[LeetCode] 26. Remove Duplicates from Sorted Array

Problem

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.

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 = [1,1,2]
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 * 10^4
• -100 <= nums[i] <= 100
• nums is sorted in non-decreasing order.

Solution

Logic

1. 배열 nums는 non-decreasing 순서이기 때문에 인접한 요소가 같다면 처리를 해주어야 한다.

2. -100 <= nums[i] <= 100의 조건이 있기 때문에 값이 같은 인접한 요소를 101로 변경하고 101을 가진 요소의 개수를 확인한다.

3. 배열 내에서 101로 변경된 값을 전부 배열의 뒤편으로 보내 배열을 정렬하면 문제를 해결할 수 있다.

   Code

   class Solution {
    public int removeDuplicates(int[] nums) {
        int beyondK = 0;
        for (int i = 0; i < nums.length - 1; i++)
            if (nums[i] == nums[i + 1]) {
                nums[i] = 101;
                beyondK++;
            }
        Arrays.sort(nums);
        return nums.length - beyondK;
    }
   }
   

   Time Complexity

   • 아래 반복문은 O(n)의 시간 복잡도를 가진다.

     for (int i = 0; i < nums.length - 1; i++)
      if (nums[i] == nums[i + 1]) {
        nums[i] = 101;
        beyondK++;
      }
     

   • 아래의 함수 호출은 O(nlog(n))의 시간 복잡도를 가진다. 따라서 본 문제는 O(nlog(n))의 시간 복잡도를 가진다.

     Arrays.sort(nums);