Leetcode 0232. Implement Queue using Stacks

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:

Easy

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]

• 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.

Solution

class MyQueue {
  Stack<Integer> stack;

  public MyQueue() {
    stack = new Stack<>();
  }

  public void push(int x) {
    Object[] temp = stack.toArray();
    while (!stack.empty()) {
      stack.pop();
    }
    stack.push(x);
    for (int i = 0; i < temp.length; i++) {
      stack.push((int)temp[i]);
    }
  }

  /** Time Complexity: BigO(1) */
  public int pop() {
    return stack.pop();
  }

  /** Time Complexity: BigO(1) */
  public int peek() {
    return stack.peek();
  }

  /** Time Complexity: BigO(1) */
  public boolean empty() {
    return stack.empty();
  }
}