gilded breach
https://www.codewars.com/kata/514b92a657cdc65150000006/train/javascript
Finish the solution so that it returns the sum of all the multiples of 3 or 5 below the number passed in. Additionally, if the number is negative, return 0 (for languages that do have them).
Note: If the number is a multiple of both 3 and 5, only count it once.```
```function solution(number){
}```
my first thought is to make an if statement
but i don't know what key words to use
do something...
}```but i don't think that's right
i think i need to use like a '%' ?
random kestrel
i think i need to use like a '%' ?
yup
vivid jewel
Looks right to me
gilded breach
but how do i nest all possible numbers into an array that's derived from 'number'
vivid jewel
Looks right to me
For example you get even numbers from number % 2 == 0 so I'm guessing its similar
gilded breach
yeah but that checks if it's true
what is the next step to list all possible division results?
random kestrel
What does your code look like?
gilded breach
no code yet
gilded breach
if (number%5==0){
let numArr[]
}
}```and generate numbers from 1 to number
gilded breach
yeah i tried a for loop, but i have no idea how to set that up
function solution(number){
for (let i = 0; ;i++){
}
}
for (let i = 0; ;i++){
}
}``` console.log(i)
}
for (let i = 0; i%5==0 || i%3==0 ;i++){
}
}```this super confuses me
why would i count from 0 up to number?
random kestrel
That just means make a variable i that is 0, if i is less than or equal to number, increment i
gilded breach
oh i guess i need to do the check
gilded breach
That just means make a variable `i` that is 0, if `i` is less than or equal to `...
i understand what it does
trying to figure out how to use it
random kestrel
That will give you all the number from 0 to number. Now, you would use that to filter the numbers that are divisible by 3 or 5
gilded breach
for (let i = 0; i<=num; i++){
if (i%5==0){
i.push();
}
}
}```something like that?
but i need to make an array to push the numbers into
gilded breach
```i.push();```
const solArr[];
for (let i = 0; i<=num; i++){
if (i%5==0){
solArr.push(i);
}
}
}``` like this?const solArr = []
gilded breach
that's what i put
oh lol
haha thanks
const solArr=[];
for (let i = 0; i<=num; i++){
if (i%5==0){
solArr.push(i);
} else if (i%3==0){
solArr.push(i);
}
}
}```so now how do i add up all the numbers inside the array?
random kestrel
You can combine the if conditions using the OR (||) operator
like you showed us before
gilded breach
You can combine the if conditions using the OR (`||`) operator
ty ty
okay so how do i add up all the numbers inside the array?
random kestrel
think how you would do it in real life
or think how you would explain it to a child
because that's pretty much what a computer is, but faster 😄
gilded breach
i need another loop statement?
random kestrel
yes
gilded breach
const solArr=[];
for (let i = 0; i<=num; i++){
if (i%5==0 || i%3==0){
solArr.push(i);
}
}
for (let i = 0; i<=...; i++){
}
}``` const solArr=[];
for (let i = 0; i<=num; i++){
if (i%5==0 || i%3==0){
solArr.push(i);
}
}
for (let i = 0; i<=solArr.length - 1; i++){
}
}```hint: my_array[0] gives you the first element, my_array[1] gives you the second element in the array
try to print all the elements in the array first
gilded breach
const solArr=[];
for (let i = 0; i<=num; i++){
if (i%5==0 || i%3==0){
solArr.push(i);
}
}
for (let i = 0; i<=solArr.length - 1; i++){
return solArr[i]++;
}
}``` xDyou got the right idea, but not quite haha
return will end the function immediately
gilded breach
wait i need to create another array?
random kestrel
return solArr[i]++; This will get the element at i index, increment it and return it
random kestrel
wait i need to create another array?
no
gilded breach
okay good
random kestrel
why do you think you need another array?
gilded breach
i don't know how to pull the numbers out
random kestrel
```return solArr[i]++;``` This will get the element at `i` index, increment it a...
👀
gilded breach
it's cycling through the array now
but where do i put the individual values
so that i can add them all
random kestrel
```function solution(number){
const solArr=[];
for (let i = 0; i<=num; i++){...
Is this your code right now?
gilded breach
const solArr=[];
for (let i = 0; i<=num; i++){
if (i%5==0 || i%3==0){
solArr.push(i);
}
}
for (let i = 0; i<=solArr.length - 1; i++){
return solArr[i] ;
}
}``` this is what i have rnyeah i'm trying to end the function at this point
add all numbers at a clean last line
random kestrel
but you haven't summed the numbers
gilded breach
i'm trying to sum them all together in the return
random kestrel
you can't do it like this
gilded breach
okay i see
random kestrel
try printing the numbers in the array
gilded breach
random kestrel
it should number instead of num
random kestrel
cool
gilded breach
it works!
random kestrel
almost there
gilded breach
almost >.<
random kestrel
how would you add them?
gilded breach
i need to take each number out individually
then add them all together
i guess i need a for loop?
i have no idea
random kestrel
Think like a computer. You can only look at one number at a time. Whereas humans can look at the whole thing. Now, let's say you have numbers 1, 4, 8. Now,
pick the first number, 1
add it to your total (1)
pick the second number, 4
add it to your total (5)
pick the third number, 8
add it to your total (13)
total is 13
total = solArr[i]
random kestrel
total = total + solArr[i]
gilded breach
solArr[i] + solArr[i];
random kestrel
solArr[i] + solArr[i];
that means you are adding the same number twice
gilded breach
const solArr=[];
for (let i = 0; i<=number; i++){
if (i%5==0 || i%3==0){
solArr.push(i);
}
}
for (let i = 0; i<=solArr.length - 1; i++){
let total = solArr[i];
solArr[i]++;
}
}```^^
gilded breach
what about this
const solArr=[];
for (let i = 0; i<=number; i++){
if (i%5==0 || i%3==0){
solArr.push(i);
}
}
for (let i = 0; i<=solArr.length - 1; i++){
let total = solArr[i];
return solArr[i]++;
}
}``` const solArr=[];
for (let i = 0; i<=number; i++){
if (i%5==0 || i%3==0){
solArr.push(i);
}
}
for (let i = 0; i<=solArr.length - 1; i++){
let total = solArr[i];
return total = total + solArr[i];
}
}```That code is basically incrementing the element at first index and returning it
gilded breach
That code is basically incrementing the element at first index and returning it
i don't understand what this means
the increment will be the next in line in the array
random kestrel
return ends the function right there. It won't let the for loop finish. So, you never reach the end of the array. You see the first element and quit
gilded breach
const solArr=[];
for (let i = 0; i<=number; i++){
if (i%5==0 || i%3==0){
solArr.push(i);
}
}
for (let i = 0; i<=solArr.length - 1; i++){
let total = total + solArr[i];
}
return total;
}```that's recurssion right?
random kestrel
No, recursion is when a function calls itself
gilded breach
oh so i'm trying to make a variable call itself, which it wont do?
random kestrel
it is fine. The right side of the assignment is executed first
and is assigned to whatever is on the left
gilded breach
random kestrel
it is fine. The right side of the assignment is executed first
yeah, that's because ^^
It is trying to execute total + solArr[i], but it doesn't know what total is
gilded breach
const solArr=[];
for (let i = 0; i<=number; i++){
if (i%5==0 || i%3==0){
solArr.push(i);
}
}
for (let i = 0; i<=solArr.length - 1; i++){
let total = solArr[i];
total = total + solArr[i];
}
return total;
}```This will reset total on each iteration
gilded breach
i have to define total outside?
random kestrel
yes!
gilded breach
const solArr=[];
let total = 0;
for (let i = 0; i<=number; i++){
if (i%5==0 || i%3==0){
solArr.push(i);
}
}
for (let i = 0; i<=solArr.length - 1; i++){
total = total + solArr[i];
}
return total;
}```it should work...
random kestrel
You did pretty good, actually
random kestrel
I just gave you some hints, and you figured it out
random kestrel
I barely have any free time 🥲
random kestrel
great
random kestrel
👍
gilded breach
oh wow
there's way simpler solutions
astral cape var sum = 0;
for(var i = 1;i< number; i++){
if(i % 3 == 0 || i % 5 == 0){
sum += i
}
}
return sum;
}```i want to be able to code like this
random kestrel
return sum(filter(lambda x: not x % 3 or not x % 5, range(3, number)))``` This is my solution in pythonThere are people who did better than me haha
random kestrel
there's way simpler solutions
looks like I overcomplicated your solution haha
gilded breach
honestly this second year has been more clarity of direction for me
random kestrel
how do i get good like that
practice and consistency
gilded breach
it's so hard to find out where to start
random kestrel
you can also learn from others
gilded breach
like i know exactly what to work on now cuz i know where it's all applicable, but i just suck at coding lol
i would like to get a job soon
gilded breach
looks like I overcomplicated your solution haha
honestly you broke it down for me, which led to my solution. now that we know what the solution is, we can probably condense it further if I had a better understanding of the language
i don't think it can be condensed any further though
what do you think?
random kestrel
I mean, JS has built in methods that will let you do stuff like that in one or two lines.
gilded breach
the one i showed you was the most upvoted
random kestrel
return number < 1 ? 0 : [...new Array(number).keys()].filter(n => n % 3 == 0 || n % 5 == 0).reduce((a, b) => a + b);
}```This is a one line solution
gilded breach
oh yeah
random kestrel
You can make it even shorter
gilded breach
i don't even know what's going on in that one
random kestrel
solution = number => {number < 1 ? 0 : [...new Array(number).keys()].filter(n => n % 3 == 0 || n % 5 == 0).reduce((a, b) => a + b);}
That should work too...
gilded breach
this is ecs6 or whatever right?
random kestrel
yeah
gilded breach
so wouldn't it be better to use the foundational method?
random kestrel
yeah, that's more readable too
gilded breach
also what are the processing speeds of the variances?
random kestrel
That has to be tested
unkempt yarrow
Nah chaining methods look better imo. You just need to format them well
unkempt yarrow
Btw i think there is an O(1) solution for this
Gimme a sec to lay it out in my phone cus lazy to program it properly with pc
random kestrel
Btw i think there is an O(1) solution for this
You can do it mathematically