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Javascript array.length returning double the value it's supposed to return. Why?

So, I'm learning Javascript through a book and it has some exercises. One of the exercises asks for you to build two functions, one that creates an array from two numbers provided in the arguments, and the other function has to sum all the numbers in the array. Here's my code:
let beg = 1;
let end = 3;
array = [];
sumNum = 0;
function range(begg, endd) {
for (let count = begg; count <= endd; count++) {
array.push(count);
}
return array;
}
console.log(range(beg, end));
function sum(arrayy) {
for (let i = 0; i <= arrayy.length - 1; i++) {
sumNum = arrayy[i] + sumNum;
console.log(sumNum);
}
console.log("\n")
console.log(arrayy.length - 1);
return sumNum / 2;
}
console.log(sum(range(beg, end)));
array2 = [1, 2, 3];
console.log("\n");
console.log(array2.length);
As I was solving the exercise I kept getting double the sum of all the numbers in the array. I started to print some information and discovered that my arrayy.length is returning double the value it's supposed to return and the loop runs double the times it should run.
Here's my output:
[ 1, 2, 3 ]
1
3
6
7
9
12
5
6
3
Sorry it this is a noob question, but my curiosity is killing me and I have not found anything on the internet, so why am I getting this result?
Thanks in advance.

As Ivan said: The "array" variable is global, so each time you call the range function you keep appending items to that shared array. You should add the array inside your function and return it. Other than that you did a pretty nice job!
function range(begg, endd) {
let array = []
for (let count = begg; count <= endd; count++) {
array.push(count);
}
return array;
}
Also: The sum function should have the "sumnum" variable inside the function to prevent it from increasing every time you call the function:
function sum(arrayy) {
let sumnum = 0;
for (let i = 0; i <= arrayy.length - 1; i++) {
sumNum = arrayy[i] + sumNum;
console.log(sumNum);
}
console.log("\n");
console.log(arrayy.length - 1);
return sumNum / 2;
}
remove the array and sumnum variables from the top of your code to get rid of the global variables.

Array.push() is not a function - attempting to push numbers to an array, but the array has changed type

I'm 3 months into Javascript, and find it has no respect for its own declarations.
I have declared an empty array, and want to push a number to it, so that it returns an array of numbers (i.e., separable). However, I am told .push() is not a function. Concat() will not work either, and of course += ruins the algorithm.
How do I get it to accept each next value into an array?
I have tried using 'new Array()', but it does not assist.
Similar to, but not the same as, Array.push throwing TypeError for array.push is not a function
In brief:
const fn = (n) => {
let factors = [];
let index = 1;
while (n % 2 == 0){
let out = 2 ** index;
factors.push(out);
index++;
n /= 2;
}
}
This returns:
Uncaught TypeError: factors.push is not a function
(I have left out a lot of code that does not affect the issue.)
Edit update: WIP. Apparently, the loop this is enclosed in has an effect, as do other variable declarations.
With any luck, I will return with a different question, having solved the initial problem.

Nested loops seems to have been the actual problem, which means others could not reproduce.
const fn = (m, n) => {
let factors = [];
let index = 1;
for (let i = m; i <= n; i++) {
while (i % 2 == 0){
let out = 2 ** index;
factors.push(out);
index++;
n /= 2;
}
}}
For unknown reasons, this causes a type mismatch. The declaration of the array appears to be the cause.
This problem has been abandoned as the initiating function was completed by other methods.

Here! its working well, what might went wrong on your side?
const fn = (n) => {
let index = 1;
let factors = [];
while (n % 2 == 0){
let out = 2 ** index;
factors.push(out);
index++;
n /= 2;
}
return factors
}
document.write(fn(400));

Javascript - For loop vs Linked List vs ES6 Set to find two matching integers

I have prepared 2 Javascript functions to find matching integer pairs that add up to a sum and returns a boolean.
The first function uses a binary search like that:
function find2PairsBySumLog(arr, sum) {
for (var i = 0; i < arr.length; i++) {
for (var x = i + 1; x < arr.length; x++) {
if (arr[i] + arr[x] == sum) {
return true;
}
}
}
return false;
}
For the second function I implemented my own singly Linked List, in where I add the complementary integer to the sum and search for the value in the Linked List. If value is found in the Linked List we know there is a match.
function find2PairsBySumLin(arr, sum) {
var complementList = new LinkedList();
for (var i = 0; i < arr.length; i++) {
if (complementList.find(arr[i])) {
return true;
} else {
complementList.add(sum - arr[i]);
}
}
return false;
}
When I run both functions I clearly see that the Linked List search executes ~75% faster
var arr = [9,2,4,1,3,2,2,8,1,1,6,1,2,8,7,8,2,9];
console.time('For loop search');
console.log(find2PairsBySumLog(arr, 18));
console.timeEnd(‘For loop search’);
console.time('Linked List search');
console.log(find2PairsBySumLin(arr, 18));
console.timeEnd('Linked List search');
true
For loop search: 4.590ms
true
Linked List search: 0.709ms
Here my question: Is the Linked List approach a real linear search? After all I loop through all the nodes, while my outer loop iterates through the initial array.
Here is my LinkedList search function:
LinkedList.prototype.find = function(data) {
var headNode = this.head;
if(headNode === null) {
return false;
}
while(headNode !== null) {
if(headNode.data === data) {
return true;
} else {
headNode = headNode.next;
}
}
return false;
}
UPDATE:
It was a good idea to go back and have another think of the problem based the comments so far.
Thanks to #nem035 comment on small datasets, I ran another test but this time with 100,000 integers between 1 and 8. I assigned 9 to the first and last position and searched for 18 to make sure the entire array will be searched.
I also included the relatively new ES6 Set function for comparison thanks to #Oriol.
Btw #Oriol and #Deepak you are right. The first function is not a binary search but rather a O(n*n) search, which has no logarithmic complexity.
It turns out my Linked List implementation was the slowest of all searches. I ran 10 iterations for each function individually. Here the result:
For loop search: 24.36 ms (avg)
Linked List search: 64328.98 ms (avg)
Set search: 35.63 ms (avg)
Here the same test for a dataset of 10,000,000 integers:
For loop search: 30.78 ms (avg)
Set search: 1557.98 ms (avg)
Summary:
So it seems the Linked List is really fast for smaller dataset up to ~1,000, while ES6 Set is great for larger datasets.
Nevertheless the For loop is the clear winner in all tests.
All 3 methods will scale linearly with the amount of data.
Please note: ES6 Set is not backward compatible with old browsers in case this operation has to be done client side.

Don't use this. Use a set.
function find2PairsBySum(arr, sum) {
var set = new Set();
for(var num of arr) {
if (set.has(num)) return true;
set.add(sum - num);
}
return false;
}
That's all. Both add and has are guaranteed to be sublinear (probably constant) in average.

You can optimize this substantially, by pre-sorting the array and then using a real binary search.
// Find an element in a sorted array.
function includesBinary(arr, elt) {
if (!arr.length) return false;
const middle = Math.floor(arr.length / 2);
switch (Math.sign(elt - arr[middle])) {
case -1: return includesBinary(arr.slice(0, middle - 1), elt);
case 0: return true;
case +1: return includesBinary(arr.slice(middle + 1), elt);
}
}
// Given an array, pre-sort and return a function to detect pairs adding up to a sum.
function makeFinder(arr) {
arr = arr.slice().sort((a, b) => a - b);
return function(sum) {
for (let i = 0; i < arr.length; i++) {
const remaining = sum - arr[i];
if (remaining < 0) return false;
if (includesBinary(arr, remaining)) return true;
}
return false;
};
}
// Test data: 100 random elements between 0 and 99.
const arr = Array.from(Array(100), _ => Math.floor(Math.random() * 100));
const finder = makeFinder(arr);
console.time('test');
for (let i = 0; i < 1000; i++) finder(100);
console.timeEnd('test');
According to this rough benchmark, one lookup into an array of 100 elements costs a few microseconds.
Rewriting includesBinary to avoid recursion would probably provide a further performance win.

first of all find2PairsBySumLog function is not a binary search, it's a kind of brute force method which parses all the elements of array and it's worst case time complexity should be O(n*n), and the second function is a linear search that' why you are getting the second method to run fastly, for the first function i.e. find2PairsBySumLog what you can do is initialize binary HashMap and check for every pair of integers in array kind of like you are doing in the second function probably like
bool isPairsPresent(int arr[], int arr_size, int sum)
{
int i, temp;
bool binMap[MAX] = {0};
for (i = 0; i < arr_size; i++)
{
temp = sum - arr[i];
if (temp >= 0 && binMap[temp] == 1)
return true;
binMap[arr[i]] = 1;
}
}

Most frequently occuring number (mode) in a list - want to get only the highest value

I'm trying to get whatever number is the most frequently occuring number in an array, so for an array containing 1,2,10,5,1 the result should be 1. The code I wrote returns me the frequency for each number, so 1 occurs twice, 2 occurs once, 10 occurs once etc. Any suggestions how I can fix my result?
function mode(arr) {
var uniqNum = {};
var numCounter = function(num, counter) {
if(!uniqNum.hasOwnProperty(num)) {
uniqNum[num] = 1;
} else {
uniqNum[num] ++;
}
};
arr.forEach(numCounter);
return uniqNum;
}

I've kept your code unchanged and added some extra statements. Here is the demo: http://codepen.io/PiotrBerebecki/pen/rrdxRo
function mode(arr) {
var uniqNum = {};
var numCounter = function(num, counter) {
if(!uniqNum.hasOwnProperty(num)) {
uniqNum[num] = 1;
} else {
uniqNum[num] ++;
}
};
arr.forEach(numCounter);
return Object.keys(uniqNum)
.sort((a,b) => uniqNum[b] - uniqNum[a]) // sort by frequency
.filter((val,ind,array) => uniqNum[array[0]] == uniqNum[val]) // leave only most frequent
.map(val => Number(val)); // convert text to number
}
console.log( JSON.stringify(mode([3,3,2,4,4])) ) // [3,4]
console.log( JSON.stringify(mode([2,4,3,3])) ) // [3]

I think it could be done only with a little modification to your forEach loop and the assistance of another auxiliary data structure:
function mode(arr) {
var freq = [], uniqNum = {}, i;
arr.forEach(function (num) {
uniqNum[num] = i = (uniqNum[num] || 0) + 1;
freq[i] = (freq[i] || []).concat(num);
});
return freq[freq.length - 1];
}
console.log(mode([1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6, 6, 7, 1, 6]));
With only one iteration over all the elements of the array we can gather enough information to print out the result:
uniqNum is the set you created to gather info about the element's frequency.
freq will be an array which last element will contain an array with the elements of higher frequency.
Fiddle. Hope it helps.

First we want to make an array where we count the number of occurrences of a certain value up to that point.
Then we use the reduce function to return an array of values read from the original array for the indexes whose values have the current max appearances. We redefine max and empty the final output array of modes (if new max is established) as we go along. We want this to be a collection in case there is a tie for maximum appearances.
Additional advantage of the below is that it doesn't require sort which is more expensive o(nlog n) and keeps the time complexity down to just linear. I also wanted to keep the functions used down to only two (map and reduce) as it is all that is need in this case.
edit: fixed a major bug uniqNum[e] += 1 instead of uniqNum[e] + 1 which went unnoticed as my initial case array was still returning expected result. Also made the syntax more concise in favor of more comments.
var arr = [1,2,10,5,1,5,2,2,5,3,3];
//global max to keep track of which value has most appearances.
var max = -1;
var uniqNum = {};
var modeArray = arr.map(function(e) {
//create array that counts appearances of the value up to that point starting from beginning of the input arr array.
if(!uniqNum.hasOwnProperty(e)) {
uniqNum[e] = 1;
return 1;
} else {
return uniqNum[e] += 1;
}
//reduce the above appearance count array into an array that only contains values of the modes
}).reduce(function (modes, e1, i) {
//if max gets beaten then redefine the mode array to only include the new max appearance value.
if(e1 > max){
//redefining max
max = e1;
//returning only the new max element
return [arr[i]];
//if its a tie we still want to include the current value but we don't want to empty the array.
}else if(e1 == max){
//append onto the modes array the co-max value
return[...modes, arr[i]];
}
return modes;
},[]);
alert(modeArray);
Here is a test you can run of my solution against #acontell. In my browser (Chrome with V8) my solution was around three-four times faster for arrays with large number of repeating values and even bigger advantage with distributions with lower number of repeating values. #acontell 's is a cleaner looking solution for sure, but definitely not faster in execution.
var arr = [];
for(var i=0; i < 100000; i++){
arr.push(Math.floor(Math.random() * (100 - 1)) + 1);
}
console.time("test");
test();
function test(){
var max = -1;
var uniqNum = {};
var modeArray = arr.map(function(e) {
//create array that counts appearances of the value up to that point starting from beginning of the input arr array.
if(!uniqNum.hasOwnProperty(e)) {
uniqNum[e] = 1;
return 1;
} else {
return uniqNum[e] += 1;
}
//reduce the above appearance count array into an array that only contains values of the modes
}).reduce(function (modes, e1, i) {
//if max gets beaten then redefine the mode array to only include the new max appearance value.
if(e1 > max){
//redefining max
max = e1;
//returning only the new max element
return [arr[i]];
//if its a tie we still want to include the current value but we don't want to empty the array.
}else if(e1 == max){
//append onto the modes array the co-max value
modes.push(arr[i])
return modes;
}
return modes;
},[]);
}
console.timeEnd("test");
console.time("test1");
test1();
function test1 () {
var freq = [],
uniqNum = {},
i;
arr.forEach(function(num) {
uniqNum[num] = i = (uniqNum[num] || 0) + 1;
freq[i] = (freq[i] || []).concat(num);
});
return freq[freq.length - 1];
}
console.timeEnd("test1");

I've tried as an exercise to solve this with native js functions.
var arr = [1,2,10,5,1];
// groupBy number
var x = arr.reduce(
function(ac, cur){
ac[cur]?(ac[cur] = ac[cur] + 1):ac[cur] = 1;
return ac;
}, {}
);
// sort in order of frequencies
var res = Object.keys(x).sort(
function(a,b){ return x[a] < x[b]}
);
res[0] has the most frequent element

proper use of the array.push method

if I have a simple test function that adds even numbers to an array:
function isEven(n){
var enumbers = [];
if (n % 2 == 0){
enumbers.push (n);
}
}
how can I increment my parameter until I have a set number of members in my array? for instance, I've tried this:
function isEven(n){
var enumbers = [];
while ( enumbers.length < 10){
if (n % 2 == 0){
enumbers.push (n);
}
console.log (enumbers);
n = n + 1;
isEven(n);
}
}
isEven(1);
but it seems to just create a new array for each number until it finally throws a range error (maximum call stack size exceeded).

It's creating that array multiple times because you're constantly calling that function with:
isEven(n);
You're also not comparing to the length of the array, just the array. Add .length to enumbers. Try changing to:
var enumbers = [];
while ( enumbers.length < 10){
if (n % 2 == 0){
enumbers.push (n);
}
console.log (enumbers);
}

I'm not sure if I understood your question.
But you shouldn't use global variables, and it is unnecessary to call your function recursively inside a while loop.
The error maximum call stack size exceeded is your browser trying to break a infinity loop.
This is what you need.
Examples here jsFiddle1 and jsFiddle2
function isEven(n) {
var enumbers = [];
while (enumbers.length < 10) {
if (n % 2 == 0) {
enumbers.push(n);
}
n++;
}
return enumbers;
}
Setup a test
var n = 1;
var evenArray = isEven(n); //call isEven function and it returns an array
document.body.innerHTML = evenArray; //2,4,6,8,10,12,14,16,18,20

The problem is that (enumber < 10) apparently always evaluates to true, causing an endless loop. But it is this comparison that is wrong, since you're comparing an integer with an array I think you're trying to get the array length?:
while (enumbers.length < 10) {
Another thing. enumbers is a local variable, so each call to isEven has it's own array. Therefore, the functions is called recursively, over and over again.

I suggest you create the array outside of is even method

I would have written something like:
function isEven(n,enumbers){
while(enumbers < 10){
if (n % 2 == 0){
enumbers.push (n);
}
console.log (enumbers);
n = n + 1;
isEven(n, enumbers);
}
}
var enumbers = [];
isEven(1,enumbers);