Trouble with an 'Odd', 'Even' array!

Question

Trying out a Kata where I have to add odd numbers to one array and even numbers to another.

Here's my code so far.

function pickIt(arr){
  var odd=[],even=[];
  //coding here

  for(var i = 0; i < arr.length; i++){
  if (arr  % 2 ) {

      even.push(arr);

  } else {
  odd.push(arr);

  }

  }
    return [odd, even];


}

And here's what the issue appears to be.

Expected: '[[1], [2]]', instead got: '[[[1, 2], [1, 2]], []]' Expected: '[[1, 3], [2]]', instead got: '[[[1, 2, 3], [1, 2, 3], [1, 2, 3]], []]' Expected: '[[3, 1], [2]]', instead got: '[[[3, 2, 1], [3, 2, 1], [3, 2, 1]], []]' Expected: '[[], [10, 20, 30]]', instead got: '[[[10, 20, 30], [10, 20, 30], [10, 20, 30]], []]' Expected: '[[11, 21, 31], []]', instead got: '[[[11, 21, 31], [11, 21, 31], [11, 21, 31]], []]' Expected: '[[1, 5, 1, 1], [8, 4, 6]]', instead got: '[[[8, 1, 5, 4, 6, 1, 1], [8, 1, 5, 4, 6, 1, 1], [8, 1, 5, 4, 6, 1, 1], [8, 1, 5, 4, 6, 1, 1], [8, 1, 5, 4, 6, 1, 1], [8, 1, 5, 4, 6, 1, 1], [8, 1, 5, 4, 6, 1, 1]], []]'

It seems to be printing the array twice and adding a further empty array.

Is the issue with the return statement?

Answer 1 · 2017-07-25T16:37:05Z

July 25, 2017 4:37pm

Hi Matt,

The overall structure looks good but you have a few issues.

In your if condition along with when you push to the arrays, you're working with the array itself arr instead of with an individual number from the array which would be arr[i]. So you would need to make that change in those 3 places.

The other thing is that your logic is the reverse of what it should be. You would actually be adding the odd numbers to the even array.

I would suggest you write it like this, arr[i] % 2 == 0

If the remainder is equal to zero then you know you have an even number and you can push it to the even array.

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Matt Jones

Matt Jones

Trouble with an 'Odd', 'Even' array!

1 Answer

Jason Anello

Jason Anello