Completed Design-1

[samikshm]

No description provided.

[super30admin]

Implement Hash Set (design-hashset.java)

Strengths:

• The solution correctly implements the required operations (add, remove, contains) with efficient double hashing.
• Lazy initialization of secondary arrays saves space until needed.
• The edge case for key=10^6 (which maps to bucket 0 and secondary index 1000) is handled appropriately by allocating an extra element in the bucket at index 0.
• Code is clean, well-commented, and follows good practices.

Areas for improvement:

• The comment about space complexity being O(1) effectively might be misleading. While the primary array has a fixed size (1000), the secondary arrays are allocated as needed, and the total space used depends on the number of keys inserted. It would be more accurate to state that the space complexity is O(n) in the worst case, but with a constant factor due to the fixed number of buckets.
• The method names getPrimaryHashKey and getSecondaryHashKey are slightly verbose. Typically, these might be named hash1 and hash2 or similar for brevity, but the current names are clear.
• There is a minor inconsistency: the variable secondaryBuckets is set to 1000, but in the secondary hash function, it is used as a divisor. Since the primaryBuckets is 1000, the secondaryBuckets should also be 1000 to ensure the secondary index fits within the allocated array. This is correct, but it might be worth noting that the secondaryBuckets value is chosen to match the primaryBuckets for even distribution.

Overall, the solution is excellent and meets all requirements.

VERDICT: PASS

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Implement Min Stack (min-stack.java)

Your solution is correct and efficient. You have successfully implemented the MinStack with constant time operations. Here are some points to consider:

1. Correctness: Your solution handles all cases correctly, including when multiple same minimum values are pushed. However, note that when pushing a value equal to the current minimum, you push the old minimum again. This is important because when you pop the current minimum, you need to restore the previous minimum (which might be the same). This is correct.

2. Time Complexity: All operations are O(1), which meets the requirement.

3. Space Complexity: The space complexity is O(n), which is acceptable. However, in the worst case (if the input is strictly decreasing), your stack will have approximately 2n elements (because for each push of a new minimum, you push two items: the old min and the new value). This is still O(n), but it uses more space than the two-stack approach (which uses exactly 2n space). But this is a trade-off: you are using one stack instead of two, which might be considered simpler.

4. Code Quality: Your code is well-structured and commented. The variable names are clear. Good job!

5. Efficiency: One minor improvement: you can initialize min to Integer.MAX_VALUE as you did, but note that if the first push is Integer.MAX_VALUE, then you push Integer.MAX_VALUE (the old min) and then push Integer.MAX_VALUE (the new value). This is correct. However, if you want to avoid pushing the old min when the stack is empty (which is the initial state), you could handle the first push differently. But it is not necessary.

Overall, your solution is excellent. It is a common and efficient approach.

VERDICT: PASS