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Design-1 Completed#2660

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CHINMAY-PRAJAPATI wants to merge 1 commit intosuper30admin:masterfrom
CHINMAY-PRAJAPATI:master
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Design-1 Completed#2660
CHINMAY-PRAJAPATI wants to merge 1 commit intosuper30admin:masterfrom
CHINMAY-PRAJAPATI:master

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Implement Hash Set (hashset.py)

Strengths:

  • The solution correctly implements the required operations.
  • The use of separate chaining with linked lists is a classic and effective approach.
  • The code is clean, well-organized, and easy to understand.
  • The student has provided complexity analysis, which shows good understanding.

Areas for improvement:

  1. The hash function could be improved by using a prime number for the number of buckets to reduce collisions. The current choice of 1000 is acceptable but not optimal. A prime number (like 1009) might distribute keys more evenly.
  2. The linked list traversal in the add and remove methods could be optimized by breaking early when the key is found. Currently, the add method traverses the entire list even if it finds a duplicate (it returns only after finding the duplicate). This is acceptable but not necessary. Similarly, the remove method could break after removal, but it already does so by returning.
  3. The remove method does not free the node (though in Python, garbage collection will handle it), but it is not a major issue.
  4. The initializer creates 1000 ListNode instances. This is acceptable, but note that each dummy node takes memory. Alternatively, you could initialize the list with None and create the dummy node only when needed. However, the current approach simplifies the code.

Overall, the solution is correct and efficient. The time and space complexities are as expected for a hash set implementation.

VERDICT: PASS


Implement Min Stack (minstack.py)

Your solution is well-structured and meets the problem requirements. Here are a few suggestions for improvement:

  1. Redundancy in getMin: The problem states that getMin will always be called on non-empty stacks, so the check if self.minstack is not necessary. You can directly return self.minstack[-1] without the condition. This simplifies the code.

  2. Initialization: It's good practice to initialize both stacks in the __init__ method. You've done that correctly.

  3. Efficiency: Your approach is efficient with O(1) time for all operations. However, you could consider if there's a way to reduce the space usage. For example, the reference solution in Java uses a single variable to track the current min, but it also uses a second stack to store the min at each push. Your solution is similar and optimal.

  4. Code Comments: You have included comments about time and space complexity, which is excellent. However, you might want to add a docstring for the class to explain its purpose.

  5. Edge Cases: Although the problem constraints ensure non-empty stacks for certain operations, it's always good to think about how the code would behave if the stack were empty. For instance, if someone were to call pop on an empty stack, it would raise an exception. But since the problem states that operations are called on non-empty stacks, this is acceptable.

Overall, your solution is correct and efficient. Keep up the good work!

VERDICT: PASS

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