[LeetCode] 80. Remove Duplicates from Sorted Array II (Python)

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80. Remove Duplicates from Sorted Array II

Medium

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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.

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Code

class Solution(object):
    def removeDuplicates(self, nums):
        i = 0
        for j in range(2, len(nums)):
            if nums[j] != nums[i]:
                nums[i+2] = nums[j] 
                i += 1
        nums[i+2:] = []

Time: 70 ms (24.13%), Space: 15.3 MB (20.62%)

이 문제는 중복되는 수를 최대 2개까지만 허용하여 nums 리스트에 저장하는 문제이다.

나는 nums[j] != nums[i]일때, nums[i+2] = nums[j]로 값을 갱신해주었다.

테스트 케이스

# 입력
[0, 0, 1, 1, 1, 1, 2, 3, 3]

# 출력
[0, 0, 1, 1, 2, 3, 3]

1. i = 0, j = 2

• nums의 맨 앞의 두 숫자는 확인하지 않고 그대로 두면 된다.
  • 두 숫자가 같든 다르든 조건을 만족하기 때문이다.
• nums[2] != nums[0] -> True
  • nums[2] = nums[2] (1)
  • i += 1

2. i = 1, j = 3

• nums[3] != nums[1] -> True
  • nums[3] = nums[3] (1)
  • i += 1

3. i = 2, j = 4

• nums[4] != nums[2] -> False

4. i = 2, j = 5

• nums[5] != nums[2] -> False

5. i = 2, j = 6

• nums[6] != nums[2] -> True
  • nums[4] = nums[6] (2)
  • i += 1

6. i = 3, j = 7

• nums[7] != nums[3] -> True
  • nums[5] = nums[7] (3)
  • i += 1

7. i = 4, j = 8

• nums[8] != nums[4] -> True
  • nums[6] = nums[8] (3)
  • i += 1

Time Complexity