Daily Interview Problem: Min Stack

Hi, here's your problem today. This problem was recently asked by Uber:

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

push(x) -- Push element x onto stack.
pop() -- Removes the element on top of the stack.
top() -- Get the top element.
getMin() -- Retrieve the minimum element in the stack.

Note: be sure that pop() and top() handle being called on an empty stack.

class minStack(object):

  def __init__(self):

    # Fill this in.



  def push(self, x):

    # Fill this in.



  def pop(self):

    # Fill this in.



  def top(self):

    # Fill this in.



  def getMin(self):

    # Fill this in.



x = minStack()

x.push(-2)

x.push(0)

x.push(-3)

print(x.getMin())

# -3

x.pop()

print(x.top())

# 0

print(x.getMin())

# -2