Remove Duplicates from Sorted Array — Problem of the Day (November 11, 2026)

Given a sorted array nums, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same.

Your function should return the new length k after removing the duplicates. Do not allocate extra space for another array; you must do this by modifying the input array in-place with O(1) extra memory.

Constraints:
- 0 <= nums.length <= 3 * 10^4
- -100 <= nums[i] <= 100
- nums is sorted in non-decreasing order.

Example 1:

Input: nums = [1,1,2]
Output: 2, nums = [1,2,_]
Explanation: Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively. It does not matter what you leave beyond the returned k (hence they are represented by underscores).

Example 2:

Input: nums = [0,0,1,1,1,2,2,3,3,4]
Output: 5, nums = [0,1,2,3,4,_,_,_,_,_]
Explanation: Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively.

Solution

Approach: This problem can be efficiently solved using a two-pointer technique.
- One pointer (j) keeps track of the position where the next unique element should be placed.
- Another pointer (i) iterates through the array to find new unique elements.
Steps:
1. Handle the edge case: If the array is empty, return 0.
2. Initialize j = 1. This pointer will point to the second position where a unique element might be written (the first element nums[0] is always considered unique in a non-empty array).
3. Iterate i from 1 to len(nums) - 1 (starting from the second element).
4. At each i, compare nums[i] with the previous element nums[i-1].
5. If nums[i] is different from nums[i-1], it means we've found a new unique element.
  - Place this unique element at nums[j].
  - Increment j to prepare for the next unique element.
6. After the loop finishes, j will represent the count of unique elements, which is also the new length of the modified array.

Code (Python):

class Solution:
    def removeDuplicates(self, nums: list[int]) -> int:
        if not nums:
            return 0
        
        # Pointer `j` marks the position for the next unique element.
        # The first element (at index 0) is always unique, so we start `j` at 1.
        j = 1
        
        # Pointer `i` iterates through the array starting from the second element.
        for i in range(1, len(nums)):
            # If the current element `nums[i]` is different from the element just before it `nums[i-1]`,
            # it means `nums[i]` is a new unique element.
            if nums[i] != nums[i-1]:
                # Place this unique element at the position `j`.
                nums[j] = nums[i]
                # Move `j` to the next available position for a unique element.
                j += 1
                
        # The value of `j` at the end represents the new length of the array
        # with duplicates removed (and also the count of unique elements).
        return j

Remove Duplicates from Sorted Array — November 11, 2026

Solution

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