What is sum(arr,n-1)+arr[n-1] means?

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  **Your code so far**

function sum(arr, n) {
// Only change code below this line
if (n<=0){
return 0;
return sum(arr,n-1)+ arr[n-1];
// Only change code above this line
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Challenge: Replace Loops using Recursion

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This is simple recursion, and best visualized with pen and paper. When you give the function sum an array arr, and the array length n, unless you have reached the beginning of the array index, sum(arr,n-1)+arr[n-1] is returned, meaning:

  • the function sum calls itself (recursion), but passes in a different index n, this function call eventually returns a value.
  • to the recursive call, sum adds the value of arr[n-1]

let us try with a test case of arr=[1,2,3,4,5] and n=5.
1st call to sum, return sum(arr,4)+arr[4],
2nd, return sum(arr,3)+arr[3],
3rd, return sum(arr,2)+arr[2],
4th, return sum(arr,1)+arr[1],
5th, return sum(arr,0)+arr[0],
6th, return 0 (since n<=0)

Then the last result of 6th call is added to each of the previous returns.
5th, return 0+1=1,
4th, return 1+2=3,
3rd, return 3+3=6,
2nd, return 6+4=10,
1st (the original call to sum) finally get 10+5=15.

Therefore running sum([1,2,3,4,5],5) returns 15, and the functionsum adds up the values within the passed array backwards from the index passed in.


Thanks a lot… This is well understood when written on paper with your explanation. :+1:

Im still struggling with something here and i feel like theres a fundamental concept that i should understand that i just missed. But when you say “5th, return 0+1=1”, i dont understand since above “5th, return sum(arr,0) + arr[0]”… im not sure how that equates 0 + 1.

My misunderstanding i think lies in how im expecting the math to work out. Does this mean sum(arr,0) is equal to 0?

Yes. Literally that function call says ‘add 0 elements from the array’. The sum of nothing is 0.

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