Roblox·Machine Learning Engineer·Technical Phone Screen·Senior

Senior

Jun 2026

Summary

Roblox ML Engineer interview with a sliding window coding problem. Pretty focused session, just the one algorithmic question from what I can tell, and the constraint to hit O(n) was the whole point of the exercise.

Questions Asked (1)

Given a list of integers, a target value, and a fixed window size k, find all contiguous windows of size k that contain at least one occurrence of the target. Return the start and end indices of each such window. You need to do this in O(n) time using a counter to track how many times the target appears in the current window.

Algorithms & Data Structures

Author's notes

The O(n) constraint is where this gets interesting.

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AI HintsAI Generated

Suggested Approach

Use a sliding window of size k and maintain a counter of how many times the target appears in the current window. Slide the window one step at a time, updating the counter by removing the element leaving the window and adding the new element entering. Whenever the counter is greater than zero, record the current window's start and end indices.

Pro tip: Clarify upfront whether the window size k can be larger than the array or if k is always valid, and mention that the counter approach avoids re-scanning the window, ensuring O(n) time. Also, discuss how you would handle multiple occurrences of the target within a window—the counter naturally handles that.

1. Initialize the first window

Compute the count of the target in the first k elements. If the count > 0, add the window [0, k-1] to the result.

2. Slide the window

For each subsequent position i from k to n-1, remove the element at i-k from the count and add the element at i. This updates the count for the new window [i-k+1, i].

3. Check and record

After each slide, if the count of the target in the current window is > 0, append the start index (i-k+1) and end index (i) to the result list.

4. Handle edge cases

If k > n or k <= 0, return an empty list. Also consider if the target is not present at all—the result will be empty.

5. Return result

After processing all windows, return the list of index pairs.

Key Points to Mention

Time complexity: O(n) because each element is added and removed from the counter at most once.
Space complexity: O(1) extra space for the counter, plus O(m) for the output where m is the number of valid windows.
The counter tracks the frequency of the target in the current window, allowing O(1) updates per slide.
Edge cases: k > n, k <= 0, target not present, multiple targets in a window.
The algorithm can be implemented in a single pass without nested loops.
Clarify that the window indices are inclusive or exclusive as per problem statement (typically inclusive start and end).

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.