Min Stack

Problem

https://leetcode.com/problems/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:

• -2:sup:`31`<= val <= 2:sup:`31`- 1

• Methods pop, top and getMin operations will always be called on non-empty stacks.

• At most 3 * 10:sup:`4` calls will be made to push, pop, top, and getMin.

Pattern

Stack, Design

Approaches

Two Stacks

Code

from collections import deque


class MinStack:

    def __init__(self):
        self.stack = deque()
        self.min_stack = deque()

    def push(self, val: int) -> None:
        self.stack.append(val)
        if len(self.min_stack) == 0:
            self.min_stack.append(val)
        else:
            self.min_stack.append(min(val, self.min_stack[-1]))

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

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

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

Test

>>> from min_stack__two_stacks import MinStack
>>> s = MinStack()
>>> s.push(-2); s.push(0); s.push(-3)
>>> s.getMin()
-3
>>> s.pop()
>>> s.top()
0
>>> s.getMin()
-2

class min_stack__two_stacks.MinStack

Bases: object

getMin() → int

pop() → None

push(val: int) → None

top() → int