Design 2

[nagasai67]

No description provided.

[super30admin]

Strengths:

• The solution correctly implements the HashMap using separate chaining with linked lists.
• The code is clean, well-commented, and easy to understand.
• The use of a dummy node in each bucket simplifies the insertion and deletion operations.

Areas for Improvement:

• In the put method, you are traversing the list to check if the key exists and then traversing again to append if it doesn't. This can be optimized by searching for the key and then updating or appending in a single pass. The reference solution uses a helper method getPrev to find the node before the target key, which allows handling both update and insertion without redundant traversal.
• Consider using a larger bucket size (e.g., 1000 or 10000) to reduce the average chain length, especially since the constraints allow up to 10^4 operations. This would improve worst-case performance.
• The remove method currently does not free the memory of the removed node. In Python, this is not critical due to garbage collection, but it's good practice to explicitly set the node to None if possible. However, the current implementation is acceptable.

Overall, the solution is correct and efficient. The improvements are minor and primarily about optimization.

[super30admin]

Your solution for the HashMap is well-implemented and follows the separate chaining approach correctly. Here are some points to consider:

Strengths:

• You used a dummy head node for each bucket, which simplifies the insertion and deletion operations.
• The code is clean and easy to understand.
• You correctly handle the operations for put, get, and remove.

Areas for Improvement:

• The initial size of the buckets array (100) might be too small for the constraints (up to 10^4 operations). Consider using a larger prime number (like 1000 or 10000) to reduce the chance of collisions and improve average performance.
• In the put method, you traverse the entire list even when you might have to update an existing key. However, you break early when you find the key for update. This is acceptable, but note that if the key is not present, you always traverse the entire list. This could be avoided by storing a tail pointer, but it's not necessary.
• The remove method does not need to set the next pointer of the removed node to None (as in the reference solution), but it is good practice to avoid memory leaks in some languages. In Python, it's less critical, but still a good habit.
• You included an unrelated solution for MinStack in the same file. In the future, please only include the solution for the problem you are solving.

Overall, your solution is correct and efficient. Well done!