Make a queue from 2 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:

• 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 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.

Solution:

My solution is fairly simple. Whenever we want to add an item into the queue, add it to the stack that has values ordered LIFO. The order of that operation is constant. When we want to peak or pop an item from this queue implementation, we need to check if there are items in the second stack, if there are not, we move the items from stack1 which were ordered in LIFO, and add them into stack2, where they will be order by last element. Finally take your value from the second stack .

class MyQueue {
    private Stack<Integer> st1;
    private Stack<Integer> st2;
    /** Initialize your data structure here. */
    public MyQueue() {
        st1=new Stack<Integer>();
        st2=new Stack<Integer>();
    }
    
    /** Push element x to the back of queue. */
    public void push(int x) {
        st1.push(x);
    }
    
    /** Removes the element from in front of queue and returns that element. */
    public int pop() {
        if(st2.isEmpty() ){
            migrateStack();
        }return st2.pop();
    }
    public void migrateStack(){
        while(!st1.isEmpty() ) st2.push( st1.pop() );
    }
    
    /** Get the front element. */
    public int peek() {
        if(st2.isEmpty() ){
            migrateStack();
        }return st2.peek();
    }
    
    /** Returns whether the queue is empty. */
    public boolean empty() {
        return st1.isEmpty() && st2.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();
 */