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62. Remove Duplicates from Sorted Array II

mediumAsked at Reddit

Allow at most two occurrences of each value in a sorted array, in-place. Reddit uses this to test slow/fast pointer with a count threshold — the same shape as rate-limiting to N per user in their abuse pipeline.

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

Source citations

Public interview reports confirming this problem appears in Reddit loops.

  • Glassdoor (2026-Q1)— Reddit phone screen, two-pointer variant.

Problem

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. Return k after placing the final result in the first k slots of nums.

Constraints

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

Examples

Example 1

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

Explanation: nums becomes [1,1,2,2,3,_].

Example 2

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

Approaches

1. Track count per value

Iterate; count consecutive duplicates; copy only when count <= 2.

Time
O(n)
Space
O(1)
function removeDuplicates(nums) {
  let k = 0, count = 0;
  for (let i = 0; i < nums.length; i++) {
    if (i > 0 && nums[i] === nums[i - 1]) count++;
    else count = 1;
    if (count <= 2) nums[k++] = nums[i];
  }
  return k;
}

Tradeoff: Clear and explicit. Works.

2. Compare against k - 2 (optimal)

Copy nums[i] to nums[k] only if k < 2 or nums[i] != nums[k - 2].

Time
O(n)
Space
O(1)
function removeDuplicates(nums) {
  let k = 0;
  for (const n of nums) {
    if (k < 2 || n !== nums[k - 2]) nums[k++] = n;
  }
  return k;
}

Tradeoff: Elegant — the k-2 lookback encodes the 'at most twice' rule directly.

Reddit-specific tips

Reddit interviewers love the k-2 lookback. Bonus signal: generalize to 'at most M occurrences' by comparing against nums[k - M]. Same shape as their per-user rate limit configurable at runtime.

Common mistakes

  • Off-by-one on k - 2 (must check k < 2 first).
  • Updating count before checking it.
  • Not iterating over n directly (passing nums.length without const n).

Follow-up questions

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

  • Remove Duplicates from Sorted Array (LC 26).
  • At most M duplicates — generalize.
  • Remove duplicates from sorted list II (LC 82) — same on linked list.

Solve it now

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Output

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FAQ

Why does k - 2 work?

If nums[i] equals nums[k - 2], then nums[k - 2] and nums[k - 1] are the same value already — adding a third would violate the rule.

Generalize to M?

Replace k < 2 with k < M and nums[k - 2] with nums[k - M].

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