[Leetcode/파이썬] 80. Remove Duplicates from Sorted Array II

Remove Duplicates from Sorted Array II

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Given an integer array nums sorted in non-decreasing order, remove some duplicates in-place such that each unique element appears at most twice. The relative order of the elements should be kept the same.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.

Return k after placing the final result in the first k slots of nums.

Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.

Custom Judge:

The judge will test your solution with the following code:

int[] nums = [...]; // Input array
int[] expectedNums = [...]; // The expected answer with correct length

int k = removeDuplicates(nums); // Calls your implementation

assert k == expectedNums.length;
for (int i = 0; i < k; i++) {
    assert nums[i] == expectedNums[i];
}

If all assertions pass, then your solution will be accepted.

 

Example 1:

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

Example 2:

Input: nums = [0,0,1,1,1,1,2,3,3]
Output: 7, nums = [0,0,1,1,2,3,3,_,_]
Explanation: Your function should return k = 7, with the first seven elements of nums being 0, 0, 1, 1, 2, 3 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

 

Constraints:

• 1 <= nums.length <= 3 * 104
• -104 <= nums[i] <= 104
• nums is sorted in non-decreasing order.

 

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        dct = defaultdict(int)
        idx_k = 0
        for idx_num in range(len(nums)) :
            if dct[nums[idx_num]] < 2 :
                nums[idx_k] = nums[idx_num]
                dct[nums[idx_num]] += 1
                idx_k += 1


        return idx_k

카운팅을 하나하나씩 해가면서 정답풀이를 하였다. 근데 이렇게 되면 O(1) extra memory에 위배되나? 좀 고민해봐야겠다. 이거 말고 더 나은 방법이 있는지 살펴보아야 겠다. 

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        k = cnt = 1

        for i in range(1, len(nums)) :
            if nums[i-1] == nums[i] :
                cnt += 1
            else :
                cnt = 1

            if cnt <= 2 :
                nums[k] = nums[i]
                k += 1

        return k

굳이 전부를 카운팅 하지 않고 현재 숫자만 카운팅 해서 하는 방법도 있구나...