# Question

Design a 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.

Example:

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin();   --> Returns -3.
minStack.pop();
minStack.top();      --> Returns 0.
minStack.getMin();   --> Returns -2.


# Solution

class MinStack(object):

def __init__(self):
"""
"""
self.stack = list()
self.min_vals = list()

def push(self, x):
"""
:type x: int
:rtype: void
"""
if len(self.min_vals) > 0:
self.min_vals.append(min(x, self.min_vals[-1]))
else:
self.min_vals.append(x)
self.stack.append(x)

def pop(self):
"""
:rtype: void
"""
self.min_vals.pop()
self.stack.pop()

def top(self):
"""
:rtype: int
"""
return self.stack[-1]

def getMin(self):
"""
:rtype: int
"""
return self.min_vals[-1]

# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()