K Closest Points to Origin
Given an array of points on a 2D plane, find the $K$ closest points to the origin $(0, 0)$. Use a Max-Heap (Priority Queue) to maintain the $K$ smallest distances in $O(n \log k)$.
Why Interviewers Ask This
Netflix values engineers who balance algorithmic efficiency with practical system constraints. This question tests your ability to optimize space complexity by using a Max-Heap instead of sorting the entire array, ensuring you understand how to maintain a dynamic subset of data efficiently without unnecessary overhead.
How to Answer This Question
1. Clarify requirements: Confirm if points can be negative and what happens if K equals or exceeds the array length.
2. Define distance metric: Explicitly state you will compare squared Euclidean distances (x^2 + y^2) to avoid expensive square root calculations while preserving order.
3. Select the data structure: Propose a Max-Heap of size K to track the K smallest elements seen so far, allowing O(1) access to the largest of the smalls for eviction.
4. Execute the loop: Iterate through each point; push it onto the heap, then pop the root if the heap size exceeds K.
5. Return result: Extract all remaining elements from the heap as the final answer.
6. Analyze complexity: Explain that this approach runs in O(N log K) time and O(K) space, which is superior to full sorting when K is small.
Key Points to Cover
- Using squared distance avoids floating-point errors and costly sqrt operations
- Selecting a Max-Heap specifically allows efficient removal of the largest element in the current window
- Achieving O(N log K) time complexity demonstrates optimization over standard O(N log N) sorting
- Handling edge cases like K being greater than or equal to the array length upfront
- Maintaining O(K) space complexity shows awareness of memory constraints
Sample Answer
To solve finding the K closest points to the origin efficiently, I would first clarify that we are looking for the K points with the smallest Euclidean distance. Since calculating square roots is computationally expensive and monotonic, I will compare the squared distances directly, which is x squared plus y squared.
My approach involves using a Max-Heap to maintain the K smallest distances encountered so far. As I iterate through the input array, I calculate the squared distance for each point. If the heap has fewer than K elements, I simply add the current point. Once the heap reaches capacity, I compare the new point's distance against the root of the Max-Heap, which holds the largest distance among the current K candidates. If the new point is closer, I remove the root and insert the new point; otherwise, I discard the new point.
This strategy ensures that at any step, the heap contains exactly the K closest points found so far. After processing all N points, the heap contains our answer. The time complexity is O(N log K) because each insertion and deletion takes logarithmic time relative to K, rather than N. This is particularly efficient for large datasets where K is much smaller than N, aligning well with Netflix's need for scalable solutions that minimize resource usage.
Common Mistakes to Avoid
- Calculating the actual square root for every point, which adds unnecessary computational overhead
- Sorting the entire array by distance, resulting in O(N log N) time complexity instead of the optimal O(N log K)
- Using a Min-Heap to store all elements, which wastes memory and fails to enforce the K-size constraint efficiently
- Forgetting to handle the case where the initial heap size is less than K during the iteration process
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