QueueUsingStack.java

// Time Complexity : For push,empty it is O(1) and for pop, peek it is amortized O(1).
// Space Complexity : O(n)
// Did this code successfully run on Leetcode : Yes
// Any problem you faced while coding this : The idea of maintaining two stacks and only using the other stack when pop or peek operation
// comes is the core idea for this problem. It helped in solving the problem at one go.

// Your code here along with comments explaining your approach

class MyQueue {
    Stack<Integer> inStack;
    Stack<Integer> outStack;

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

    public void push(int x) {
        inStack.push(x);
    }

    public int pop() {
        if(empty()){
            return -1;
        }
        peek();
        return outStack.pop();
    }

    public int peek() {
        if(outStack.isEmpty()){
            while(!inStack.isEmpty()){
                outStack.push(inStack.pop());
            }
        }
        return outStack.peek();
    }

    public boolean empty() {
        if(inStack.isEmpty() && outStack.isEmpty()){
            return true;
        }
        return false;
    }
}

/**
 * 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();
 */