80. Remove Duplicates from Sorted Array II

Question

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 * 10 4
-10 4 <= nums[i] <= 10 4
nums is sorted in non-decreasing order.

Solutions

int removeDuplicates(vector<int>& nums) {
	// two pointer approach:
    // slow pointer points to the position waiting for replaced or not.
    // fast pointer iterates through the array.
    // [0, slow) ensures only have 2 duplicated elements
    // Edge cases:
    // 1. when no elements in nums -> directly return 0;
    // 2. when only one elements -> no duplicated
    // 3. when n == 2 -> directly return no matter duplicated or not.
    // Time Complexity: O(N)
    // Space Complexity: O(1)
    int n = static_cast<int>(nums.size());
    if (n <= 2)
      return n;

    int slow = 2;
    for (int fast = 2; fast < n; ++fast) {
      if (nums[fast] != nums[slow - 2]) {
        nums[slow] = nums[fast];
        ++slow;
      }
    }
    return slow;
}