[LeetCode] 26. Remove Duplicates from Sorted Array (Python)

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26. Remove Duplicates from Sorted Array

Easy

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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.

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 = [1,1,2]
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).

 

Constraints:

• 1 <= nums.length <= 3 * 104
• -100 <= nums[i] <= 100
• nums is sorted in non-decreasing order.

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Code

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

Time: 2 ms (70.13%), Space: 14 MB (8.52%)

.pop() 등의 함수를 사용하면, 시간 복잡도가 높아지므로 i 변수를 따로 두어 해결해주었다.

nums[i:] = []를 통해 중복되는 원소들을 nums 리스트의 앞쪽에 정리해주고 나서, 그 뒤의 원소들을 없애준다.

Time Complexity

$O(N)$