NURO·Machine Learning Engineer·Technical Phone Screen·Intermediate

Intermediate

Apr 2026

Summary

Second round coding interview at Nuro for an ML engineer role. Pretty standard algorithmic problem but the implementation detail tripped me up a bit.

Questions Asked (1)

Given n children each with some value, find the maximum. The twist is you need to track the top two maximums using a single loop.

Algorithms & Data Structures

Author's notes

My first instinct was to just sort and grab the top two, which they pushed back on immediately.

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AI HintsAI Generated

Suggested Approach

Clarify the problem constraints (e.g., n >= 2, distinct values) and then walk through a single-pass algorithm that maintains two variables for the largest and second-largest values. Emphasize edge cases and the O(n) time, O(1) space complexity.

Pro tip: Mention that this pattern generalizes to finding the top k elements with a heap, but for k=2 a simple two-variable approach is optimal. Also, proactively discuss how you'd handle duplicates or negative numbers.

1. Clarify requirements and edge cases

Ask if n can be less than 2, if values can be negative or duplicate, and whether the second maximum must be strictly less than the maximum. This shows attention to detail.

2. Initialize two variables

Set max1 and max2 to negative infinity (or the smallest possible value) to handle negative numbers and ensure correct updates.

3. Iterate through the array once

For each value, if it's greater than max1, update max2 = max1 and max1 = value; else if it's greater than max2 and not equal to max1, update max2 = value.

4. Return the second maximum

After the loop, max2 holds the second largest value. If max2 remains negative infinity, handle the case where no second maximum exists (e.g., all elements equal).

5. Analyze complexity and test

State that time complexity is O(n) and space is O(1). Walk through a small example and edge cases to verify correctness.

Key Points to Mention

Single-pass algorithm with O(n) time and O(1) space
Handling duplicates: decide whether second maximum can equal the maximum
Edge cases: n < 2, all elements equal, negative numbers
Comparison logic: using else-if to avoid unnecessary checks
Initialization with negative infinity or first two elements
Generalization to top-k using a min-heap for larger k

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