Skip to main content

4. Remove Duplicates from Sorted Array

easyAsked at Snowflake

In-place deduplicate a sorted array, returning the new length. Snowflake asks this to test two-pointer mechanics and to set up a follow-up on dictionary-encoded columnar deduplication.

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

Source citations

Public interview reports confirming this problem appears in Snowflake loops.

  • Glassdoor (2026-Q1)— Snowflake intern phone screens use this as a two-pointer warm-up.
  • LeetCode Discuss (2025-08)— Recurring at Snowflake SDE-I screens.

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. Then return the number of unique elements in nums.

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. Set + rebuild

Insert into a Set, then write back. Violates the in-place requirement.

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

Tradeoff: Violates the O(1) extra space requirement. Mention only to show you read the constraint.

2. Two-pointer in-place (optimal)

Slow pointer writes unique values; fast pointer scans. When nums[fast] != nums[slow], advance slow and copy.

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

Tradeoff: Linear, O(1) extra space — and it's exactly the loop that builds a run-length-encoded column.

Snowflake-specific tips

Snowflake interviewers want to hear you connect this to dictionary encoding and run-length encoding in columnar storage. Bonus signal: extend to 'count duplicates per value' and discuss how RLE saves bytes for sorted low-cardinality columns.

Common mistakes

  • Returning the array instead of the length (the problem asks for length and mutates in-place).
  • Starting slow at 1 instead of 0 — off-by-one in the return value.
  • Comparing nums[fast] to nums[fast-1] instead of nums[slow] — works for sorted input but breaks if you generalize to nearly-sorted.

Follow-up questions

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

  • Allow each value to appear at most twice (LC 80).
  • Remove duplicates from unsorted array preserving order.
  • Build a run-length-encoded representation in-place.

Solve it now

Free. No sign-up. Python and JavaScript run instantly in your browser.

Output

Press Run or Cmd+Enter to execute

FAQ

Why O(1) space matters here?

In columnar storage, you cannot afford to allocate a second array the size of a column. In-place compaction is the realistic constraint.

Why does sorted input matter?

Sorted input means duplicates are adjacent. That's also why Snowflake's micro-partitions sort on clustering keys — it makes downstream dedup, RLE, and predicate pushdown cheap.

Practice these live with InterviewChamp.AI

Drill Remove Duplicates from Sorted Array and other Snowflake interview questions under real-loop conditions with instant feedback on your reasoning, complexity claims, and code.

Practice these live with InterviewChamp.AI →