5. Remove Element

Q: Why is splice() bad here?

Each splice is O(n) and the indices shift, so you'd either skip elements or process some twice. Two pointers avoids both issues.

Q: When is swap-from-end better?

When val is rare. With swap-from-end, each removal does one write; with the standard approach, every keep does one write.

easyAsked at Asana

Remove all instances of a value from an array in-place. Asana asks this to test whether you understand the two-pointer write/read separation that scales to their bulk-task-mutation pipelines.

By Alex Chen, Founder, InterviewChamp.AI · Last verified 2026-05-20

Source citations

Public interview reports confirming this problem appears in Asana loops.

Glassdoor (2026-Q1)— Asana phone-screen warmup.

Problem

Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The order of the elements may be changed. Return the number of elements in nums which are not equal to val.

Constraints

0 <= nums.length <= 100
0 <= nums[i] <= 50
0 <= val <= 100

Examples

Example 1

Input

nums = [3,2,2,3], val = 3

Output

2, nums = [2,2,_,_]

Example 2

Input

nums = [0,1,2,2,3,0,4,2], val = 2

Output

5, nums = [0,1,4,0,3,_,_,_]

Approaches

1. Filter and copy

Build a new array with .filter, then write back.

Time: O(n)
Space: O(n)

function removeElement(nums, val) {
  const keep = nums.filter(x => x !== val);
  for (let i = 0; i < keep.length; i++) nums[i] = keep[i];
  return keep.length;
}

Tradeoff: O(n) extra space. Asana wants the in-place version.

2. Two-pointer in-place

Write pointer advances only on non-val elements.

Time: O(n)
Space: O(1)

function removeElement(nums, val) {
  let w = 0;
  for (let r = 0; r < nums.length; r++) {
    if (nums[r] !== val) nums[w++] = nums[r];
  }
  return w;
}

Tradeoff: O(1) space. If most elements equal val, swap-from-end is faster — fewer writes.

Asana-specific tips

Asana cares about the variation: if val is rare, swap-from-end minimizes writes; if val is common, the standard two-pointer wins. Mention both and pick based on the expected distribution — interviewers love that you think about real workloads.

Common mistakes

Splicing in a loop — O(n^2) and shifts indices unpredictably.
Returning the array length instead of the count of kept elements.
Off-by-one in the write pointer.

Follow-up questions

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

What if order doesn't matter — can you do fewer writes?
Remove multiple values at once.
Apply to a linked list instead of an array.

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Output

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FAQ

Why is splice() bad here?

Each splice is O(n) and the indices shift, so you'd either skip elements or process some twice. Two pointers avoids both issues.

When is swap-from-end better?

When val is rare. With swap-from-end, each removal does one write; with the standard approach, every keep does one write.

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