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# 232. Implement Queue using Stacks

## LeetCode [232. Implement Queue using Stacks](https://github.com/txfs19260817/LeetCode-Go/blob/master/solutions/title/README.md)

### 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.

**Follow-up:** Can you implement the queue such that each operation is [**amortized**](https://en.wikipedia.org/wiki/Amortized_analysis) `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.

**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.

### Tags

Stack, Design

### Solution

We use 2 stacks `a` and `b`, for receive and emit elements, respectively.

* Push: push `x` into `a`;
* Peek: When `b` is empty, pop all elements from `a` and push them into `b`, then return the top of `b`;
* Push: do Peek first, then pop the top of `b`;
* Empty: return true if no elements exist in both `a` and `b`.

### Complexity

* Time complexity: $$O(n)$$ for Pop and Peek; $$O(1)$$ for others;
* Space complexity: $$O(n)$$

### Code

```go
type MyQueue struct {
	a, b []int
}

// Constructor initialize your data structure here.
func Constructor() MyQueue {
	return MyQueue{make([]int, 0, 100), make([]int, 0, 100)}
}

// Push element x to the back of queue.
func (this *MyQueue) Push(x int) {
	this.a = append(this.a, x)
}

// Pop removes the element from in front of queue and returns that element.
func (this *MyQueue) Pop() int {
	val := this.Peek()
	this.b = this.b[:len(this.b)-1]
	return val
}

// Peek gets the front element.
func (this *MyQueue) Peek() int {
	if len(this.b) == 0 {
		for len(this.a) > 0 {
			this.b = append(this.b, this.a[len(this.a)-1])
			this.a = this.a[:len(this.a)-1]
		}
	}
	return this.b[len(this.b)-1]
}

// Empty returns whether the queue is empty.
func (this *MyQueue) Empty() bool {
	return len(this.a) == 0 && len(this.b) == 0
}
```