Removing Elements from an Array based on their Position for every 3 Elements

Say I have an array called elements_n:

    elements_n = [0, 1];

I have another array called Arr:

   Arr = [0, 1, 3, 1, 2, 4];

If any of the first 3 elements of the array Arr are equal to the first element of elements_n which is 0, I would like to delete that element from the array called Arr. I then repeat the same process for the next 3 elements of the Array Arr. So to explain myself better, I will use the following example:

Compare the first 3 elements of array Arr which are 0, 1, 3 to the first element of elements_n which is 0. Since Arr[1] == elements_n[1]. I delete Arr[1] from the array Arr.

Compare the next 3 elements of array Arr which are 1, 2, 4 to the second element of elements_n which is 1. Since Arr[4] == elements_n[2]. I delete Arr[4] from the array Arr. So the elements that should be left in the array Arr are:

   Arr = [1, 3, 2, 4];

Answer 1

[Adam]

If you are using R2015a or later then this will work:

elements_n = repelem( elements_n, 3 );
Arr( Arr == elements_n ) = [];

[CharlesB]

Thanks :)