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Given the following text description, write Python code to implement the functionality described below step by step
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Description:
 Implement One by Other(s)
Implement Queue by Two Stacks
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Step1:  Implement Stack by Two Queues
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Step2:  Min Stack
Implement a stack with min() function, which will return the smallest number in the stack.
It should support push, pop and min operation all in O(1) cost.
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Python Code:
class MyQueue:
    def __init__(self):
        self.stack1 = []
        self.stack2 = []
        
    def push(self, element):
        # write your code here
        self.stack1.append(element)
    def top(self):
        # write your code here
        # return the top element
        if len(self.stack2) == 0:
            while self.stack1:
                self.stack2.append(self.stack1.pop())
        return self.stack2[-1]
        
    def pop(self):
        # write your code here
        # pop and return the top element
        if len(self.stack2) == 0:
            while self.stack1:
                self.stack2.append(self.stack1.pop())
        return self.stack2.pop()
        
Explanation: Implement One by Other(s)
Implement Queue by Two Stacks
http://www.lintcode.com/en/problem/implement-queue-by-two-stacks/
End of explanation
class Stack:
    # initialize your data structure here.
    def __init__(self):
        self.queue = []
    # @param x, an integer, push a new item into the stack
    # @return nothing
    def push(self, x):
        # Write your code here
        self.queue.append(x)
    # @return nothing, pop the top of the stack
    def pop(self):
        # Write your code here
        n = len(self.queue)
        for i in range(n-1):
            self.queue.append(self.queue.pop(0))
        return self.queue.pop(0)
        
    # @return an integer, return the top of the stack
    def top(self):
        # Write your code here
        top = None
        n = len(self.queue)
        for i in range(n):
            top = self.queue.pop(0)
            self.queue.append(top)
        return top
    # @return an boolean, check the stack is empty or not.
    def isEmpty(self):
        # Write your code here
        return self.queue == []
Explanation: Implement Stack by Two Queues
http://www.lintcode.com/en/problem/implement-stack-by-two-queues/
End of explanation
class MinStack(object):
    def __init__(self):
        # do some intialize if necessary
        self.stack = []
    def push(self, number):
        # write yout code here
        curMin = self.min()
        if curMin is None or number < curMin:
            curMin = number
        self.stack.append([number, curMin])
    def pop(self):
        # pop and return the top item in stack
        pop = self.stack.pop()
        return pop[0]
    def min(self):
        # return the minimum number in stack
        if len(self.stack) == 0:
            return None
        else:
            return self.stack[-1][1]
Explanation: Min Stack
Implement a stack with min() function, which will return the smallest number in the stack.
It should support push, pop and min operation all in O(1) cost.
http://www.lintcode.com/en/problem/min-stack/
https://leetcode.com/problems/min-stack/#/description
End of explanation
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