MinStack.py

# Time Complexity : O(1)
# Space Complexity : O(n)
# Did this code successfully run on Leetcode : Yes
# Any problem you faced while coding this : No
# Approach : Only when a new min appears, we store the previous min in the stack,
# so that when the current min is removed, we can restore the old one in constant time.

class MinStack:

    def __init__(self):
        self.stack = []
        self.min = float('inf')

    def push(self, val: int) -> None:
        if val <= self.min:
            self.stack.append(self.min)
            self.min = val
        self.stack.append(val)

    def pop(self) -> None:
        if self.min == self.stack.pop():
            self.min = self.stack.pop()

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

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