Sure, here is a step-by-step solution in Python:

1. First, we need to check if the length of the array is less than or equal to 2. If it is, we return the length of the array because in this case, all elements are unique or appear at most twice.

```python
def removeDuplicates(nums):
if len(nums)

Question

Sure, here is a step-by-step solution in Python:

1. First, we need to check if the length of the array is less than or equal to 2. If it is, we return the length of the array because in this case, all elements are unique or appear at most twice.

```python
def removeDuplicates(nums):
    if len(nums) <= 2:
        return len(nums)
```

2. If the array length is more than 2, we initialize a pointer `i` at the second element (index 1). We then start a loop from the third element to the end of the array.

```python
    i = 1
    for j in range(2, len(nums)):
```

3. In the loop, we check if the current element is the same as the element two positions before it in the array. If it is not, we move the element at the current position to the position after the pointer and increment the pointer.

```python
        if nums[j] != nums[i - 1]:
            i += 1
            nums[i] = nums[j]
```

4. After the loop, we return the position of the pointer plus one, which is the number of unique elements that appear at most twice.

```python
    return i + 1
```

Here is the complete function:

```python
def removeDuplicates(nums):
    if len(nums) <= 2:
        return len(nums)
    i = 1
    for j in range(2, len(nums)):
        if nums[j] != nums[i - 1]:
            i += 1
            nums[i] = nums[j]
    return i + 1
```

This function works in-place, meaning it does not create a new array and only uses O(1) extra memory. It also maintains the relative order of the elements.

Knowee AI · Accepted Answer