/**
 * 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.
 * @see <a href="https://leetcode.com/problems/implement-queue-using-stacks/">LeetCode</a>
 */
class MyQueue() {
    private val inputStack = ArrayDeque<Int>()
    private val outputStack = ArrayDeque<Int>()

    fun push(x: Int) {
        inputStack.add(x)
    }

    fun pop(): Int {
        prepareOutputStack()
        return outputStack.removeLast()
    }

    fun peek(): Int {
        prepareOutputStack()
        return outputStack.last()
    }

    fun empty(): Boolean {
        return inputStack.isEmpty() && outputStack.isEmpty()
    }

    private fun prepareOutputStack() {
        if (outputStack.isEmpty()) {
            while (inputStack.isNotEmpty()) {
                outputStack.addLast(inputStack.removeLast())
            }
        }
    }
}

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

/**
 * 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.
 * @see <a href="https://leetcode.com/problems/implement-queue-using-stacks/">LeetCode</a>
 */
class MyQueue() {
    private val inputStack = ArrayDeque<Int>()
    private val outputStack = ArrayDeque<Int>()

    fun push(x: Int) {
        inputStack.add(x)
    }

    fun pop(): Int {
        prepareOutputStack()
        return outputStack.removeLast()
    }

    fun peek(): Int {
        prepareOutputStack()
        return outputStack.last()
    }

    fun empty(): Boolean {
        return inputStack.isEmpty() && outputStack.isEmpty()
    }

    private fun prepareOutputStack() {
        if (outputStack.isEmpty()) {
            while (inputStack.isNotEmpty()) {
                outputStack.addLast(inputStack.removeLast())
            }
        }
    }
}

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