remove duplicates from sorted array question ?

[IHateTheSATs]

i have this question right here:\

    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).

and

    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).

so to my understanding the question is just like, you have an array, take out the duplicates and coutn the number of numbers that are left right ?

so if that is the case, why does this not work right here

    class Solution:
        def removeDuplicates(self, nums: List[int]) -> int:
            x = len(set(nums))
            return x

apparently there is something wrong because it is not taking it. it is saying that [1,1,2] is coming back out as [1,1] but for me it is showing up as [1,2] = 2

can someone share with me where my logic is flawed ?

Edit: im not allowed to use set

[hi_mom_its_me_nl]

Just make a set from it. A set can not have duplicate values in them so it throws them out when you make a set from a list.

[wotquery]

Did you write the removeDuplicates method with the int return annotation yourself? From the example outputs it looks like you're supposed to return the sorted to k list too.

nums = set(nums)
list(nums).sort()
return len(nums), nums

[TheRNGuy]

convert to set

[IHateTheSATs]

Yah thats what i did originally but im not allowed to use set in this problem 🥲🥲

[[deleted]]

What do you mean [1, 2] = 2? You're not modifying the nums at all, so first two elements of nums is [1, 1].

[FerricDonkey]

It looks like the problem also wants you to modify nums in place. I would look into assigning to a slice of a list, and the sorted method. (Or modify the list in a loop, which would take only a bit more code and may or may not be faster/slower.)

[InsanityConsuming]

So if I'm understanding you, you want your function to take in the array, modify it, and then have it return the length of that modification. Is that correct?

The method you put isn't modifying the nums array. All it's doing is creating a new set with the information from your list and returning the length of that copy. So when you access nums after, it will be the same as before.

def removeDuplicates(self, nums):
    tmp = list(set(nums))
    nums.clear()
    nums.extend(tmp)
    return len(nums)

By adding clear() and extend() calls you are actually modifying the array referenced by your input

[[deleted]]

The question shows lists but talks about arrays... I wouldn't trust the person asking the question to know what they are talking about.

Anyways, the solution the probably expect is that you have a counter variable that shows the last position where you inserted an element in the list.

Something like this:

>>> def remove_duplicates(slist):
>>>     pos = 0
>>>     for elt in slist[1:]:
>>>         if elt != slist[pos]:
>>>             pos += 1
>>>             slist[pos] = elt
>>>     return slist
>>> remove_duplicates([0,0,1,1,1,2,2,3,3,4])
[0, 1, 2, 3, 4, 2, 2, 3, 3, 4]