# Question

Implement the following operations of a stack using queues.

• push(x) – Push element x onto stack.
• pop() – Removes the element on top of the stack.
• top() – Get the top element.
• empty() – Return whether the stack is empty.

Example:

MyStack stack = new MyStack();

stack.push(1);
stack.push(2);
stack.top();   // returns 2
stack.pop();   // returns 2
stack.empty(); // returns false


Notes:

• You must use only standard operations of a queue – which means only push to back, peek/pop from front, size, and is empty operations are valid.
• Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
• You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

# Solution

Use the two queues to find the last item.

from collections import deque

class MyStack(object):

def __init__(self):
"""
"""
self.inque = deque()
self.outque = deque()

def push(self, x):
"""
Push element x onto stack.
:type x: int
:rtype: void
"""
self.inque.append(x)

def pop(self):
"""
Removes the element on top of the stack and returns that element.
:rtype: int
"""
while len(self.inque) > 1:
n = self.inque.popleft()
self.outque.append(n)

out = self.inque.popleft()
self.inque, self.outque = self.outque, self.inque
return out

def top(self):
"""
Get the top element.
:rtype: int
"""
if len(self.inque) == 0:
raise IndexError
else:
return self.inque[-1]

def empty(self):
"""
Returns whether the stack is empty.
:rtype: bool
"""
return len(self.inque) == 0

# Your MyStack object will be instantiated and called as such:
# obj = MyStack()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.top()
# param_4 = obj.empty()