
subarray.push(smallest)
}
// Once that loop has run, either we ran out of elements,
// or we hit the max. Either way, our job is done.
console.log("Original array:", original);
console.log("Maximal subset:", subarray);
console.log("Maximal subset length:", subarray.length);
Finally, if you want to get fancy, you can even do this with a single .reduce call:
const original = [3, 1, 2, 1];
const maxsum = 4;
const maximalSubset = original.slice().sort().reduce((subset, current) => {
if (subset.reduce((s, c) => s + c, 0) + current <= maxsum) subset.push(current);
return subset;
}, []);
console.log("Orignal:", original);
console.log("Maximal subset:", maximalSubset);
console.log("Maximal subset length:", maximalSubset.length);
Although while shorter, that second snippet has the downside that we have to iterate the entire array before we get the result, whereas the first one will stop once the maximum is reached.
EDIT:
It turns out that the subarray needs to be a continuous piece of the original, so changing the order of the original won't work since we need to make sure the result is a continuous slice of the original.
To do that, instead, just check each subslice of the array, and keep the best one:
let original = [74,659,931,273,545,879,924,710,441,166,493,43,988,504,328,730,841,613,304,170,710,158,561,934,100,279,817,336,98,827,513,268,811,634,980,150,580,822,968,673,394,337,486,746,229,92,195,358,2,154,709,945,669,491,125,197,531,904,723,667,550];
const maxsum = 22337;
function arraySum(arr) {
return arr.reduce((p, c) => p + c, 0);
}
// Double for loop will do the trick.
let bestSoFar = [];
for (let i = 0; i < original.length; i++) {
for (let j = i+1; j < original.length; j++) {
if (j-i > bestSoFar.length && arraySum(original.slice(i, j)) < maxsum) {
bestSoFar = original.slice(i, j);
}
}
}
console.log("Longest continuous subarray is:", bestSoFar.length);

A brute force approach is likely the best solution to this problem. Start at each entry and see how far you can go before arriving at a sum > i and if it's better than the best you've seen so far save it. I provided a sample Java solution below, I haven't actually run it so one or two of my indices may be off but I think you can get the gist. Runtime is O(n^2), memory is O(n) (both occurring with, for example, #getMaxSubArray(new int[] {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, 1000000))
private int[] getMaxSubArray(int[] a, int i) {
int minIndex = 0;
int maxIndex = -1;
for (int j = 0; j < a.length; j+=1) {
if (a[j] > i) {
continue;
}
int sum = a[j];
for (int k = j + 1; k < a.length; k++) {
if (sum + a[k] > i) {
if ((k - 1) - j > maxIndex - minIndex) {
maxIndex = k - 1;
minIndex = j;
}
break;
}
sum += a[k];
}
}
if (maxIndex - minIndex < 0) {
return null;
}
int[] result = new int[maxIndex - minIndex + 1];
for (int p = minIndex; p <= maxIndex; p += 1) {
result[p - minIndex] = a[p];
}
return result;
}

This is my solution. It will return the max length of the subarray. Can you please take a look, guys?
function maxLength(a, k) {
const sortedArray = a.sort((i,j) => i - j);
let sum = 0;
let length = 0;
const subArray = [];
for (let i=0; i < sortedArray.length; i++) {
sum = sum + sortedArray[i];
if (sum <= k) {
length++;
subArray.push(sortedArray[i]);
} else {
return length;
}
}
return length;
}

More efficient solution than many loop iterations with JavaScript

I have written an algorithm that returns the smallest common multiple of two integers that can be evenly divided by both, and can also be evenly divided by all sequential integers in the range between the initial two integers.
My algorithm works well for numbers where the smallest common multiple is a reasonably low number. For numbers where the smallest common multiple is a very large number (ex for [1,13] its 360360) my algorithm breaks because it is dependant on running more and more for loops.
I have posted my code below. It seems very long, but the problem section is near the beginning, and the rest of the code is added for context.
Is there a more efficient solution that running more and more loop iterations to return larger numbers?
// If mystery failure occurs, examine i & j loop limits.
function smallestCommons(arr) {
var firstArray = [];
var secondArray = [];
var commonNums = [];
var sequential = [];
// Sort the array from lowest to highest
arr.sort(function(a,b) {
return a-b;
});
// Assing arr values to vars so that original data stays intact
var array0 = arr[0];
var array1 = arr[1];
//console.log(arr);
// Find all multiples of both inputs.
// THIS IS THE PROBLEM SECTION
for (var i = 0; i < 1000000; i++) {
firstArray.push(arr[0] * (i+1));
}
for (var j = 0; j < 1000000; j++) {
secondArray.push(arr[1] * (j+1));
}
// THAT WAS THE PROBLEM SECTION
// Find common multiples.
for (var k = 0; k < firstArray.length; k++) {
for (var l = 0; l < secondArray.length; l++) {
if (firstArray[k] == secondArray[l]) {
commonNums.push(firstArray[k]);
} // End if.
} // End l loop.
} // End k loop.
// Find sequential numbers in range between input parameters.
for (var m = 0; m < array1; m++) {
if (array0 + 1 !== array1) {
sequential.push(array0 += 1);
} // End of if.
} // End of m loop.
// Find commonNums divisible by arr && sequential to produce a whole number.
// commonNums [ 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75]
// arr [1,5]
// sequential [ 2, 3, 4 ]
for (var n = 0; n < commonNums.length; n++) {
var match = true;
for (var o = 0; o < sequential.length; o++) {
for (var p = 0; p < arr.length; p++) {
if (commonNums[n] % arr[p] !== 0 || commonNums[n] % sequential[o] !== 0) {
match = false;
}
}
}
if(match) {
console.log(commonNums[n]);
}
} // End of n loop.
} // End function.
smallestCommons([1,13]);

What you do does not make a lot of sense and is really hard to read. All you need to know to solve this problem is a little bit of math. And this math tells you that:
lcm(a, b) = a * b / gcd(a, b)
lcm(a,b,c) = lcm(a,lcm(b,c))
which translates into something like this:
function gcd(a, b){
while (b !== 0) {
var tmp = a;
a = b;
b = tmp % b;
}
return a
}
function lcm(a, b){
return a * b / gcd(a, b);
}
function lcm_arr(arr){
var res = 1;
for (var i = 0; i < arr.length; i++){
res = lcm(res, arr[i]);
}
return res;
}
which gives you your results in less than a second:
lcm_arr([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13])

Sounds like a good place to use recursion, because you have a solid base case when the number difference is one (in your example smallestCommons([12,13]); or smallestCommons([1,2]); depending on the direction), and the rest (sequentials) would be incrementing the smaller or decrementing the bigger number in recursive calls. That could save you quite a few CPU cycles.

Implementing merge sort in Javascript

So I'm working on Khan Academy's Algorithms course, and am trying to implement a recursive merge sort in Javascript. Here is my code so far:
var mergeSort = function(array, p, r) {
if(r>p) {
var q = floor(r/2);
mergeSort(array, p, q);
mergeSort(array, q+1, r);
merge(array, p, q, r);
}
};
merge is a function provided by Khan Academy to merge the subarrays back together. It is giving me the error: 'Uncaught RangeError: Maximum call stack size exceeded'.
EDIT: More details: I am fairly sure the error is in my code, there code is purposefully obfuscated and unreadable because the user needs to implement it themselves in a later challenge.
Here is the code that actually calls the mergeSort function initially and declares the array:
var array = [14, 7, 3, 12, 9, 11, 6, 2];
mergeSort(array, 0, array.length-1);
println("Array after sorting: " + array);
Program.assertEqual(array, [2, 3, 6, 7, 9, 11, 12, 14]);
And here is the code for the merge function, although it is obfuscared as I mentioned above:
var merge = function(array, p, q, r) {
var a = [],
b = [],
c = p,
d, e;
for (d = 0; c <= q; d++, c++) {
a[d] = array[c];
}
for (e = 0; c <= r; e++, c++) {
b[e] = array[c];
}
c = p;
for (e = d = 0; d < a.length && e < b.length;) {
if (a[d] < b[e]) {
array[c] = a[d];
d++;
} else {
array[c] = b[e];
e++;
}
c++;
}
for (; d < a.length;) {
array[c] = a[d];
d++;
c++;
}
for (; e < b.length;) {
array[c] = b[e];
e++;
c++;
}
};
They also require my code inside of the mergeSort function be of the form:
if (____) {
var ____ = ____;
mergeSort(____,____,____);
mergeSort(____,____,____);
merge(____,____,____,____);
}

Mergesort is a divide and conquer algorithm which splits the range of indices to sort in two, sorts them separately, and then merges the results.
Therefore, middle variable should be the arithmetic mean of from and to, not the half of to.
I have renamed the variables to make it more understandable:
var mergeSort = function(array, from, to) {
if(to > from) {
var middle = Math.floor( (from+to)/2 ); // Arithmetic mean
mergeSort(array, from, middle);
mergeSort(array, middle+1, to);
merge(array, from, middle, to);
}
};

q is supposed to be the half way point between p and r, but you've failed to take into account that the starting point (i.e. p) might not be 0 when you do this:
var q = floor(r/2);
You need to do something like:
var q = floor((r-p)/2) + p;
Although as #Oriol points out the middle point is actually exactly the same as the arithmetic mean and so the calculation can be simplified.

Just for the sake of implementation of the merge sort in JS
function merge(arr){
if(arr.length <= 1) return arr;
let mid = Math.floor(arr.length/2);
let left = merge( arr.slice(0, mid));
let right = merge(arr.slice(mid))
function mergeSort(arr1, arr2) {
let result = [];
let i=0;
let j=0;
while(i< arr1.length && j < arr2.length) {
if(arr1[i] < arr2[j]){
result.push(arr1[i])
i++;
} else {
result.push(arr2[j])
j++;
}
}
while(i < arr1.length) {
result.push(arr1[i])
i++;
}
while(j < arr2.length){
result.push(arr2[j])
j++;
}
return result
}
return mergeSort(left,right)
}
console.log(merge([1,4,3,6,2,11,100,44]))

High level strategy
the merge sort works on the principle that it's better to sort two numbers than to sort a large list of numbers. So what it does is that it breaks down two lists into their individual numbers then compares them one to the other then building the list back up. Given a list of say 1,3,2, it'll split the list into 1 and 3,2 then compare 3 to 2 to get 2,3 and then compare the list of 2,3 to 1. If 1 is less than the first element in list of 2,3 it simply places 1 in front of the list. Just like that, the list of 1,3,2 is sorted into 1,2,3.
pseudocode-steps
1.take first member of first array
2.compare to first member of second array
3.if first member of first array is less than first member
of second array
4.put first member into sorted array
5.now merge first array minus first element
with second array
6.else take first member of second array
merge first array with remaining portion of second array
7.return sorted array
function mergesorted(list1, list2) {
let sorted = [];
if (list1.length === 0 || list2.length === 0) {
return list1.length === 0 ? list2.length === 0 ? [] : list2 : list1;
}
if (list1[0] < list2[0]) {
sorted.push(list1[0])
return sorted.concat(mergesorted(list1.slice(1), list2))
} else {
sorted.push(list2[0])
return sorted.concat(mergesorted(list1, list2.slice(1)))
}
}
console.log(mergesorted([1, 2], [3, 4])) //should: [1,2,3,4]
console.log(mergesorted([1,2], [3])) //should: [1,2,3]

Merge sort implementation, stable and in place
function sort(arr, start, end) {
if (start >= end-1) return;
var mid = start + ~~((end-start)/2);
// after calling this
// left half and right half are both sorted
sort(arr, start, mid);
sort(arr, mid, end);
/**
* Now we can do the merging
*/
// holding merged array
// size = end-start
// everything here will be copied back to original array
var cache = Array(end-start).fill(0);
var k=mid;
// this is O(n) to arr[start:end]
for (var i=start, r=0;i<mid;r++,i++) {
while (k<end && arr[k] < arr[i]) cache[r++] = arr[k++];
cache[r] = arr[i];
}
// k marks the position of the element in the right half that is bigger than all elements in the left
// effectively it tells that we should only copy start~start+k element from cache to nums
// because the rests are the same
for (var i=0;i<k-start;i++) arr[i+start]=cache[i];
}

A nice solution:
const merge = (left, right) => {
const resArr = [];
let leftIdx = 0;
let rightIdx = 0;
while (leftIdx < left.length && rightIdx < right.length) {
left[leftIdx] < right[rightIdx]
? resArr.push(left[leftIdx++])
: resArr.push(right[rightIdx++]);
}
return [...resArr, ...left.slice(leftIdx), ...right.slice(rightIdx)];
};
const mergeSort = arr =>
arr.length <= 1
? arr
: merge(
mergeSort(arr.slice(0, Math.floor(arr.length / 2))),
mergeSort(arr.slice(Math.floor(arr.length / 2)))
);

try this
const mergeTwoSortedArr = (arr1 = [], arr2 = []) => {
let i = 0,
j = 0,
sortedArr = [];
while(i < arr1.length || j < arr2.length) {
if(arr1[i] <= arr2[j] || (i < arr1.length && j >= arr2.length)) {
sortedArr.push(arr1[i]);
i++;
}
if(arr2[j] <= arr1[i] || (j < arr2.length && i >= arr1.length)) {
sortedArr.push(arr2[j]);
j++;
}
}
return sortedArr;
}
const mergeSort = (arr) => {
if(arr.length === 0 || arr.length === 1) {
return arr;
}
var mid = Math.floor(arr.length / 2);
var left = mergeSort(arr.slice(0, mid));
var right = mergeSort(arr.slice(mid));
return mergeTwoSortedArr(left, right);
}
when trying this
mergeSort([5, 2, 1, 4]) //[1, 2, 4, 5]
Time Complexity O(nlogn)
logn --> decomposition
n comaprison
Time complexity O(nlogn)