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###### tags: `Leetcode`
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# Leetcode 155. Min Stack

[link](https://leetcode.com/problems/min-stack/)

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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.

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The provided code defines a class called MinStack that implements a stack data structure with an additional feature to retrieve the minimum element in constant time. The class maintains two stacks: stack and minStack. Here's a breakdown of the methods:

__init__(self): This is the constructor method that initializes the two stacks.

push(self, val: int) -> None: This method adds a new element to the stack. It appends the value to the stack and also updates the minStack. If minStack is not empty, it appends the minimum of the current minimum (minStack[-1]) and the new value (val). Otherwise, it appends the new value itself.

pop(self) -> None: This method removes the top element from both the stack and the minStack.

top(self) -> int: This method returns the top element of the stack without modifying the stack itself.

getMin(self) -> int: This method returns the minimum element of the stack in constant time by accessing the top element of the minStack.

The usage of the minStack allows for efficient retrieval of the minimum element at any point, as the minimum value is always stored at the top of the minStack.

#### Solution 1

class MinStack:

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

    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(val)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()