Implemented Design-Hashset

[LeelaTotapally]

No description provided.

[super30admin]

Implement Hash Set (Design-Hashset.java)

Your solution is correct and efficient for the given constraints. It is simple and easy to understand. However, consider the following points for improvement:

1. Space Efficiency: The array size is fixed to 1,000,001, which always allocates memory for the entire range of possible keys (0 to 10^6), even if only a few keys are stored. This might be inefficient in terms of memory usage. The reference solution uses a dynamic approach with double hashing to reduce the memory usage by only allocating buckets as needed.
2. Scalability: If the key range were to increase (e.g., up to 10^9), your solution would not be feasible due to the large array size. The double hashing technique can be adapted to handle larger ranges by adjusting the bucket sizes.
3. Best Practices: Your code is clean and well-commented. However, you could consider using a more flexible design that doesn't rely on a fixed array size if you want to handle larger key ranges or optimize for memory.

For future problems, think about how to balance time and space complexity. While your solution is optimal for time, there are trade-offs with space. The double hashing method is a common technique to achieve O(1) time complexity with better space efficiency in practice.

VERDICT: PASS

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Implement Min Stack (Min-Stack-155.java)

Your solution correctly implements the push, pop, and top operations with O(1) time complexity using an ArrayList. However, the getMin function currently runs in O(n) time because it scans the entire list to find the minimum. This does not meet the problem's requirement of constant time for each function.

To improve, consider using an auxiliary data structure to keep track of the minimum values. One common approach is to use two stacks: one for storing all the elements and another for storing the current minimum at each push operation. This way, when you push a value, you also push the current minimum (which is the minimum between the new value and the previous minimum) to the min stack. When you pop, you also pop from the min stack, and the top of the min stack will always be the current minimum. This ensures that getMin runs in O(1) time.

Here's a brief outline of how you can modify your solution:

• Initialize two stacks: stack for the main elements and minStack for the minimums.
• In push(val), push val to stack. Then, push the minimum between val and the current top of minStack (if minStack is not empty) to minStack. If minStack is empty, push val.
• In pop(), pop from both stack and minStack.
• top() returns the top of stack.
• getMin() returns the top of minStack.

This approach maintains O(1) time for all operations and O(n) space, which is optimal for this problem.

Your current code is clean and well-commented, which is good. Keep up the good work on code quality, but focus on optimizing the time complexity for getMin.

VERDICT: NEEDS_IMPROVEMENT