We Keep Coding

Finding the sum of a "counter" variable loop that ran ten times then was pushed into the "numbers" array. Each way I tried resulted with a list

I'm asking for help to find the sum of an array with elements that were pushed from a counter variable that had previously looped 10 times. I'm new to Javascript and was practicing for an assessment, and I've tried several different ways to do it and have only resulted with just a list of the elements within the numbers array.
var counter = 10;
var numbers = [];
for (i = 1; i <= 10; i ++) {
counter = [i + 73];
numbers.push(counter);
}
console.log(numbers);
function sum(arr) {
var s = 0;
for(var i = 0; i < arr.length; i++) {
s = s += arr[i];
}
return s;
}
console.log(sum([numbers]));
function getArraySum(a) {
var total = 0;
for (var i in a) {
total += a[i];
}
return total;
}
var numbers = getArraySum([numbers]);
console.log(numbers);

you should push only the value of counter without the brackets and then make a reduce to have the sum of each number in the array
var counter = 10;
var numbers = [];
for (i = 1; i <= 10; i++) {
counter = i + 73;
numbers.push(counter);
}
console.log(numbers.reduce((a,b) => a+b));

You had a couple of typos in the code:
Typos
You were wrapping the sum in square brackets:
counter = [i + 73];
You should just remove the brackets like:
counter = i + 73;
2. You were wrapping a value that is already an array in square brackets while passing it as an argument to a function:
sum( [numbers] )
// ...
getArraySum( [numbers] );
You should remove the brackets, like this:
sum( numbers );
// ...
getArraySum( numbers );
Fix
I updated the code that you shared to fix the above-mentioned things:
var numbers = [];
// Loop 10 times and push each number to the numbers array
for (var i = 1; i <= 10; i ++) {
var sumNumbers = i + 73;
numbers.push(sumNumbers);
}
console.log(numbers);
function sum(arr) {
var total = 0;
for(var i = 0; i < arr.length; i++) {
total += arr[i];
}
return total;
}
// Call the function by passing it the variable numbers, holding an array
var result1 = sum(numbers);
console.log( result1 );
function getArraySum(a) {
var total = 0;
for (var i in a) {
total += a[i];
}
return total;
}
var result2 = getArraySum(numbers);
console.log(result2);

JavaString loop through string and break string into sub strings

I want to loop through set of strings and break them into sub strings by removing two characters and storing them into an array.
lets say two stings are returned, the number of characters will always be multiple of 2.
AADDFFCC
GGDD
The first string will give me AADDFF,AADD,AA
The second string will give me GG.
i want to store all sub string into a single array.
So based upon above example i should end up with.
subString = ["AADDFF","AADD","AA","GG"]
This is my incomplete attempt.
var StringArray = ["AADDFFCC","GGDD"] //this could be indefinite length
var subString = [];
var StringArrayLength = StringArray.length;
var loopCurrentString;
var loopCurrentSubString
for (var i = 0; i < StringArrayLength; i += 2) {
loopCurrentString = StringArray[i];
loopCurrentSubString = loopCurrentString.substring(0, loopCurrentString.length - 2);
}

for (var i = 0; i < StringArrayLength; i ++) {
//get element
loopCurrentString = StringArray[i];
//add substrings in the array
while(loopCurrentString.length>2){
loopCurrentSubString = loopCurrentString.substring(0, loopCurrentString.length - 2);
substring.push(loopCurrentSubString);
}
}

Here is an implementation with nested for loops:
const data = ["AADDFFCC", "GGDD"]
const endResult = []
for (var i = 0; i < data.length; i++) {
for (var j = 0; j < data[i].length; j += 2) {
if (j != 0)
endResult.push(data[i].slice(0, data[i].length - j))
}
}
console.log(endResult)
You could do this by using reduceRight to chunk the array into the desired pieces starting from the right side:
const data = ["AADDFFCC","GGDD"]
const chunkBy = (arr, by=2) => [...arr].reduceRight((r,c,i,a) =>
((((a.length-1) - i)%by == 0 && a.length-1 != i) ?
r.push(a.slice(0, i+1).reduce((r,c) => `${r}${c}`)) :
0, r),[])
console.log(data.reduce((r,c) => [...chunkBy(r), ...chunkBy(c)]))
And on the end stitch them together via ES6 spread operator.

JavaScript neglecting the else statement

I created a function which takes in two values..
Both are numbers represented by n & p. What the function does is that it gets the number n and splits it up then squares it to the value of p and sums them in an increasing order like this: n^p + n^(p+1) + n^(p+2) + ...
Here is the function
function digPow(n, p) {
// ...
let num = n.toString();
let pow = p;
let arrn = [];
let arrp = [];
for (let i = 0; i < num.length; i++) {
arrn.push(JSON.parse(num[i]));
}
let index = arrn.join('');
let sindex = index.split('');
for (let j = 0; j < sindex.length; j++) {
let power = p + j;
let indexs = sindex[j];
let Mathpow = Math.pow(indexs, power);
arrp.push(Mathpow);
}
let total = 0;
for (let m in arrp) {
total += arrp[m]
}
let secondVal = total / n;
let totals = total / secondVal;
let mx = [-1]
if (totals.length == n.length) {
return secondVal
} else {
return -1
}
}
Now i created variables and arrays to store up the values and then the if part is my problem.. The if/else statement is meant to let the program check if a particular variable totals is equal to n which is the input.. if true it should return a variable secondVal and if not it should return -1..
So far its only returning secondVal and i'snt returning -1 in cases where it should like:
digPow(92, 1) instead it returns 0.14130434782608695
What do i do?

totals and n are both numbers. They don't have a .length property, so both totals.length and n.length evaluate to undefined. Thus, they are equal to each other.
There are plenty of other weird things going on in your code, too. I'd recommend finding a good JavaScript tutorial and working through it to get a better feel for how the language (and programming in general) works.

Let's start by stripping out the redundant variables and circular-logic code from your function:
function digPow(n, p) {
let num = n.toString();
// let pow = p; // this is never used again
// let arrn = []; // not needed, see below
// let arrp = []; // was only used to contain values that are later summed; can instead just sum them in the first place
// this does the same thing as num.split(''), and isn't needed anyway:
//for (let i = 0; i < num.length; i++) {
// arrn.push(JSON.parse(num[i]));
//}
// this is the same as the original 'num' variable
// let index = arrn.join('');
// This could have been num.split(), but isn't needed anyway
// let sindex = index.split('');
let total = 0; // moved this line here from after the loop below:
for (let j = 0; j < num.length; j++) { // use num.length instead of the redundant sindex
let power = p + j;
// The only reason for the sindex array was to get individual characters from the string, which we can do with .charAt().
//let indexs = sindex[j];
let indexs = num.charAt(j);
let Mathpow = Math.pow(indexs, power);
//arrp.push(Mathpow); // No need to collect these in an array
total += Mathpow; // do this instead
}
// No need for this loop, since we can sum the total during the previous loop
// let total = 0;
//for (let m in arrp) {
// total += arrp[m]
//}
let secondVal = total / n;
// let totals = total / secondVal;
// The above is the same thing as total / total / n, which is:
let totals = 1/n;
// This is never used
//let mx = [-1]
// This was totals.length and n.length, which for numbers would always be undefined, so would always return true
if (totals == n) {
return secondVal
} else {
return -1
}
}
So the above reduces to this functionally identical code:
function digPow(n, p) {
let num = n.toString();
let total = 0;
for (let j = 0; j < num.length; j++) {
let power = p + j;
let indexs = num.charAt(j);
let Mathpow = Math.pow(indexs, power);
total += Mathpow;
}
let secondVal = total / n;
let totals = 1 / n;
if (totals == n) {
return secondVal
} else {
return -1
}
}
Now let's talk about the logic. The actual output will always be -1, unless the input is 1, due to what's clearly a logic error in the totals variable: the only case where 1/n == n is true is when n==1.
Setting that aside, and looking only at the secondVal variable, some examples of what it's calculating for a given input would be
digPow(123,1) --> (1^1 + 2^2 + 3^3) / 123 --> 14/123
digPow(321,2) --> (3^2 + 2^3 + 1^4) / 321 --> 21/321
digPow(92, 1) --> (9^1 + 2^2) / 92 --> 13/92
I'm pretty sure from your description that that's not what you intended. I'm not at all sure from your description what you did intend, so can't be much help in correcting the function beyond what I've done here.
What I'd suggest is to sit down and think through your algorithm first; make sure you know what you're trying to build before you start building it. There were some syntax problems with your code, but the real issues are with the logic itself. Your original function shows clear signs of "just keep throwing more lines of code at it until something happens" rather than any planned thinking -- that's how you wind up with stuff like "split a string into an array, then join it back into a string, then split that string into another array". Write pseudocode first: break the problem down into steps, think through those steps for some example inputs and make sure it'll produce the output you're looking for. Only then should you bust out the IDE and start writing javascript.

How to get string from odd's array ?

I need help... I'm learning JavaScript, and it seems easy, but I may just be overlooking... everything... My problem is I need to return a string of all the even numbers between 0 and num inclusive; ex 7 gets you 0246, etc. I've gotten:
function stringOfEvens(num) {
for (let i = 0; i <= num.length; ++i) {
if (i % 2 === 0 ); {
return i;
}
}
}
I know the whole outlook is to run a for loop that goes from 0 to the number in question appending each time and if that number % 2 = 0, return that number, but something is a miss... I may even be overthinking and approaching this whole thing wrong... It is figuratively driving me mad...

function theJob(limit) {
var res = ''; //initialize as a string so that the other numbers will be appended instead of added
for (i = 0; i <= limit; i += 2) { // step increase by 2 to skip odd numbers
res += i;
}
return res; // returning the resulting string
}
console.log(theJob(10));

You can do this without using the modular function, by simply starting the loop at zero and incrementing by 2 each time the loop iterates...
function stringOfEvens(num) {
var result = "";
for (let i = 0; i <= num; i += 2) {
result += i; // append this even number to the result string
}
return result;
}
console.log(stringOfEvens(10));

You're returning the first number you find. a better approach would be to build the string in the loop then return after that string after the loop. Also no need for num.length if num is an int.
function stringOfEvens(num) {
var stringToReturn = "";
for (let i = 0; i <= num; i++) {
if (i % 2 === 0) {
stringToReturn = stringToReturn + i;
}
}
return stringToReturn;
}

function stringOfEvens(num) {
var str= ' ';
for(var i = 0; i <= num; i = i + 2){
str += i;
}
return str;
}
console.log(stringOfEvens(10))

Just for fun, since it's not particularly efficient:
const stringOfEvens = n => Array(~~(n/2)+1).fill().map((_, i) => 2*i).join("")
or annotated:
const stringOfEvens = n => // arrow function definition, returning:
Array(~~(n / 2) +1) // a sparse array of the required length
.fill() // made unsparse, so .map works
.map((_, i) => 2 * i) // then containing the even numbers
.join("") // and converted to a string
or alternatively (and more efficiently) using Array.from which can call an explicit map function in place:
const stringOfEvens = n => Array.from(Array(~~(n/2)+1), (_, i) => 2*i).join("")

Splitting Numbers and adding them all - JavaScript

I have a function that returns the sum of all its digits For both POSITIVE and NEGATIVE numbers.
I used split method and converted it to string first and then used reduce to add them all. If the number is negative, the first digit should count as negative.
function sumDigits(num) {
var output = [],
sNum = num.toString();
for (var i = 0; i < sNum.length; i++) {
output.push(sNum[i]);
}
return output.reduce(function(total, item){
return Number(total) + Number(item);
});
}
var output = sumDigits(1148);
console.log(output); // --> MUST RETURN 14
var output2 = sumDigits(-316);
console.log(output2); // --> MUST RETURN 4
Instead of returning the sum, it returned 4592 -1264
Am I doing it right or do I need to use split function? Or is there any better way to do this?
Sorry newbie here.

I think you'll have to treat it as a string and check iterate over the string checking for a '-' and when you find one grab two characters and convert to an integer to push onto the array. Then loop over the array and sum them. Of course you could do that as you go and not bother pushing them on the array at all.
function sumDigits(num) {
num = num + '';
var output = [];
var tempNum;
var sum = 0;
for (var i = 0; i < num.length; i++) {
if (num[i] === '-') {
tempNum = num[i] + num[i + 1];
i++;
} else {
tempNum = num[i];
}
output.push(parseInt(tempNum, 10));
}
for (var j = 0; j < output.length; j++) {
sum = sum + output[j];
}
return sum;
}
var output = sumDigits(1148);
console.log(output); // --> MUST RETURN 14
var output2 = sumDigits(-316);
console.log(output2); // --> MUST RETURN 4