Good evening! Here's our prompt for today.

Recall that a stack is an abstract data type modeling a collection of elements. Its primary operations are push (which adds an element to the top of the stack) and pop (which removes the most newest element).

Traditionally, a stack can easily be implemented in many dynamic languages using an array (and its built-in methods).

1stack = []
2
3stack.append(5)
4stack.append(6)
5stack.append(7)
6print(stack.pop())
7# 7
8print(stack.pop())
9# 6

However, let's say we wanted to implement a stack with the following interface, requiring the following methods to be defined. The most important being the last one, min() - a method that lets us obtain the minimum element at any given time.

push(val) - add an element on to the top of the stack.

pop(val) - remove the element at the top of the stack and return it.

peek(val) - see the element at the top of the stack without removing it.

min() - get minimum element in stack.

How would you do it, and can you implement it via a MinStack class? The class should have the following methods. Work off this skeleton:

1class MinStack {
2  constructor() {
3  }
4
5  push(val) {
6  }
7  
8  pop() {
9  }
10
11  peek() {
12  }
13
14  min() {
15  }
16}

Can you do call min() and retrieve it in O(1) time?

Constraints

• Total number of operations in the stack <= 100000
• The values will be in the range -1000000000 and 1000000000
• Expected time complexity : O(1)
• Expected space complexity : O(n)

Try to solve this here or in Interactive Mode.

How do I practice this challenge?

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​

class MinStack:

    def __init__(self):

        self._stack = []

​

    def push(self, val):

        return

​

    def pop(self):

        return

​

    def peek(self):

        return

​

    def min(self):

        return

​

​

import unittest

​

​

class Test(unittest.TestCase):

    def test_1(self):

        minstack = MinStack()

        minstack.push(4)

        minstack.push(7)

        minstack.push(3)

        minstack.push(2)

        minstack.push(6)

        min1 = minstack.min()

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