Remove All Instance

Instance.py

def remove_all_instances(nums, val):
	"""
	This returns the number of values in num that isn't equal to val
	its time complexity is O(n/2) because I used two pointers to navigate from left to right
	and from right to left the both meet at p = n/2.  while space O(1)
	"""

	if nums == None: return 0
	if val == None: return 0

	res = 0

	ptr1 = 0
	ptr2 = len(nums) - 1

	while ptr1 <= ptr2:
		if ptr1 == ptr2: #once the two pointers meet at the center for odd lengths we only use the value for one of the pointers to check.
			if nums[ptr1] != val:
				res += 1
				break

		if nums[ptr1] != val:
			res += 1
		if nums[ptr2] != val:
			res += 1

		ptr1 += 1
		ptr2 -= 1

	return res

Thank you very much, I'm very grateful 😊.