When n <= 0 sum(arr, n) returns 0 . What I don’t understand is why should sum be equal to zero, if an array can be non-zero? Sure, n is 0 in this case… but is it really a sum if an array is, actually, say 5? Wouldn’t sum be 0+5=5 in that case? Here’s a link to the challenge: [image] …

[image] SergeSolkatten: Wouldn’t sum be 0+5=5 in that case? Yes, in the end the sum will be 0 + 5 = 5. Only the case when n will get to zero or less will return 0 but this will be summed up with the rest. Let’s say we have an array [1, 2, 3] and we pass it to our function sum(arr, n) like so…

sum(arr, n) here, n refers to the index of arr… freeCodeCamp Challenge Guide: Replace Loops using Recursion When n is larger than 0 sum(arr, n) returns sum(arr, n - 1) + arr[n - 1] when n==1 it’ll return first element + zero(n-1==0), thus terminating the recursion

This was a perfect explanation! I had the right answer, but was still not getting how it worked. This made everything click. Thanks!

could you write the explanation with ([2,3,4,5],3)

why don’t you try it yourself? we can tell you how well you are doing

this explained it well - thank you

Hi @ILM …I’m going to try! function sum(arr, n) { if n<=0 { return 0; } else { return sum(arr, n-1) + arr[n-1]; } } for the base case, where n<=0, it returns 0 - so whenever n gets to this point, it returns zero and stops. for larger values of n (that is, greater than or equal to 1) the func…

you get to the right result but you are missing what the single things mean [image] newby: sum([2,3,4,5] , 2) + 2 this is not correct the return statement of sum([2,3,4,5],3) is sum([2,3,4,5],2 + arr[2] and arr[2] is equal to 4 so that line is sum([2,3,4,5],2) + 4 see if you can correct t…

I have had the same problem but my first approach was using another approach where you actually start counting numbers from 0,1,2,3 so this will equal to [2,3,4,5]. This does not add up but it sounds more logical since we are dealing with arrays. What do you think???

ok… I think I’ve got it now (if I’m right, what I was missing is really obvious now) sum ([ 2, 3, 4, 5] , 3) return sum ([ 2, 3, 4, 5] , 2) + arr[2] //result is 5 + 4 (because the 2nd index of arr is 4) sum (([2,3,4,5], 1) + arr[1]) + 4 //result is 2 + 3 + 4 sum ((([2,3,4,5], 0) + arr[0]) +…

for me, if I finally got it right, it came down to the arr[ n-1] part. I was totally misunderstanding that part, and once I got it…well, it all fell into place - regardless of what numbers you pass in. If I got it wrong, however…then I’m back to square one

An unclear thing about one challenge

JavaScript

mukeshgurpude October 11, 2020, 6:49am 2

sum(arr, n) here, n refers to the index of arr…

when n==1 it’ll return first element + zero(n-1==0), thus terminating the recursion

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An unclear thing about one challenge

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