Count of Smaller Numbers After Self
Given an integer array `nums`, return an array `counts` where `counts[i]` is the number of elements to the right of `nums[i]` that are strictly smaller than `nums[i]`. Use a modified Merge Sort or Fenwick Tree/Segment Tree.
Why Interviewers Ask This
Tesla evaluates this question to assess a candidate's ability to optimize brute-force solutions for real-time systems. They specifically look for mastery of divide-and-conquer strategies or advanced data structures like Fenwick Trees, which are critical for handling high-volume sensor data processing efficiently within their autonomous driving stacks.
How to Answer This Question
1. Clarify constraints immediately: Ask about array size and value ranges to determine if O(n^2) is acceptable or if O(n log n) is mandatory.
2. Propose the optimal strategy: Explicitly state you will use Merge Sort or a Binary Indexed Tree (Fenwick Tree) to achieve logarithmic time complexity per element.
3. Explain the logic clearly: For Merge Sort, describe how counting smaller elements happens during the merge phase by comparing left and right subarrays. For Fenwick Trees, explain coordinate compression and updating indices from right to left.
4. Walk through a dry run: Use a small example like [5, 2, 6, 1] to trace the algorithm step-by-step, showing exactly how the count updates.
5. Discuss trade-offs: Briefly mention space complexity versus implementation complexity, demonstrating awareness of memory constraints relevant to Tesla's embedded environments.
Key Points to Cover
- Explicitly identifying the need for O(n log n) complexity over O(n^2)
- Explaining the specific mechanism of counting during the merge phase
- Demonstrating understanding of coordinate compression if using a Fenwick Tree
- Providing a clear, step-by-step dry run with a concrete example array
- Acknowledging space complexity implications for embedded systems
Sample Answer
To solve the Count of Smaller Numbers After Self problem efficiently, I would avoid the naive O(n^2) nested loop approach, as it won't scale for large datasets typical in automotive telemetry. Instead, I recommend a modified Merge Sort algorithm that achieves O(n log n) time complexity. The core insight is that during the merge step of sorting, we can simultaneously count how many elements from the right subarray are smaller than the current element in the left subarray. When merging two sorted halves, if an element from the right half is smaller than the current element in the left half, it implies that all remaining elements in the left half are also larger than this right element. We increment our counter accordingly before placing the right element into the merged result. Alternatively, a Fenwick Tree could be used after coordinate compression to handle updates and queries in logarithmic time. I would start by implementing the Merge Sort variant because it doesn't require preprocessing the values for range mapping. Let's trace [5, 2, 6, 1]: splitting down to single elements, then merging [2, 5] with no counts, then [1, 6], and finally merging those results while tracking that 5 has one smaller number to its right (the 1), and 2 has one (the 1). This approach balances performance and code readability, fitting well within strict interview time limits.
Common Mistakes to Avoid
- Implementing a brute force O(n^2) solution without discussing optimization due to time constraints
- Forgetting to handle duplicate values correctly when strictly smaller comparisons are required
- Confusing the direction of iteration, failing to process elements from right to left for tree-based approaches
- Skipping the explanation of why standard sorting algorithms alone cannot solve the problem without modification
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