26. Remove Duplicates from Sorted Array (Easy)

Remove Duplicates from Sorted Array is an Easy level problem.

Given an array of integers sorted in non-decreasing (aka, increasing) order, remove all the duplicate values IN PLACE and return the number of unique values in the list.

My first solution to this was: 

class Solution:

def removeDuplicates(self, nums: list[int]) -> int:

dupeCount = 0

origLength = len(nums)

i = 0

while (i < len(nums) - 1):

thisElem = nums[i]

nextElem = nums[i+1]

while thisElem == nextElem:

dupeCount += 1

if (i<len(nums)-2):

nums[i+1:] = nums[i+2:]

else:

nums[i+1:] = []

if i+1 < len(nums):

nextElem = nums[i+1]

else:

nextElem = None

i += 1

return origLength - dupeCount

That solution worked, but it was super slow, beating only 5.07% of users in Runtime and 23.85% of users in Memory use.

My original solution was complicated and messy. I wrote it, start to finish, without really considering the problem. My first implementation was buggy and missed easy edge cases (like walking off the start of the array, for example).

I re-evaluated the problem and white boarded up the skeleton of a solution.

Thinking through the problem really helped with solving it. My second solution had a mistake in the shifting of the end index, at first, but once I fixed that it ran great. This solution beat 98.86% of users in Runtime and 57.6% of users in memory use.

class Solution:
    def removeDuplicates(self, nums: list[int]) -> int:
        end = len(nums)-1
        beg = end
        dupeCount = 0
        origCount = len(nums)
        while (end > 0):
            val = nums[end]
            lookBack = val
            while (beg > 0 and lookBack == val):
                lookBack = nums[beg-1]
                if lookBack == val:
                    beg -= 1
            
            # beg is now the index of the first occurence of val
            if beg != end:
                # Truncate nums by removing all the dupes, leaving just
                # the first occurence of val
                nums[beg+1:] = nums[end+1:]
                dupeCount += end-beg
            
            # Shift the end pointer
            end = beg - 1
            beg = end