LeetCode - Implement Queue using Stacks

Problem description

description

Implement the following operations of a queue using stacks.

• push(x) – Push element x to the back of queue.
• pop() – Removes the element from in front of queue.
• peek() – Get the front element.
• empty() – Return whether the queue is empty. Example: ```text MyQueue queue = new MyQueue();

queue.push(1); queue.push(2);
queue.peek(); // returns 1 queue.pop(); // returns 1 queue.empty(); // returns false

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, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
- You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).

## Analysis

The first thought is to use two stacks, one save the forward direction, and one save the backward direction.

```java
class MyQueue {
    
    Stack<Integer> forward;
    Stack<Integer> backward;

    /** Initialize your data structure here. */
    public MyQueue() {
        forward = new Stack<>();
        backward = new Stack<>();
    }
    
    void ftob(){
        while (!forward.isEmpty()){
            backward.push(forward.pop());
        }
    }
    
    void btof(){
        while (!backward.isEmpty()){
            forward.push(backward.pop());
        }
    }
    
    /** Push element x to the back of queue. */
    public void push(int x) {
        if (forward.isEmpty()){
            btof();
        }
        forward.push(x);
    }
    
    /** Removes the element from in front of queue and returns that element. */
    public int pop() {
        if (backward.isEmpty()){
            ftob();
        }        
        return backward.pop();
    }
    
    /** Get the front element. */
    public int peek() {
        if (backward.isEmpty()){
            ftob();
        }        
        return backward.peek();
    }
    
    /** Returns whether the queue is empty. */
    public boolean empty() {
        return forward.isEmpty() && backward.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();
 */

However, this solution is not that efficient. We don’t need to move the items in backward back to forward since what we need is pop item, and we can pop from the backward. If it’s empty, we can move items in forward into backward stack.

import java.util.Stack;
import java.util.EmptyStackException;

class MyQueue {

    Stack<Integer> pushStack;
    Stack<Integer> popStack;
    
    /** Initialize your data structure here. */
    public MyQueue() {
        pushStack = new Stack<Integer>();
        popStack = new Stack<Integer>();
    }
    
    /** Push element x to the back of queue. */
    public void push(int x) {
        pushStack.add(x);
    }
    
    /** Removes the element from in front of queue and returns that element. */
    public int pop() {
        this.peek();
        return popStack.pop();        
    }
    
    /** Get the front element. */
    public int peek() {
        if (this.empty())
            throw new EmptyStackException();
        
        if (popStack.isEmpty()){
            while (!pushStack.isEmpty()){
                popStack.add(pushStack.pop());
            }
        }
        return popStack.peek();
    }
    
    /** Returns whether the queue is empty. */
    public boolean empty() {
        return pushStack.isEmpty() && popStack.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();
 */

What to improve