232. Implement Queue using Stacks

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.

  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.

Solution

class MyQueue {
    private Deque<Integer> left;
    private Deque<Integer> right;
    // Initialize your data structure here.
    public MyQueue() {
        left = new ArrayDeque<>();
        right = new ArrayDeque<>();
    }

    // Push element x to the back of queue.
    public void push(int x) {
        while (!right.isEmpty()) {
            left.add(right.pop());
        }
        left.add(x);
    }

    // Removes the element from in front of queue and returns that element.
    public int pop() {
        while (!left.isEmpty()) {
            right.add(left.pop());
        }
        return right.pop();
    }

    // Get the front element.
    public int peek() {
        while (!left.isEmpty()) {
            right.add(left.pop());
        }
        return right.peek();
    }

    // Returns whether the queue is empty.
    public boolean empty() {
        return right.isEmpty() && left.isEmpty();
    }
}

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

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).