Basic JavaScript - Replace Loops using Recursion , Why n - 1?

JavaScript
1

Tell us what’s happening:
Describe your issue in detail here.
I was able to complete this challenge only by looking at the example given and substituting the multiply(*) operator with the sum(+) operator and changing the function name from multiply to sum. I can’t after much thought and several videos figure out what is exactly happening. Let’s use this example —
sum ([2, 3, 4, 5] , 3 ) — I know that what the problem asks is for the sum of the first ‘n’ elements in the array, which is in this case 3 so that means 2 + 3 + 4 = 9. I don’t understand how sum (arr, n-1) + arr [ n - 1 ]; gets me there. If I replace n with 3 in this case doesn’t that then leave me with sum (arr, 2) + arr[ 2 ] ? If I use 0 indexing doesn’t that equate to 4 + 4? I need to understand what is happening in sum( arr, n-1) + arr[ n - 1 ] when I switch in a “arr” and “n” as in the function sum(arr, n) ie. what does — ‘sum(arr, n-1)’ — represent and what does— 'arr[ n-1]-- represent. If someone could please help me understand this with a detailed explanation I would be very grateful. Thanks in advance.
Your code so far

function sum(arr, n) {
  // Only change code below this line
if (n <= 0) {
  return 0;
} else {
  return sum(arr, n-1) + arr[n - 1];
}
  // Only change code above this line
}

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Challenge: Basic JavaScript - Replace Loops using Recursion

Link to the challenge:

2

arr[2] does equal 4. So you can convert this to

sum(arr, 2) + 4

Now what does sum(arr, 2) equal?

3

I know it equals 5, but I don’t know why. Why does sum (arr, 2) equal 2+3 ? I also still don’t understand why we need to use n-1 ?

4

You are jumping the gun here.

Let’s review. You correctly said above that if n = 3 and so we call the function as sum(arr, 3) then it will return sum(arr, 2) + arr[ 2 ] and replacing arr[2] with its value gives you sum(arr, 2) + 4.

Then I asked you what sum(arr, 2) returns. I need you to do the same thing you did above for sum(arr, 3) except this time for sum(arr, 2).

What is the return statement for sum(arr, 2)?

5

I am assuming (arr, 1) , but if I take away 1 from (arr, 2) don’t I have to take away another 1 from arr[2] changing the number? is the “2” in (arr, 2) literally just the 2 elements “2” + “3”, so no 0 index.

6

Does (arr, n-1) keep repeating like so (arr, 3-1) ==2 (3) + 4, then (arr, 2-1)==1(2) +(3) + ([arr, 2] (4) ==9. (then (arr, n-1)(0) ==return 0? Essentialy [arr, n-1] stays at 4 and (arr, n-1) repeats until n==0?

7

You are still jumping ahead.

Step through the code one line at a time.

Only talk about sum(arr, 2). Don’t talk about anything else. What happens when we call sum(arr, 2)?

1 Like
8

it returns sum(arr, 1)

9

Does it? Only sum(arr, 1) and nothing else at all?

10

Ok, I left out the value of [arr, 2] which is 4 , so I’m left with sum(arr 2)+(arr, 1) + 4

11

I don’t know what this means. It does not look like the line of code I highlighted.

You started with sum([2, 3, 4, 5], 3 )

You said sum([2, 3, 4, 5], 3) will return sum([2, 3, 4, 5], 2) + 5.

Good!

Now, sum([2, 3, 4, 5], 2) returns what exactly?

12

sum([2, 3, 4, 5, ] , 2) + 4. with 0 index 4 will be 2nd index

13

Not quite, sum([2, 3, 4, 5], 2) cannot return sum([2, 3, 4, 5], 2) + 4 because then we’d be stuck making the same call back to sum([2, 3, 4, 5], 2) forever and ever and ever and ever.

14

I have to decrease “n” by 1 so sum([2 , 3, 4, 5] , 1) + 4

15

Ack! Indexing typo - but you are getting the pattern.

Lets fix that indexing typo

sum([2, 3, 4, 5], 3) will return sum([2, 3, 4, 5], 2) + 4

sum([2, 3, 4, 5], 2) will return sum([2, 3, 4, 5], 1) + 3

What does sum([2, 3, 4, 5], 1) return?

16

Ha ha, yeah that typo had my head spinnin for a minute, so we are left
with sum([2, 3, 4, 5], 0) + 2.

17

Right, so

sum([2, 3, 4, 5], 3) will return sum([2, 3, 4, 5], 2) + 4

sum([2, 3, 4, 5], 2) will return sum([2, 3, 4, 5], 1) + 3

sum([2, 3, 4, 5], 1) will return sum([2, 3, 4, 5], 0) + 2

and sum([2, 3, 4, 5], 0) …

18

if n <= 0 returns 0 and stops calling the function and adds 4+3+2=9;

19

Nope, slow down!

sum([2, 3, 4, 5], 3) will return sum([2, 3, 4, 5], 2) + 4

sum([2, 3, 4, 5], 2) will return sum([2, 3, 4, 5], 1) + 3

sum([2, 3, 4, 5], 1) will return sum([2, 3, 4, 5], 0) + 2

and sum([2, 3, 4, 5], 0) will return 0

So… sum([2, 3, 4, 5], 1) will …

20

so sum([2, 3, 4, 5 ] , 1) will return 9 because it is the last element in the array.