155. Min Stack | LeetCode | Solution

This problem is all about designing a stack data structure. Here is the problem link. See the detailed explanation there.

To solve this problem, we have to write the most common stack operations. A stack is an abstract data structure. It’s Follow the LIFO (Last In First Out) rules. I hope you already know about the stack. There are descriptions of which operations to write.

We use a list to illustrate the stack. For that, in the constructor function, we will declare an empty list. A class already created for us in LeetCode editor. We have to write the missing parts.

def __init__(self):
	"""
	initialize your data structure here.
	"""
	self.stack = []

Next, we have to write the push() method. When we store new data in a stack, we add it at the end of all. Data will be pass through the function argument. Let’s see the code.

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

Next, we will write the pop() method. This method removes the last element from the stack.

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

Our next method is top(). It returns the top value (the last element in the list) from the stack but does not delete it.

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

And the last method is getMin(). Here we will return the minimum value from our stack. Remember, we have to return value, not the element. To get the minimum value, we will use the min() function. Let’s see the code.

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

Operations will always be called on non-empty stacks. That’s why we don’t need to check anything.

See the full code below.

class MinStack:

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = []
        

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

    def pop(self) -> None:
        self.stack.pop()    
    
    def top(self) -> int:
        return self.stack[-1]

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

If you submit the code in LeetCode, it will be accepted.