CODEWITHSUNDEEP
javascriptrecursion
Alan Richards
Alan Richards

Reputation: 283

Using Recursion

I came across this problem on FreeCodeCamp.org (link to the problem below) and I was wondering if someone could help me better understand why this equals 2 when calling with sum([2, 3, 4], 1); I sat and looked at this for a bit but just feel blocked mentally in understanding how this works. Simple developer trying to understand Recursion in Javascript.

Any help would be much appreciated! Thanks internet!

function sum(arr, n) {
  if(n <= 0){
    return 0;
  }else {
    return sum(arr, n - 1) + arr[n - 1];
  }
}
sum([2, 3, 4], 1) // Returns 2

Here is a link to the problem: https://www.freecodecamp.org/learn/javascript-algorithms-and-data-structures/basic-javascript/replace-loops-using-recursion

Upvotes: 0

Views: 3660

Answers (3)

Swetank Poddar
Swetank Poddar

Reputation: 1291

sum(arr, 1-1) evaluates to 0 because of the if statement that checks if n is <=0

and

arr[n-1] is equal to 2 as you are essentially trying to access the arr[1-1] which is arr[0].

Therefore, it is returning 2 because return sum(arr, n-1) + arr[n-1]; evaluates to 2.

Upvotes: 0

Danoz
Danoz

Reputation: 1097

sum(arr, n) is a recursive function that returns the sum of the first n elements of an array arr.

In your example, you provide sum([2, 3, 4], 1) which basically says, compute the sum of the first element (i.e. the value of n in this example is 1).

So it would go through the function as such...

// the first time through 
function sum([2, 3, 4], 1) {
  if(1 <= 0){ // false this time
    return 0; 
  }else { // this is where we end up
    return sum([2, 3, 4], 0) + 2; // sum will be the result of the recurse back into the function, plus 2
  }
}

// the second time through 
function sum([2, 3, 4], 0) {
  if(0 <= 0){ // true this time 
    return 0; // send this result back up to the first run through
  }else {
    // not relevant this time
  }
}

// back in the the first time through, 
// we now have a value to work with below
// remember, this isn't the 'third' time through,
// it is back in the first time run through
// just re-printed here so you could see 
// where the value gets returned from the second run
function sum([2, 3, 4], 1) {
  if(1 <= 0){ // false this time
    return 0; 
  }else { // this is where we end up
    return sum(0 + 2); // we got a result from the second run through, sum is now 2
  }
}

Upvotes: 1

ZER0
ZER0

Reputation: 25332

Actually, with n equals to 1 is quite easy. Let's check step by step. Here your sum function; let's add some line to make it easier to refer:

1: function sum(arr, n) {
2:  if(n <= 0){
3:    return 0;
4:  }else {
5:    return sum(arr, n - 1) + arr[n - 1];
6:  }
7: }

Now, let's see what's happening step by step when you execute:

sum([2, 3, 4], 1)

The function sum is called with arr equals to [2, 3, 4], and n equals to 1.

Since n is not less or equals than 0 (line 2), we go to the else block, at line 5.

Now, here is where the recursion happens: we call again the function sum, passing the same arr, but not the same n, instead we pass n - 1.

So we call sum again, this time with arr equals to [2, 3, 4] but with n equals to 0. Since n is 0 this time, at the line 2 check we proceed to line 3 and returns 0.

Now, the function exit, with the value 0, that we gave to the caller.

And the caller of sum([2, 3, 4], 0) was the execution sum([2, 3, 4], 1), specifically at line 5:

5:    return sum(arr, n - 1) + arr[n - 1];

Since it returned 0, we can imaging like:

5:    return 0 + arr[n - 1];

And remember that n is 1, so:

5:    return 0 + arr[0];

Since arr[0] is equals to 2:

5:    return 0 + 2;

And then why sum([2, 3, 4], 1) returns 2.

I'm not sure if it's clearer now, but I hope so. :)

Upvotes: 5

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