Min Stack 20 XP Medium

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Design a stack that supports push, pop, top, and retrieving the minimum element — all in O(1) time.

min_stack = MinStack()
min_stack.push(-2)
min_stack.push(0)
min_stack.push(-3)
min_stack.getMin()  # → -3
min_stack.pop()
min_stack.top()     # → 0
min_stack.getMin()  # → -2

The trick: maintain two stacks:

Main stack: Stores all values normally
Min stack: Stores the current minimum at each level
When pushing, the min stack stores min(val, current_min)
When popping, pop from both stacks

This way, getMin() just peeks at the top of the min stack — always O(1)!

📋 Instructions

Implement a MinStack class with push, pop, top, and getMin. Test with the provided sequence and print the results.

Keep a second stack (min_stack) that always has the current minimum on top. When pushing value x, push min(x, min_stack[-1]) onto min_stack. Pop from both stacks together.

🧪 Test Cases

Input

ms.getMin()

Expected

-3

Test case 1

Input

ms.top()

Expected

0

Boundary value

Input

ms.getMin()

Expected

-2

Test case 3

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main.py

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Output

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