RemoveDuplicatesFromSortedArrayII

Problem

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

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:sup:`4`

• -10:sup:`4`<= nums[i] <= 10:sup:`4`

• nums is sorted in non-decreasing order.

Solution

Position i = 2 to start from the third element of nums. If the current element nums[i] equals the element two indices before nums[i - 2], look ahead to the next element by incrementing an offset. If the current element nums[i + offset] is unequal to nums[i - 2], then copy nums[i + offset] to nums[i] and increment i. Repeat until i + offset is out of bounds.

Pattern

Array, Two Pointers

Code

from typing import List


def removeDuplicates(nums: List[int]) -> int:
    """Removes duplicates from the sorted array ``nums`` in-place such that
    each element appears at most twice.
    """
    offset = 0
    i = 2
    while i + offset < len(nums):
        if nums[i + offset] == nums[i - 2]:
            offset += 1
        else:
            nums[i] = nums[i + offset]
            i += 1

    return len(nums) - offset

Test

>>> from RemoveDuplicatesFromSortedArrayII import removeDuplicates
>>> nums = [1, 1, 1, 2, 2, 3]
>>> k = removeDuplicates(nums)
>>> nums[:k]
[1, 1, 2, 2, 3]
>>> nums = [0, 0, 1, 1, 1, 1, 2, 3, 3]
>>> k = removeDuplicates(nums)
>>> nums[:k]
[0, 0, 1, 1, 2, 3, 3]

RemoveDuplicatesFromSortedArrayII.removeDuplicates(nums: List[int]) → int

Removes duplicates from the sorted array nums in-place such that each element appears at most twice.