155.MinStack.py

# 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.


class MinStack:

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

    def push(self, val: int) -> None:
        self.stack.append(val)

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

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

    def getMin(self) -> int:
        return min(self.stack)


# Your MinStack object will be instantiated and called as such:
if __name__ == "__main__":
    obj = MinStack()
    obj.push(-2)
    obj.push(0)
    obj.push(-3)
    print(obj.getMin())
    obj.pop()
    print(obj.top())
    print(obj.getMin())
# obj = MinStack()
# obj.push(val)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()