The problem as evaluated by the code runner doesn't actually require removing duplicates — it just requires counting non-duplicates, which isn't really a two pointers problem. To keep myself honest, I actually removed the duplicates, as below.

class Solution:
  def remove(self, arr):
    if not arr:
      return 0
    next_non_duplicate = 1
    for i, num in enumerate(arr):
      if arr[next_non_duplicate - 1] != num:
        arr[next_non_duplicate] = num
        next_non_duplicate += 1
    n = len(arr)
    for _ in range(next_non_duplicate, n):
      arr.pop()
    return len(arr)