minimum_stack

Solution

	
	"""
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.

"""

class MinStack:
    def __init__(self):
        self.stack = []
        self.min = []

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

    def pop(self) -> None:
        if not self.stack:
            raise IndexError("stack is empty")
        self.stack.pop()
        self.min.pop()

    def top(self) -> int:
        if not self.stack:
            raise IndexError("stack is empty")
        return self.stack[-1]

    def getMin(self) -> int:
        if not self.min:
            raise IndexError("stack is empty")
        return self.min[-1]

	

Tests

	
	import pytest

from publicmatt.leet.problems.minimum_stack import MinStack


@pytest.fixture()
def stack():
    s = MinStack()
    s.push(5)
    s.push(4)
    s.push(3)
    s.push(2)
    s.push(1)
    return s


def test_min(stack):
    assert stack.getMin() == 1
    stack.pop()
    assert stack.getMin() == 2
    stack.pop()
    assert stack.getMin() == 3
    stack.pop()
    assert stack.getMin() == 4
    stack.pop()
    assert stack.getMin() == 5


def test_stack_elements(stack):
    assert stack.stack == [5, 4, 3, 2, 1]
    assert stack.min == [5, 4, 3, 2, 1]
    stack.push(5)
    assert stack.stack == [5, 4, 3, 2, 1, 5]
    assert stack.min == [5, 4, 3, 2, 1, 1]


def test_empty():
    s = MinStack()
    with pytest.raises(IndexError):
        min = s.getMin()