Problem statement

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.

Problem statement taken from: https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii

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.

Explanation

We have seen a similar problem in our previous blog post Remove Duplicates from Sorted Array. The only difference in this problem, we should keep the occurrences of unique elements at most twice.

If we observe the previous blog post algorithm, we are comparing the current element ith index with the i-1th index element.

int i = 0;

for(int j = 1; j < nums.size(); j++){
    if(nums[j] != nums[i]){
        i++;
        nums[i] = nums[j];
    }
}

The condition if(nums[j] != nums[i]) compares the two neighbouring elements, which resolves to if(nums[i - 1] != nums[i]). Since we can keep at most two similar elements, the condition will be similar to if(nums[i - 1] != nums[i] || nums[i - 2] != nums[i]).

Let's check the algorithm to get a clear picture.

- set k = 2, n = nums.size()

- if n <= 2
  - return n

- loop for i = 2; i < n; i++
  - if nums[i] != nums[k - 2] || nums[i] != nums[k - 1]
    - nums[k] = nums[i]
    - k++

- return k

We keep an integer k that updates the kth index of the array only when the current element does not match either of the two previous indexes. If the kth index matches the k-1th and k-2th elements, we keep moving forward in the array. Let's check out our solutions in C++, Golang, and Javascript.

C++ solution

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int k = 2;
        int n = nums.size();

        if(n <= 2) {
            return n;
        }

        for(int i = 2; i < n; i++){
            if(nums[i] != nums[k - 2] || nums[i] != nums[k - 1]){
                nums[k] = nums[i];
                k++;
            }
        }

        return k;
    }
};

Golang solution

func removeDuplicates(nums []int) int {
    k := 2
    n := len(nums)

    if n <= 2 {
        return n
    }

    for i := 2; i < n; i++ {
        if nums[i] != nums[k - 2] || nums[i] != nums[k - 1] {
            nums[k] = nums[i]
            k++
        }
    }

    return k
}

Javascript solution

var removeDuplicates = function(nums) {
    let k = 2;
    let n = nums.length;

    if(n <= 2) {
        return n;
    }

    for(let i = 2; i < n; i++) {
        if(nums[i] != nums[k - 2] || nums[i] != nums[k - 1]) {
            nums[k] = nums[i];
            k++;
        }
    }

    return k;
};

Dry-run of our C++ approach for this input nums = [1, 1, 1, 2, 2, 3] looks as below:

Input: nums = [1, 1, 1, 2, 2, 3]

Step 1: k = 2
        n = nums.size()
          = 6

Step 2: if n <= 2
           6 <= 2
           false

Step 3: loop for i = 2; i < n;
          2 < 6
          true

          if nums[i] != nums[k - 2] || nums[i] != nums[k - 1]
             nums[2] != nums[0] || nums[2] != nums[1]
             1 != 1 || 1 != 1
             false

          i++
          i = 3

Step 4: loop i < n
          3 < 6
          true

          if nums[i] != nums[k - 2] || nums[i] != nums[k - 1]
             nums[3] != nums[0] || nums[3] != nums[1]
             2 != 1 || 2 != 1
             true

             nums[k] = nums[i]
             nums[2] = nums[3]
             nums[2] = 2

             k++
             k = 3

             nums = [1, 1, 2, 2, 2, 3]

          i++
          i = 4

Step 5: loop i < n
          4 < 6
          true

          if nums[i] != nums[k - 2] || nums[i] != nums[k - 1]
             nums[4] != nums[1] || nums[4] != nums[2]
             2 != 1 || 2 != 2
             true

             nums[k] = nums[i]
             nums[3] = nums[4]
             nums[3] = 2

             k++
             k = 4

             nums = [1, 1, 2, 2, 2, 3]

          i++
          i = 5

Step 6: loop i < n
          5 < 6
          true

          if nums[i] != nums[k - 2] || nums[i] != nums[k - 1]
             nums[5] != nums[2] || nums[5] != nums[3]
             3 != 2 || 3 != 2
             true

             nums[k] = nums[i]
             nums[4] = nums[5]
             nums[4] = 3

             k++
             k = 5

             nums = [1, 1, 2, 2, 3, 3]

          i++
          i = 6

Step 7: loop i < n
          6 < 6
          false

Step 8: return k

So we return the answer as 5, and the array till the 5th index is [1, 1, 2, 2, 3].