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Min Stack

Asked at:AmazonGoogleBloomberg

Design a stack that supports push, pop, top, and retrieving the minimum element, all in O(1) time. Implement the MinStack class with these four operations.

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Problem

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

Input

A sequence of operations: `MinStack()` constructs the stack; `push(val)` pushes an integer; `pop()` removes the top element; `top()` returns the top element; `getMin()` returns the current minimum.

Output

The return values of `top()` and `getMin()` calls (push, pop, and constructor return nothing).

Examples

Input: MinStack ms; ms.push(-2); ms.push(0); ms.push(-3); ms.getMin(); ms.pop(); ms.top(); ms.getMin();

Output: [-3, 0, -2]

getMin() after three pushes → -3. After pop(), top() → 0 (the -3 was removed). getMin() now → -2.

The brute-force approach

Use a regular stack for push/pop/top. For getMin(), scan the entire stack every time to find the smallest element.

class MinStack:
    def __init__(self):
        self.stack = []

    def push(self, val): self.stack.append(val)
    def pop(self):       self.stack.pop()
    def top(self):       return self.stack[-1]

    def get_min(self):
        return min(self.stack)    # ← O(n) scan every call

getMin() scanning the whole stack on every call is O(n). The problem requires all four operations to be O(1).

Time: O(n) for getMinSpace: O(n)

Spotting the pattern

This is a Stack problem. The key question to ask yourself:

Can I store the current minimum alongside each pushed element so I always know it without scanning?

Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.

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