Problem

Follow up for Remove duplicates from Sorted Array I: What if duplicates are allowed at most twice?

OR

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];
}

Examples

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).

Solution

Method 1 - Updating the Original Array

We can not change the given array’s size, so we only change the first k elements of the array which has duplicates removed. Flag is used to check if we have already allowed 2 occurences of duplicate element.

public int removeDuplicates(int[] A) {
    if (A == null || A.length == 0)
        return 0;

    int pre = A[0];
    boolean flag = false;
    int count = 0;

    // index for updating
    int j = 1;

    for (int i = 1; i < A.length; i++) {
        int curr = A[i];

        if (curr == pre) {
            if (!flag) {
                flag = true;
                A[j++] = curr;
                continue;
            } else {
                count++;
            }
        } else {
            pre = curr;
            A[j++] = curr;
            flag = false;
        }
    }

    return A.length - count;
}

Method 2 - Use 2 Pointers in Original Array

public int removeDuplicates(int[] A) {
    if (A.length <= 2)
        return A.length;

    int prev = 1; // point to previous
    int curr = 2; // point to current

    while (curr < A.length) {
        if (A[curr] == A[prev] && A[curr] == A[prev - 1]) {
            curr++;
        } else {
            prev++;
            A[prev] = A[curr];
            curr++;
        }
    }

    return prev + 1;
}