solution to both problems

[avishwak]

No description provided.

[super30admin]

1. Correctness:

   • For Problem 1 (Implement Queue using Stacks), the solution correctly uses two stacks to implement a queue, maintaining the FIFO order. The approach of transferring elements from inStack to outStack when outStack is empty is standard and correct.
   • For Problem 2 (Design HashMap), the solution uses an array of linked lists with a hash function to handle collisions, which is a correct approach. The use of a dummy head node simplifies insertion and deletion operations.

2. Time Complexity:

   • For Problem 1, the time complexity is correctly stated as O(1) for push and O(n) for pop and peek in the worst case. This is accurate because transferring elements from inStack to outStack can take O(n) time.
   • For Problem 2, the time complexity is correctly stated as O(1) for put, get, and remove operations, assuming a good hash function and uniform distribution of keys.

3. Space Complexity:

   • For Problem 1, the space complexity is correctly stated as O(n) where n is the number of elements in the queue.
   • For Problem 2, the space complexity is correctly stated as O(n) where n is the number of unique keys added to the HashMap.

4. Code Quality:

   • The code is well-structured and readable. Comments are provided to explain the approach, which is helpful.
   • For Problem 1, the method names (push, pop, peek, empty) are standard and appropriate.
   • For Problem 2, the use of a nested Node class and helper methods (find, hash_func) improves modularity and readability.
   • One minor improvement could be to add more detailed comments in the find method of Problem 2 to explain its purpose and return value.

5. Efficiency:

   • For Problem 1, the solution is efficient and follows the standard approach. No further optimizations are needed.
   • For Problem 2, the solution is efficient. However, the hash function uses a fixed size (10000), which might not be optimal for all cases. A dynamic resizing mechanism could be considered for better performance with varying numbers of keys.

Potential Edge Cases:

• For Problem 1, edge cases like popping from an empty queue are handled implicitly by the empty check in peek.
• For Problem 2, edge cases like removing a non-existent key or getting a non-existent key are handled correctly.