80. Remove Duplicates from Sorted Array II

1. Description

Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length.
Do not allocate extra space for another array; you must do this by modifying the input array in-place with O(1) extra memory.
Clarification:
Confused why the returned value is an integer, but your answer is an array?
Note that the input array is passed in by reference, which means a modification to the input array will be known to the caller.
Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

2. Example

Example 1:
Input: nums = [1,1,1,2,2,3]
Output: 5, nums = [1,1,2,2,3]
Explanation: Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively. It doesn’t matter what you leave beyond the returned length.

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 length = 7, with the first seven elements of nums being modified to 0, 0, 1, 1, 2, 3 and 3 respectively. It doesn’t matter what values are set beyond the returned length.

3. Constraints

• 1 <= nums.length <= $3 * 10^4$
• $-10^4$ <= nums[i] <= $10^4$
• nums is sorted in ascending order.

4. Solutions

Two Pointers

n = m_nums.size()
Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    int removeDuplicates(vector<int>& m_nums) {
        int index = 0, limit = 2;
        for (auto num : m_nums) {
            if (index < limit || num > m_nums[index - limit]) {
                m_nums[index] = num;
                index++;
            }
        }

        return index;
    }
};