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4. Remove Duplicates from Sorted Array

easyAsked at Plaid

In-place de-duplicate a sorted array and return the new length. Plaid asks this because de-duping near-duplicate transactions (same id, same amount, slight timestamp drift) is a daily reality on their ingestion pipeline.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Plaid loops.

  • Glassdoor (2025)— Plaid SWE I onsite warm-up.
  • LeetCode Discuss (2026)— Frames as dedup of bank-feed entries.

Problem

Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same. Return k, the number of unique elements.

Constraints

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

Examples

Example 1

Input
nums = [1,1,2]
Output
2, nums = [1,2,_]

Example 2

Input
nums = [0,0,1,1,1,2,2,3,3,4]
Output
5, nums = [0,1,2,3,4,_,_,_,_,_]

Approaches

1. Use a Set then copy back

Push everything into a Set, sort, write back.

Time
O(n log n)
Space
O(n)
function removeDuplicates(nums) {
  const u = [...new Set(nums)].sort((a,b)=>a-b);
  for (let i = 0; i < u.length; i++) nums[i] = u[i];
  return u.length;
}

Tradeoff: Allocates a Set and an array. Wasteful given the input is already sorted.

2. Two-pointer in-place write

A write pointer trails a read pointer. Advance the writer only when the current value differs from the previous unique.

Time
O(n)
Space
O(1)
function removeDuplicates(nums) {
  if (nums.length === 0) return 0;
  let write = 1;
  for (let read = 1; read < nums.length; read++) {
    if (nums[read] !== nums[read - 1]) {
      nums[write++] = nums[read];
    }
  }
  return write;
}

Tradeoff: Linear time, constant space. The 'compare to nums[read-1]' trick is what saves you from tracking last-seen as a separate variable.

Plaid-specific tips

Plaid grades this on whether you exploit the sortedness you were given. State the precondition explicitly before coding. Bonus signal: explain how the two-pointer pattern is exactly what they use when de-duping consecutive webhook deliveries with the same idempotency key.

Common mistakes

  • Comparing nums[read] to nums[write] (off-by-one) instead of nums[read-1].
  • Returning the array instead of k — re-read the prompt.
  • Allocating a new array — wastes the in-place requirement.

Follow-up questions

An interviewer at Plaid may pivot to one of these next:

  • Allow each unique element at most twice (LC 80).
  • Same problem on a sorted linked list (LC 83).
  • What if the array is too large to fit in memory? Stream + writer pointer over a chunked file.

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Output

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FAQ

Why compare to the previous element instead of tracking 'last unique'?

Because the array is already sorted, the previous-position approach implicitly tracks last-unique with zero extra state.

Does this work if nums is unsorted?

No — the algorithm relies on duplicates being adjacent. For unsorted input use a hash set, which costs O(n) space.

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