Leetcode - Implement Stack using Queues Solution

Implement a last in first out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal queue (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 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 such that each operation is amortized O(1) time complexity? In other words, performing n operations will take overall O(n) time even if one of those operations may take longer. You can use more than two queues.

Solution in python

class MyStack:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.stack = deque()

    def push(self, x: int) -> None:
        """
        Push element x onto stack.
        """
        self.stack.append(x)
        

    def pop(self) -> int:
        """
        Removes the element on top of the stack and returns that element.
        """
        return self.stack.pop()
        

    def top(self) -> int:
        """
        Get the top element.
        """
        return self.stack[-1]

    def empty(self) -> bool:
        """
        Returns whether the stack is empty.
        """
        return False if self.stack else True
        


# 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()