Design-1 solution

[shakthinandana]

No description provided.

[super30admin]

Implement Hash Set (hashset.py)

Your solution is well-implemented and follows the double hashing approach correctly. You have correctly identified the time and space complexity. The code is clean and easy to read.

One minor point: in the remove method, you don't need to check if the primary bucket is not None because if it is None, the method returns without doing anything. However, your current code does this check correctly. Alternatively, you could have written:

if self.hashset[primary_bucket] is None:
    return
self.hashset[primary_bucket][secondary_index] = False

But your current code is also correct and clear.

Another note: you have defined buckets and bucket_size as class variables. This is acceptable, but note that if multiple instances of MyHashSet are created, they will share these constants, which is fine since they are constants. However, it might be slightly clearer to define them as instance variables in __init__ if you anticipate needing to change them per instance (though not necessary here).

Overall, excellent job!

VERDICT: PASS

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Implement Min Stack (min-stack.py)

Your solution is excellent and meets all the requirements of the problem. Here are some strengths and a minor suggestion:

Strengths:

• You have correctly implemented the two-stack approach, which is a standard solution for this problem.
• The code is clean, readable, and well-commented.
• All operations are O(1) in time and space complexity is O(n), which is optimal.

Suggestion:

• In the push method, you check if the minStack is empty to decide whether to push the value directly. Instead, you could initialize the minStack with a large value (like float('inf')) in the init method. This would allow you to avoid the conditional check in push, making the code slightly more efficient and consistent. However, this is optional and your current solution is correct.

Example of the alternative initialization:

def __init__(self):
    self.stack = []
    self.minStack = [float('inf')]  # Initialize with a large value

def push(self, val):
    self.stack.append(val)
    self.minStack.append(min(val, self.minStack[-1]))

This change would simplify the push method by removing the if-else condition. But note that when the stack is empty, getMin would return float('inf'), which is acceptable only if the problem constraints guarantee that getMin is called on non-empty stacks (as stated in the constraints). However, since the constraints say that operations are called on non-empty stacks, this is safe.

Overall, your solution is correct and efficient. Well done!

VERDICT: PASS