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

Hello @Fredpwol, thank you for participating in Week 2 of Algorithm Fridays and congratulations, you are the winner of the $20 award 🎉🎉.

Your solution was selected because it is most optimal in terms of memory and time complexity. It is also robust and takes care of edge cases such as when any of the input values is None.

I have made a blog post here about the different solutions and about you as the winner for Week 2 of Algorithm Fridays.

We will contact you in less than 24 hours for your award.

Congratulations once again!

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