My interview question was that I need to return the length of an array that removed duplicates but we can leave at most 2 duplicates.

For example, [1, 1, 1, 2, 2, 3] the new array would be [1, 1, 2, 2, 3]. So the new length would be 5. I came up with an algorithm with O(2n) I believe. How can I improve that to be the fastest.

def removeDuplicates(nums):if nums is None:return 0if len(nums) == 0:return 0if len(nums) == 1:return 1new_array = {}for num in nums:new_array[num] = new_array.get(num, 0) + 1new_length = 0for key in new_array:if new_array[key] > 2:new_length = new_length + 2else:new_length = new_length + new_array[key]return new_lengthnew_length = removeDuplicates([1, 1, 1, 2, 2, 3])
assert new_length == 5

My first question would be is my algorithm even correct?

Your logic is correct however he is a simpler method to reach the goal you had mentioned in your question.

Here is my logic.

myl = [1, 1, 1, 2, 2, 3, 1, 1, 1, 2, 2, 3, 1, 1, 1, 2, 2, 3]newl = []for i in myl:if newl.count(i) != 2:newl.append(i)print newl
[1, 1, 2, 2, 3, 3]

Hope this helps.