232. Implement Queue using Stacks

Description

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.

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.

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.

Submitted Code

class MyQueue(object):

    def __init__(self):
        self.stack_main = []
        self.stack_temp = []

    def push(self, x):
        """
        :type x: int
        :rtype: None
        """
        while len(self.stack_main) > 0:
            self.stack_temp.append(self.stack_main.pop())
        
        self.stack_temp.append(x)

        while len(self.stack_temp) > 0:
            self.stack_main.append(self.stack_temp.pop())

    def pop(self):
        """
        :rtype: int
        """
        return self.stack_main.pop()

    def peek(self):
        """
        :rtype: int
        """
        return self.stack_main[-1]

    def empty(self):
        """
        :rtype: bool
        """
        return len(self.stack_main) == 0

# 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()

Runtime: 0 ms | Beats 100.00%
Memory: 12.49 MB | Beats 56.06%

push를 할 때 순서를 정렬하는 방법으로 구현해봤다.

Other Solutions

1st

class MyQueue:

    def __init__(self):
        self.input = []
        self.output = []

    def push(self, x: int) -> None:
        self.input.append(x)

    def pop(self) -> int:
        self.peek()  # Ensure output stack has the current front
        return self.output.pop()

    def peek(self) -> int:
        if not self.output:  # Transfer elements if output stack is empty
            while self.input:
                self.output.append(self.input.pop())
        return self.output[-1]

    def empty(self) -> bool:
        return not self.input and not self.output

time complexity: 𝑂(1): push, empty / amortized 𝑂(1): pop, peek
space complexity: 𝑂(𝑛)

제출했던 답변은 push가 자주 일어날 경우 효율이 좋지 않기 때문에, 이것처럼 각 원소가 input에서 output으로 단 한번만 이동하는 방식이 더 좋은 성능을 낼 수 있다.

push(1), push(2), push(3), pop(), push(4), pop()

            input         output
push(1)     [1]           []
push(2)     [1, 2]        []
push(3)     [1, 2, 3]     []
pop()       []            [3, 2, 1] → [3, 2] (remove 1)
push(4)     [4]           [3, 2]
pop()       [4]           [3] (remove 2)