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

## Question

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

Implement the `MinStack` class:

• `MinStack()` initializes the stack object.
• `void push(int val)` pushes the element `val` onto the stack.
• `void pop()` removes the element on the top of the stack.
• `int top()` gets the top element of the stack.
• `int getMin()` retrieves the minimum element in the stack.

You must implement a solution with `O(1)` time complexity for each function.

Example 1:

```Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]

Output
```

[null,null,null,null,-3,null,0,-2]

Explanation MinStack minStack = new MinStack(); minStack.push(-2); minStack.push(0); minStack.push(-3); minStack.getMin(); // return -3 minStack.pop(); minStack.top(); // return 0 minStack.getMin(); // return -2

Constraints:

• `-231 <= val <= 231 - 1`
• Methods `pop``top` and `getMin` operations will always be called on non-empty stacks.
• At most `3 * 104` calls will be made to `push``pop``top`, and `getMin`.

## Python Solution

```class MinStack:

def __init__(self):
"""
"""
self.stack = []

def push(self, x: int) -> None:
cur_min = self.getMin()
if cur_min==None or cur_min > x:
cur_min = x
self.stack.append([x,cur_min])

def pop(self) -> None:
self.stack.pop()

def top(self) -> int:
if len(self.stack)==0:
return None
return self.stack[-1][0]

def getMin(self) -> int:
if len(self.stack)==0:
return None
return self.stack[-1][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()```