I knew this one but still fumbled explaining the why behind the auxiliary stack.
Use an auxiliary stack that tracks the minimum value at each level of the main stack. When pushing, push the new value onto the main stack and push the new minimum onto the auxiliary stack. When popping, pop from both stacks. The min operation simply returns the top of the auxiliary stack.
Pro tip: Mention that this approach uses O(n) extra space, but you can optimize to O(1) extra space by storing the difference between the value and the current minimum, or by using a single stack with encoded values. Also, clarify that the min operation should return the minimum element in the current stack, not remove it.
Confirm that all operations (push, pop, min) must be O(1) time, and that min returns the minimum element without removing it. Ask about constraints like memory usage or thread safety if relevant.
Explain that you will maintain two stacks: one for the actual elements and one for the minimum values. The auxiliary stack's top always holds the current minimum.
For push: push value onto main stack; if auxiliary stack is empty or value <= current min, push value onto auxiliary stack. For pop: pop from main stack; if popped value equals auxiliary stack's top, pop from auxiliary stack as well.
The min operation simply returns the top of the auxiliary stack (or None if stack is empty). This is O(1).
Mention the O(1) space optimization using difference encoding or a single stack with a min variable. Handle edge cases like popping from an empty stack or pushing duplicate minimums.
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