📌 Problem

Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push, peek, pop, and empty).

Implement the MyQueue class:

• void push(int x) Pushes element x to the back of the queue.
• int pop() Removes the element from the front of the queue and returns it.
• int peek() Returns the element at the front of the queue.
• boolean empty() Returns true if the queue is empty, false otherwise.

Notes:

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

Follow-up : Can you implement the queue 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.

Example 1:

Input
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
[null, null, null, 1, 1, false]

Explanation
MyQueue myQueue = new MyQueue();
myQueue.push(1); // queue is: [1]
myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
myQueue.peek(); // return 1
myQueue.pop(); // return 1, queue is [2]
myQueue.empty(); // return false

Constraints:

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

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📝 Solution

class MyQueue:
    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.__push_stack=[]
        self.__pop_stack=[]
        

    def push(self, x: int) -> None:
        """
        Push element x to the back of queue.
        """
        self.__push_stack.append(x)
        

    def pop(self) -> int:
        """
        Removes the element from in front of queue and returns that element.
        """
        if self.__pop_stack:
            return self.__pop_stack.pop()
        else:
            while self.__push_stack: # Move push_stack to pop_stack
                temp=self.__push_stack.pop() 
                self.__pop_stack.append(temp) 
            return self.__pop_stack.pop()
            
            
        
        

    def peek(self) -> int:
        
        """
        Get the front element.
        """
        if self.__pop_stack:
            return self.__pop_stack[-1]
        else:
            
            while self.__push_stack: # Move push_stack to pop_stack
                temp=self.__push_stack.pop()
                self.__pop_stack.append(temp)
            return self.__pop_stack[-1]
            
        

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


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

• push 의 경우 push_stack에 담는다.
• pop & peek 의 경우 push_stack 에 있는 데이터를 pop_stack 에 옮겨서 순서를 바꿔준 후 pop 한다.

  

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📚 Reference

The figure for header # Solution