# Leetcode 26. Remove Duplicates from Sorted Array

Description:

Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same. Then return the number of unique elements in nums.

Consider the number of unique elements of nums to be k, to get accepted, you need to do the following things:

Change the array nums such that the first k elements of nums contain the unique elements in the order they were present in nums initially. The remaining elements of nums are not important as well as the size of nums.
Return k.

Example 1:

> Input: nums = [1,1,2]
> Output: 2, nums = [1,2,_]
> Explanation: Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively.
> It does not matter what you leave beyond the returned k (hence they are underscores).

Example 2:

> Input: nums = [0,0,1,1,1,2,2,3,3,4]
> Output: 5, nums = [0,1,2,3,4,_,_,_,_,_]
> Explanation: Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively.
> It does not matter what you leave beyond the returned k (hence they are underscores).

Solution: two pointers start at index 1
* left: record the index for storing the unique number
* right: traverse the array and compare current number with previous number(the reason why starting at index 1), if current number != previous number, then it's unique

Code:

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        # empty case
        if len(nums) == 0:
            return None
        
        # normal case
        # initialize two pointers
        left = 1
        for right in range(1, len(nums)):
            # current != previous -> unique
            # move to where left pointer is
            # then increment left pointer
            if nums[right] != nums[right - 1]:
                nums[left] = nums[right]
                left += 1
                
        # for each loop, right will automatically increment 1, so no need to add 1
        
        # after iteration, return left since it represent the number of unique number
        return left

* Time Complexity: O(n)
* Space Complexity: O(1) (in place)