Sample.java

// Time Complexity : Amortized time complexity for operations - push, pop, peek , Empty is O(1).
// Time Complexity : Worst case time complexity for operations - pop and peek is O(n) when the outStack is Empty.
// Space Complexity : 2 * O(n) = O(n)
// Did this code successfully run on Leetcode : Yes
// Any problem you faced while coding this : No


// Your code here along with comments explaining your approach
class MyQueue {
    private Stack<Integer> inStack;
    private Stack<Integer> outStack;

    public MyQueue() {
        this.inStack = new Stack<>();
        this.outStack = new Stack<>();
    }

    //Time Complexity - O(1)
    public void push(int x) {
        inStack.push(x);
    }

    //Time Complexity - O(1)
    public int pop() {
        if(outStack.isEmpty()){
            while(!inStack.isEmpty()){
                outStack.push(inStack.pop());
            }
        }
        return outStack.pop();
    }

    //Time Complexity - O(1)
    public int peek() {
        if(outStack.isEmpty()){
            while(!inStack.isEmpty()){
                outStack.push(inStack.pop());
            }
        }
        return outStack.peek();
    }

    //Time Complexity - O(1)
    public boolean empty() {
        return inStack.isEmpty() && outStack.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();
 */