## QUESTIONS

1. Suppose the DblListnode<E> class is implemented as in the readings.
The following class implements the ListADT interface using a circular doubly-linked chain of nodes:

``````

// Reference to the first node (position 0) of a circular doubly-linked
// chain of nodes. Recall that the next reference of the last node
// references the first node and the prev reference of the first node
// references the last node making the chain of nodes circular.
private DblListnode<E> items;
// The number of items in the list
private int numItems;

// Initially the list is empty
items = null;
numItems = 0;
}

// Return the item at position pos if pos is valid, otherwise throw
// IndexOutOfBoundsException. This method traverses the fewest nodes
// possible while locating the node at pos.
public E get(int pos) {
// IMPLEMENT THIS
}
}
``````

Complete the get() method according to its specification. For full credit, your solution:

• must locate the node at pos in the most efficient manner possible (i.e., call getNext() and getPrev() as few times as possible).
• must not modify the DblListnode class, the CircularDblLinkedList class's methods or fields declarations.
• must not change the contents or ordering of the nodes in the list during execution.

2. Assume you are comparing three different algorithms to solve some problem and have determined the worst-case time equations for each:

• Algorithm 1: T(N) = 1500N + 500
• Algorithm 2: T(N) = 2N2 + N + 20
• Algorithm 3: T(N) = 30Nlog2N + 3N

1. What is the worst-case time complexity for each algorithm?
2. Which is the algorithm that has the lowest worst-case time complexity?
3. Should the algorithm with the lowest complexity always be used? Briefly explain why/why not?

3. Assume the following:
• The StackADT<E> interface and the ArrayStack<E> class implementation as in readings.

Consider the following secret method:
```
public static void secret(String secretString) {

int n = secretString.length();

for (int i = 0; i < n; i++) {
stack.push(secretString.charAt(i));
}

while(!stack.isEmpty()) {
Character ch = stack.pop();
System.out.print(ch);
}
}
```
Briefly describe what secret(...) method does.