Completed Design 2

[Sanket-S-Kale]

No description provided.

[super30admin]

Create Queue using Stacks (Problem1.py)

Strengths:

• Well-documented code with clear comments explaining time complexity and logic
• Good variable naming (inStack, outStack)
• Correct implementation of the core two-stack queue algorithm
• Proper handling of edge cases in empty() method

Areas for Improvement:

1. Bug in peek() method: When outStack is empty, directly returning self.inStack[0] fails after elements have been transferred to outStack. The peek() method should mirror the pop() logic by transferring elements from inStack to outStack when outStack is empty, then return the top of outStack.
2. Consistency: The pop() method correctly transfers elements before popping, but peek() doesn't follow the same pattern, leading to potential inconsistency.

Suggested fix for peek():

def peek(self) -> int:
    if not self.outStack:
        while self.inStack:
            self.outStack.append(self.inStack.pop())
    return self.outStack[-1]

VERDICT: NEEDS_IMPROVEMENT

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Implement Hash Map (Problem2.py)

EVALUATION

Correctness

The solution correctly implements a HashMap using chaining with linked lists. Let me trace through the key operations:

1. put: Correctly handles both updating existing keys and inserting new nodes. The logic to find the previous node and update/insert is sound.

2. get: Returns the value if found, otherwise returns -1. The check previous.next and previous.next.key == key ensures correct retrieval.

3. remove: Properly unlinks the node from the chain by adjusting pointers.

4. findPrevious: This function has a subtle issue. It returns None as previous when the key is found at the head node (index 0 of the chain), which could cause issues in certain edge cases. However, since dummy nodes are used at the head of each chain, this is handled correctly.

Time Complexity

• Average case: O(1) - With 10,000 buckets and at most 10,000 operations, the linked lists remain short (max ~1-2 nodes on average).
• Worst case: O(n) - If all keys hash to the same bucket, traversal becomes O(n).
• The solution uses dummy nodes (initialized with key=-1, val=-1), which is a good practice to simplify edge case handling.

Space Complexity

• O(n + buckets): Where n is the number of entries and buckets (10,000) is the size of the storage array.
• The reference solution uses 1,000 buckets while this solution uses 10,000 buckets, resulting in slightly higher space overhead but still within acceptable bounds.

Code Quality

• Strengths:

  • Clean class structure with nested Node class
  • Good use of type hints
  • Well-documented with comments explaining the approach
  • Consistent naming conventions

• Areas for improvement:

  • The findPrevious function could be simplified - the previous = None initialization at the start is unnecessary since it's always set in the first iteration
  • Comments mention "Complexity: O(1)" but this is only amortized/average case, not guaranteed

Efficiency

• The solution is efficient for the given constraints (≤10⁴ operations, keys/values ≤10⁶)
• Using 10,000 buckets is slightly more than necessary but provides good balance between space and time
• The dummy node approach eliminates many edge cases, making the code cleaner

FEEDBACK

Strengths:

1. Excellent use of dummy head nodes in each bucket - this is a best practice that simplifies insertion and removal logic significantly
2. Clean, well-structured code with clear separation of concerns
3. Good variable naming and readable structure
4. Proper handling of both new key insertion and existing key updates in put

Areas for Improvement:

1. The findPrevious function has slightly redundant logic - the initial previous = None assignment is never used since current starts at the dummy head
2. Consider adding a check in get to verify the key matches (currently it checks previous.next.key == key which is good, but could be more explicit)
3. The comment stating "Complexity: O(1)" should clarify this is amortized time complexity

Minor observations:

• The solution uses more buckets (10,000) than the reference (1,000), which is a valid design choice but uses more memory
• The Python | type syntax for None requires Python 3.10+; for broader compatibility, Optional from typing could be used

The solution correctly implements a HashMap with chaining that handles all required operations (put, get, remove) properly. It follows good practices including the use of dummy nodes and achieves O(1) amortized time complexity for the given constraints. The code is well-structured and readable.

VERDICT: PASS