Develop Reference

Loops & Control Flow

When I run this I can't seem to get the rest of the values.
Write a function mergingTripletsAndQuints which takes in two arrays as arguments. This function will return a new array replacing the elements in array1 if they are divisible by 3 or 5. The number should be replaced with the sum of itself added to the element at the corresponding index in array2.
function mergingTripletsAndQuints(array1, array2) {
let result = [];
let ctr = 0;
let x = 0;
for (let i = 0; i < array1.length; i++) {
for (let j = 0; j < array2.length; j++) {
ctr = array1[i] + array2[j];
if (ctr % 3 === 0 || ctr % 5 === 0) {
result.push(ctr);
} else {
return array1[i];
}
}
}
return result;
}
console.log(mergingTripletsAndQuints([1, 2, 3, 4, 5, 15], [1, 3, 6, 7, 8, 9])); // expected log [1, 2, 9, 4, 13, 24]
console.log(mergingTripletsAndQuints([1, 1, 3, 9, 5, 15], [1, 2, 3, 4, 5, 6])); // expected log [1, 1, 6, 13, 10, 21]
It is only logging [1], [1]

I'm not sure, but I suppose there is a typo returning array1[i] in nested loop. I suppose you mean result.push(array1[i]) instead.
I think it should be something like this:
function mergingTripletsAndQuints(array1, array2) {
let result = [];
for (let i = 0; i < array1.length; i++) {
if (array1[i]% 3 === 0 || array1[i]% 5 === 0) {
result.push(array1[i] + array2[i]);
} else {
result.push(array1[i]);
}
}
return result;
}
console.log(mergingTripletsAndQuints([1, 2, 3, 4, 5, 15], [1, 3, 6, 7, 8, 9])); // expected log [1, 2, 9, 4, 13, 24]
console.log(mergingTripletsAndQuints([1, 1, 3, 9, 5, 15], [1, 2, 3, 4, 5, 6])); // expected log [1, 1, 6, 13, 10, 21]

A nested for loop is not necessary, look at this code:
function mergingTripletsAndQuints(array1, array2) {
let sum = [];
for (let i = 0; Math.max(i < array1.length, i < array2.length); i++) {
if (array1[i] % 3 == 0 || array1[i] % 5 == 0) {
sum.push(array1[i] + array2[i])
} else {
sum.push(array1[i])
}
}
return sum;
}

How to log how many possibilities fulfil the if statement javascript

const grades = [9, 8, 5, 7, 7, 4, 9, 8, 8, 3, 6, 8, 5, 6];
for (let i = 0; i < grades.length; i++) {
if (grades[i] >= 8) {
console.log(grades[i])
}
}
I'm trying to log how many items from the array fulfil the condition. the output I'm looking for is : 6 (because 6 of the numbers are equal or greater than 8)
tried
let count = 0;
for (let i = 0; i < grades.length; i++) {
if (grades[i]>= 8){
count++
console.log(count)
}
}

function countGreaterThan8(grades){
// initialize the counter
let counter = 0;
for (let i = 0; i < grades.length; i++) {
// if the condition satisfied counter will be incremented 1
if (grades[i] >= 8) {
counter++;
}
}
return counter;
}
const grades = [9, 8, 5, 7, 7, 4, 9, 8, 8, 3, 6, 8, 5, 6];
console.log(countGreaterThan8(grades)); // 6

You can call Array.filter to create a new array containing only items that fulfill the condition. You can then make use of the length of the array however you want. Like this
const grades = [9, 8, 5, 7, 7, 4, 9, 8, 8, 3, 6, 8, 5, 6];
const gradesThatPassCondition = grades.filter(grade => grade > 6);
console.log(gradesThatPassCondition.length);

Comparing Elements Values In An Array, and returning Boolean

Can anyone help I am nearly there I can get the code to return true for the first 2 test below, but not return false for the next 2 tests, what I need to do is to check that all the numbers in the array are in ascending order so every element number is greater than the last for the length of the array. Any ideas? Kind regards Jon.
Test.expect(inAscOrder([1, 2, 4, 7, 19])
Test.expect(inAscOrder([1, 2, 3, 4, 5])
Test.expect(!inAscOrder([1, 6, 10, 18, 2, 4, 20])
Test.expect(!inAscOrder([9, 8, 7, 6, 5, 4, 3, 2, 1])
function inAscOrder(arr) {
let result = false;
for (let i = 0; i <= arr.length; i++) {
for (let k = 0; k <= arr.length; k++) {
if (arr[i] < arr[k+1]) {
result = true;
}
}
}
return result;
}

You were very close to answer
function inAscOrder(arr) {
let result = true;
for (let i = 0; i <= arr.length; i++) {
if (arr[i] > arr[i+1]) {
result = false;
break;
}
}
return result;
}

Try this,
You can use the functional way instead.
const inAscOrder = arr => arr.slice(1).every((elem,i) => elem > arr[i]);
console.log(inAscOrder([1,2,5]));

The easiest way to compare two arrays, is to convert them to a strings and then compare the strings.
const isAscending = (arr) => arr.join("") == arr.sort((a, b) => a - b).join("");
console.log(isAscending([1, 2, 4, 7, 19]));
console.log(isAscending([1, 2, 3, 4, 5]));
console.log(isAscending([1, 6, 10, 18, 2, 4, 20]));
console.log(isAscending([9, 8, 7, 6, 5, 4, 3, 2, 1]));

For every element in the array, return true if it's the first element or it's greater than the previous element.
const isAscending = (arr) => arr.every((el, i, arr) => i === 0 || el > arr[i - 1]);
console.log(isAscending([1, 2, 4, 7, 19]));
console.log(isAscending([1, 2, 3, 4, 5]));
console.log(isAscending([1, 6, 10, 18, 2, 4, 20]));
console.log(!isAscending([9, 8, 7, 6, 5, 4, 3, 2, 1]));

How to summize every second number in an Array?

I'm having trouble summize every second/other number in an Array in Javascript. Any suggestions? (Using Norwegian functions, sorry for that!)
Such as: 2, 4, 6, 8, 10 = 30..
My function for the second/other number is
function tall(nummer) {
var tall = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15];
if (nummer == 5) {
partall = tall.filter((_,i) => i&1);
document.getElementById("tall").innerHTML = partall;
}
and for the final sum:
if (nummer == 9) {
partall = tall.filter((_,i) => i&1);
partall += tall;
document.getElementById("tall").innerHTML = partall;
}

Try the following:
var tall = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15];
let sum = 0;
for (let i = 0; i < tall.length; i += 2){
sum += tall[i];
}
console.log(sum)
Instead of loop over all the number you increment i by 2 thus only looping over the odd numbers.

The simplest way I can see is to use reduce (good for summing) and just don't add in the values for the indexes you don't want to contribute to the sum:
const tall = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15];
const sum = tall.reduce((sum, val, i) => sum + (i & 1 ? val : 0), 0);
console.log(sum);
or
const tall = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15];
const sum = tall.reduce((sum, val, i) => (i & 1 ? sum + val : sum), 0);
console.log(sum);

Array.prototype.reduce is certainly the way to go here:
var sum = [0,1,2,3,4,5,6,7,8,9,10].reduce((sum, n, i) => {
if (i % 2 === 0)
sum += n;
return sum;
}, 0);
console.log(sum);

Adding an if condition to a for loop

I'm trying to make it work on every number divisible by three.
here's my code:
var number = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
for(var i = 0; i<10 ; i++) {
if (i % 3 === 0) {
console.log(color[i]);
}
}

I think you need to use number[i] % 3 === 0. And what is the color there in your code? Change it into number.
var number = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
for(var i = 0; i<10 ; i++) {
if (number[i] % 3 === 0) {
console.log(number[i]);
}
}

Try this
var number = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
for(var i = 0; i<10 ; i++) {
if (i % 3 === 0) {
console.log(i);
}
}