Google·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jul 2026

Summary

Got a Google coding round with a sliding window problem that had a twist. Not your standard max/min window setup, which threw me a bit.

Questions Asked (1)

Given an array of numbers, a window size w, and an integer k, slide the window across the array one step at a time. For each window position, remove the k largest elements and return the average of the remaining w minus k elements. Do this efficiently, not brute force.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

My first instinct was just sorting each window, which is obviously too slow and I knew it while saying it out loud.

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AI HintsAI Generated

Suggested Approach

Clarify the problem constraints and edge cases, then propose an efficient solution using a sliding window with a data structure that supports fast insertion, deletion, and order statistics (e.g., two heaps or a balanced BST). Explain how to maintain the sum of the smallest w-k elements dynamically as the window slides, and analyze the time and space complexity.

Pro tip: Discuss the trade-offs between different data structures (e.g., two heaps vs. balanced BST vs. Fenwick tree) and mention that in an interview, you might start with a simpler approach and then optimize, showing your thought process.

1. Understand the problem and constraints

Ask clarifying questions about input size, whether k is fixed, if the array can contain duplicates, and what to return if w-k <= 0. Confirm that the window slides one step at a time.

2. Design an efficient data structure

Choose a data structure that maintains the window elements and can efficiently remove the k largest and compute the sum of the rest. Consider two heaps (min-heap for smallest w-k, max-heap for largest k) or a balanced BST with subtree sums.

3. Handle sliding window updates

When the window slides, remove the outgoing element and add the incoming element. Update the data structure to maintain the partition of smallest w-k and largest k, adjusting the sum accordingly.

4. Compute and return averages

For each window position, after updating the data structure, compute the average of the smallest w-k elements by dividing the maintained sum by (w-k). Handle division by zero if w-k = 0.

5. Analyze complexity and edge cases

State the time complexity per window step (e.g., O(log w) with heaps) and overall O(n log w). Discuss edge cases like k >= w, empty array, and large inputs.

Key Points to Mention

Sliding window technique to avoid recomputing from scratch for each window.
Use of two heaps (min-heap and max-heap) to maintain the smallest w-k and largest k elements, with lazy deletion or a balanced BST for efficient updates.
Maintaining a running sum of the smallest w-k elements to compute averages in O(1) per window.
Time complexity: O(n log w) with heaps or O(n log w) with balanced BST, compared to O(n * w log w) brute force.
Space complexity: O(w) for the data structures.
Edge cases: w-k <= 0, k >= w, duplicates, and negative numbers.

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