225. Implement Stack using Queues

Easy


Implement a last-in-first-out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal stack (push, top, pop, and empty).

Implement the MyStack class:

  • void push(int x) Pushes element x to the top of the stack.
  • int pop() Removes the element on the top of the stack and returns it.
  • int top() Returns the element on the top of the stack.
  • boolean empty() Returns true if the stack is empty, false otherwise.

Notes:

  • You must use only standard operations of a queue, which means that only push to back, peek/pop from front, size and is empty operations are valid.
  • Depending on your language, the queue may not be supported natively. You may simulate a queue using a list or deque (double-ended queue) as long as you use only a queue's standard operations.

 

Example 1:

Input
["MyStack", "push", "push", "top", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
[null, null, null, 2, 2, false]

Explanation
MyStack myStack = new MyStack();
myStack.push(1);
myStack.push(2);
myStack.top(); // return 2
myStack.pop(); // return 2
myStack.empty(); // return False

 

Constraints:

  • 1 <= x <= 9
  • At most 100 calls will be made to push, pop, top, and empty.
  • All the calls to pop and top are valid.

 

Follow-up: Can you implement the stack using only one queue?





 class MyStack:
    """
    Using One Queue
    """

    def __init__(self):
        self.q1 = []
        self.top_val = None

    def push(self, x: int) -> None:
        self.q1.append(x)
        self.top_val = x


    def pop(self) -> int:
        size  = len(self.q1)
        while size > 1:
            item = self.q1.pop(0)
            self.q1.append(item)
            self.top_val = item
            size = size - 1

        return self.q1.pop(0)

    def top(self) -> int:
        return self.top_val

    def empty(self) -> bool:
        return len(self.q1)==0


#     """
#     Using Two Queue
#     """

#     def __init__(self):
#         self.q1 = []
#         self.q2 = []
#         self.top_val = None

#     def push(self, x: int) -> None:
#         self.q1.append(x)
#         self.top_val = x


#     def pop(self) -> int:

#         while len(self.q1) > 1:
#             # pop element and add to q2
#             item = self.q1.pop(0)
#             self.top_val = item
#             self.q2.append(item)
#         # performing stack pop
#         val = self.q1.pop()
#         # copying q2 to q1
#         self.q1 = self.q2.copy()
#         self.q2 = []
#         return val

#     def top(self) -> int:
#         return self.top_val

#     def empty(self) -> bool:
#         return len(self.q1) == 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()

Random Note


From python 3.7 dict guarantees that order will be kept as they inserted, and popitem will use LIFO order but we need FIFO type system. so we need OrderedDict which have popIten(last = T/F) for this req. One thing, next(iter(dict)) will return the first key of the dict