80. Remove Duplicates from Sorted Array II

Problem Statement

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

Input: nums = [1,1,1,2,2,3] Output: 5, nums = [1,1,2,2,3,_]

Input: nums = [0,0,1,1,1,1,2,3,3] Output: 7, nums = [0,0,1,1,2,3,3,_,_]

Solution Approach

Leverage a two-pointer technique to monitor and manage allowable duplicates in a sorted array.

Algorithm

1. Two-pointers: Use i to scan nums and j to mark placements.
2. Check Duplicates: If nums[i] differs from nums[j-2], move nums[i] to position j.
3. Move Forward: Increment both i and j post-placement.

Implement

    class Solution:
      def removeDuplicates(self, nums: List[int]) -> int:
        i = 0
        for n in nums:
            if i < 2 or n != nums[i - 2]:
                nums[i] = n
                i += 1
        return i

 

    
    

Twitter  facebook  Email