Leetcode 155. Min Stack

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.

The key for this problem is determining the current minimal value in the stack. We can leverage a min_stack which will restore the minimal value while the stack is stacking up.

class MinStack:

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

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

        if self.min_stack:
            self.min_stack.append(min(self.min_stack[-1], val))
        else:
            self.min_stack.append(val)
        

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

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

    def getMin(self) -> int:
        return self.min_stack[-1]