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The solution correctly implements the required methods (push, pop, peek, empty) using two stacks.
The time and space complexity analysis in the comments is accurate and demonstrates understanding.
The code is clean, well-organized, and follows Python naming conventions (using underscores for variable names).
The comments explain the approach clearly, which is good for maintainability.
Areas for Improvement:
While the code is correct, it's important to note that the pop method should ideally handle the case when the queue is empty. The problem states that all calls to pop and peek are valid, so it's acceptable to assume non-empty queues. However, adding a check in pop (like calling empty) might be more robust for real-world scenarios, though it's not required here.
The comments at the top are helpful, but they could be expanded to explain why the time complexity is amortized O(1) (e.g., because each element is moved from in_stack to out_stack exactly once).
The code includes a commented-out test section at the bottom, which is good for testing but should be removed in a production or submission context to avoid clutter.
Overall, this is a strong solution that meets all requirements.
VERDICT: PASS
Implement Hash Map (Hashmap.py)
Your solution is correct and handles the requirements of the problem. Here are some points to consider for improvement:
Bucket Size: You chose 1001 as the bucket size. While this is a prime number (which is good for reducing collisions), the problem constraints (keys up to 10^6 and 10^4 operations) might benefit from a larger bucket size to minimize collisions. The reference solution uses 1000, but you could consider a larger prime number (e.g., 10007) to spread the keys more evenly.
Data Structure for Buckets: You are using a list of lists, where each inner list stores key-value pairs as two-element lists. This works, but it might be more efficient to use a linked list of nodes (as in the reference) or even a list of tuples. However, for the given constraints, your approach is acceptable. Alternatively, you could use a list of dictionaries for each bucket, but that might be overkill.
Efficiency in Removal: In the remove method, you are iterating through the bucket and then using pop(i), which has a time complexity of O(n) for the inner list. Since the inner lists are small, this is acceptable. However, if you expect many collisions, you might consider using a linked list to make removal O(1) after finding the node. But for most cases, your current method is fine.
Code Comments: Your comments are good, but you could add more details about why you chose 1001 as the size and how the hash function works.
Edge Cases: Your solution handles non-existent keys correctly in get and remove. However, consider what happens if the key is negative? The problem states that keys are non-negative, so it's not an issue, but it's good to note.
Performance: The average time complexity is O(1) for each operation, which meets the requirements. The worst-case O(n) is acceptable given the constraints.
Overall, your solution is solid. Keep up the good work!
VERDICT: PASS
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