Sliding Window Maximum
Given an array `nums` and a sliding window size $k$, find the maximum number in each window. Use a Deque (Double-Ended Queue) for an $O(n)$ solution.
Why Interviewers Ask This
Google engineers ask this to evaluate your ability to optimize time complexity beyond naive solutions. They specifically test if you can leverage a Deque to maintain state efficiently, demonstrating mastery of data structures and sliding window patterns essential for high-throughput systems.
How to Answer This Question
1. Clarify constraints: Confirm input size and whether k is valid. 2. Propose the brute force O(n*k) solution first to establish a baseline, then immediately pivot to the optimal approach. 3. Explain the Deque strategy: Describe how indices are stored so the front always holds the maximum index, and elements smaller than the current number are removed from the back. 4. Walk through the logic: Detail how you slide the window, remove expired indices (those outside the current k range), and add the new element while maintaining order. 5. Analyze complexity: Explicitly state why each element is added and removed at most once, proving O(n) time and O(k) space. 6. Implement carefully: Write clean code handling edge cases like empty arrays or k=1.
Key Points to Cover
- Demonstrating knowledge of the Deque data structure for monotonic queue optimization
- Explaining the O(n) time complexity proof based on single-pass operations
- Handling edge cases like invalid k values or empty input arrays gracefully
- Clearly articulating the logic for removing expired indices from the front
- Maintaining the descending order invariant within the Deque during iteration
Sample Answer
To solve the Sliding Window Maximum problem efficiently, I would start by acknowledging that a brute force approach checking every window takes O(n*k) time, which is inefficient for large datasets Google often handles. I…
Common Mistakes to Avoid
- Using a priority queue which results in O(n log k) instead of the required O(n)
- Forgetting to remove indices that have slid out of the window from the front
- Storing actual values instead of indices, making it impossible to check window boundaries
- Implementing the solution without explaining why the monotonic property holds
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