Roblox·Machine Learning Engineer·Technical Phone Screen·Senior

Senior

May 2026

Summary

Roblox ML engineer interview had at least one coding problem that looked deceptively simple but required you to think carefully about efficiency. The sliding window framing was clear enough but getting the implementation right under pressure was a different story.

Questions Asked (1)

Given a list of integers, a target value, and a fixed window size k, find the starting index of the first window of size k that contains the most occurrences of the target. Break ties by choosing the smallest starting index, and solve it in linear time.

Algorithms & Data Structures

Author's notes

The O(n) constraint is what makes this non-trivial.

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AI HintsAI Generated

Suggested Approach

Use a sliding window of size k to maintain the count of the target in the current window. Iterate through the list, updating the count by adding the new element and removing the old element, and track the maximum count and the earliest starting index where it occurs. This achieves O(n) time and O(1) extra space.

Pro tip: Clarify edge cases upfront, such as when k > n or when the target is not present, and confirm the expected return value (e.g., -1). Also, mention that you're optimizing for linear time by avoiding recomputing counts for each window.

1. Understand the problem and edge cases

Restate the problem to ensure clarity: find the first window of size k with the most occurrences of target, breaking ties by smallest index. Discuss edge cases like k > n, empty list, or target not in list.

2. Initialize the first window

Compute the count of target in the first window (indices 0 to k-1). Set this as the current maximum count and record the starting index 0 as the best index.

3. Slide the window across the list

For each subsequent starting index i from 1 to n-k, update the count by subtracting the element leaving the window (at i-1) and adding the element entering (at i+k-1). Compare the new count with the maximum; if greater, update max and best index. If equal, keep the earlier index (so no update).

4. Return the result

After processing all windows, return the best starting index. If no window exists (k > n), return -1 or as specified.

Key Points to Mention

Sliding window technique for O(n) time complexity
Maintaining a running count of the target in the current window
Updating the count in O(1) per window by adding and removing one element
Tracking the maximum count and the earliest index for tie-breaking
Handling edge cases such as k > n or empty input
Space complexity O(1) beyond the input

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