5. Remove Element

Q: Does order matter?

The problem says order can change. If order DID matter, the two-pointer write approach above preserves relative order anyway.

Q: When would you swap-with-end instead?

When val is rare and most elements are kept. Swap-with-end performs fewer writes than copy-forward.

easyAsked at Datadog

Given an array and a value, remove all occurrences in place and return the new length. Datadog uses this as a hello-world for in-place stream filtering — same pattern as their per-tag drop-rule applied to incoming logs.

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

Source citations

Public interview reports confirming this problem appears in Datadog loops.

Glassdoor (2026-Q1)— Reported as part of Datadog OA pool.
LeetCode Discuss (2025-10)— Cited in Datadog tagged problem set.

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. Then 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 + rebuild

Filter out val, then copy back.

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

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

Tradeoff: Allocates a new array. Datadog will fail you for not doing it in-place when explicitly asked.

2. Two-pointer in-place (optimal)

k tracks the next write slot. Scan with i; whenever nums[i] != val, copy to nums[k] and increment k.

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

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

Tradeoff: Single pass, O(1) extra space. Direct mapping to streaming filter — same shape as a Datadog pipeline transform.

Datadog-specific tips

Datadog will probe ordering: 'Do you preserve relative order? What if you didn't have to?' Show you know the swap-with-end variant (faster when val is rare). Mention how this generalizes to a streaming filter operator.

Common mistakes

Skipping increment of i after a swap-with-end — could miss checking the newly-swapped element.
Returning nums or nums.length instead of k.
Allocating a new array when in-place was required.

Follow-up questions

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

Remove Duplicates from Sorted Array (LC 26).
Move Zeroes — same pattern but val is implicit (LC 283).
Partition array by predicate — generalization.

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Output

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FAQ

Does order matter?

The problem says order can change. If order DID matter, the two-pointer write approach above preserves relative order anyway.

When would you swap-with-end instead?

When val is rare and most elements are kept. Swap-with-end performs fewer writes than copy-forward.

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