Seek and Destroy algorithm - Question

Seek and Destroy algorithm - Question
0

#1

Hi,

I’m working on the Seek and Destroy algorithm, but there’s one thing I don’t understand.

Here’s my solution (which works):

function destroyer(arr) {
  var args = Array.from(arguments);

  for (var i = 0; i < arr.length; i++) {
    for (var j = 1; j < args.length; j++) {
      if (arr[i] === args[j]) {
        delete arr[i];
      }
    }
  }

  return arr.filter(Boolean);
}

Here I use j = 1 to iterate through the arguments, except the first one which as far as I understand is the same as arr.

But I saw other solutions that have j = 0 and the algorithm also works. I don’t understand this because in the first time we run the for loop it will compare arr with args[0], which are both [1, 2, 3, 1, 2, 3]. Therefore, all the elements in arr should be deleted.

Could anyone help me understand why this solution (with j = 0) is also correct?

Thanks


#2

I agree that what you describe shouldn’t work. Do you have any examples you could provide?


#4

It makes sense now. Thanks a lot @P1xt !