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It appears there has been a significant mix-up. The solution you provided is for the "Min Stack" problem, not the "Design HashSet" problem.
To correctly solve the Design HashSet problem, you need to:
Create a MyHashSet class
Implement three methods: add(key), remove(key), and contains(key)
Use a hashing strategy (like the reference solution's two-level bucketing approach, or a simpler approach using a large boolean array since keys are bounded by 10^6)
A simple approach could be:
Use a boolean array of size 10^6 + 1
add(key): set array[key] = True
remove(key): set array[key] = False
contains(key): return array[key]
Or a more space-efficient approach using bucketing as shown in the reference solution.
Please resubmit with the correct solution to the Design HashSet problem.
VERDICT: NEEDS_IMPROVEMENT
Min Stack
Strengths:
Correct implementation of the two-stack approach
Excellent documentation with detailed docstrings explaining the algorithm
Good variable naming (main_stack, min_stack)
Handles edge cases (empty stack pop)
Time and space complexity match the reference solution
Areas for Improvement:
Remove debug print statements: The numerous print() statements throughout the code are debugging artifacts. In production code, these should be removed or replaced with proper logging.
Redundant length variable: The self.length variable is unnecessary since len(self.main_stack) provides the same information. This adds complexity without benefit.
Simplify top() method: The method can be simplified to just return self.main_stack[-1] since the problem guarantees non-empty stacks. The current implementation with val = None and conditional logic is unnecessarily complex.
Consider using len() directly: Instead of maintaining self.length, use len(self.main_stack) for emptiness checks.
The run_min_stack() function: This is fine for testing but should be in a separate test file or under if __name__ == "__main__": guard for better code organization.
VERDICT: PASS
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