LC155 - Min Stack

Problem

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

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

Solution

In this solution, we utilize two stacks to implement the MinStack. One to hold the elements of the actual stack, and another to hold the minimum values of the stack corresponding to the current stack state. We use $O(2n)$ space and all functions take $O(1)$ time.

class MinStack:

	def __init__(self):
		self.stack = []
		self.minVal = []
	
	def push(self, val: int) -> None:
		self.stack.append(val)
		if self.minVal:
			minVal = min(val, self.minVal[-1])
		else:
			minVal = val
		self.minVal.append(minVal)
	
	def pop(self) -> None:
		self.stack.pop()
		self.minVal.pop()
	
	def top(self) -> int:
		return self.stack[-1]
	
	def getMin(self) -> int:
		return self.minVal[-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()