Chris medisas interview

gistfile1.txt

     5
   /  \
  3     7
 / \   / \
1   4 6   9
 \
  2

//find the next highest node

//ex. given 5

1. check the right side child 
    exisits
        rightside child..traverse left most ancestor child..to give us the next highest
        
// ex. given 4

4. right side child doesnt exist so go to parent 
    is the starter node a right side child? yes
        go up again. 



capacity: 5
get (key) O(1)
set (key, value) O(1)

// If space set it
// If no space, kick out least recently used

LinkedList list = new LinkList();


Node front = null;
Node back = null;


public void set(String input) {
  
    if (list.size() < 5){
        if (front == null){
            Node newNode = new Node(input);
            front = newNode;
            back = newNode;
        }else{
            Node newNode = new Node(input);
            front.next = newNode;
            back = newNode;      
        }
    }else{
        //find the least used object aka last 
        list.delete_tail
        list.insert_head
    }


}

//it is in the list

//it is not in the list