Intermediate Algorithm Scripting - Diff Two Arrays

JavaScript
1

Should i have !== -1 set to <=0 or is something else the issue? maybe the way i coded the indexOf part of it?

Your code so far

function diffArray(arr1, arr2) {
  const newArr = [];
  let a = arr1.concat(arr2)
  for(let i = 0; i < a.length; i++) {
    if(arr1.indexOf(a[i]) && arr2.indexOf(arr[i]) !== -1 ) {
      newArr.push(a[i])
    }
  }
  return newArr;
}

diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);

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Challenge: Intermediate Algorithm Scripting - Diff Two Arrays

Link to the challenge:

2

What does this mean?

Each side of the && needs to be an independent statement.

4

arent arr1 and arr2 already defined since theyre used as arguments?

5

would i just want to return arr1 and arr2 before looping then? or do something like let a1 = arr1.slice() and same for arr2?

6

arr1 and arr2 are defined. a and newArr are defined. arr is not.

8

You should really, really read this.

9

Ok i see it now haha :slight_smile: Thanks, totally missed it

10

if(arr1.indexOf(a[i]) === -1 && arr2.indexOf(a[i]) === -1 ) Ok so like this but with different equality operators right?

11

Right, you need two separate comparisons, but you need to make sure the comparisons are expressing the logic you need.

12

I thought === -1 is the same as saying if one of the iterations is not present in arr1/arr2

13

or setting it equal to < 0 to specify it as not in the array

14

Right, so this says "a[i] is not in arr1 and a[i] is not in arr2.

Is that what you want?

15

No because it has to be in one but not the other so i can identify the missing element and solve it

16

Right, so you probably need a different logical combination of those two conditions.

17

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