"""
155. Min Stack
Difficulty: Medium
https://leetcode.com/problems/min-stack/

──────────────────────────────────────────────────

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 value) pushes the element value 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:

        • -2^31 <= val <= 2^31 - 1

• Methods pop, top and getMin operations will always be called on
non-empty stacks.

        • At most 3 * 10^4 calls will be made to push, pop, top, and getMin.
"""


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

    def push(self, value: int) -> None:
        self.stack.append(value)
        value = min(value, self.minStack[-1] if self.minStack else value)
        self.minStack.append(value)

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

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

    def getMin(self) -> int:
        return self.minStack[-1]


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