CS307 Fall 2009 Final Exam
Suggested Solution and Criteria:

1 1.5 Each, differences in capitalization and spacing okay. On Big O no O() okay.

A:      37
       /  \
      13   42
      /      \
    -12        101
     
B: 17, 19, 25, 10, 20, 30, 7, 15

C: 19, 10, 25, 17, 20, 7, 30, 15

D: 10, 25, 19, 7, 15, 30, 20, 17

E: No.

F: valid binary search tree using values between 1 and 10. No repeats.

G: 130

H: ANI

I: O(N^2)

J: O(N)

K: O(N)

L: O(N^2)

M: O(NlogN)

N: O(N^2)

O: O(N^2)

P: Answers in range 30 - 50 inclusive are correct.

Q: 9 11

R: 6

S: HashSet Add, access, and remove are average case O(1). With TreeSet they are average case O(logN). (-1 if HashSet is answer, but bad explanation.)

T: Variables must have a box and arrow.


2. Suggested Solution:

Using look ahead

public boolean removeFirst(E obj) {
	int oldSize = size;
	if(first != null) {
		// special case, first contains obj
		if(obj.equals(first.getValue()){
			first = first.getNext();
			size--;
		}
		else { 
			// general case
			Node<E> temp = first;
			while(temp.getNext() != null 
				&& obj.equals(temp.getNext().getValue()))
				
				temp = temp.getNext();
			if(temp.getNext() != null){
				// found it!
				temp.setNext(temp.getNext().getNext())
				size--;
			}
		}
	}
	return oldSize != size;	
}

Using trailer

public boolean removeFirst(E obj) {
	int oldSize = size;
	if(first != null) {
		// special case, first contains obj
		if(obj.equals(first.getValue()){
			first = first.getNext();
			size--;
		}
		else { 
			// general case
			Node<E> leader = first.getNext();
			Node<E> trailer = first;
			while( leader != null ){
				if( obj.equals(leader.getValue()) ){
					// found it!
					trailer.setNext(leader.getNext());
					size--;
				}
				else {
					trailer = leader;
					leader = leader.getNext();
				}
			}
		}
	}
	return oldSize != size;	
}

Two pass algorithm

public boolean removeFirst(E obj) {
	int oldSize = size;
	Node<E> temp = first;
	int position = 0;
	while(temp != null && !obj.equals(temp.getValue()){
		position++;
		temp = temp.getNext();
	}
	if( temp != null ){
		// found it so remove it
		if(position == 0)
			first = first.getNext();
		else{
			temp = first;
			for(int i = 1; i < position; i++)
				temp = temp.getNext();
			// now temp on node before first occurrence of obj
			temp.setNext(temp.getNext().getNext());
		}
		size--;
	}
	return oldSize != size;	
}


Grading:

Update first correctly if first node contains obj: 2 points

search list for obj:
	attempt: 2 points
	correct: 5 points (-2 if OBOE or NPE or doesn't stop at correct location, -5 if doesn't actually move)

check elements equal to obj: 2 points (-1 if ==)

remove first node that contains obj from list if present: 3 

return correct answer: 1 point


3.
public MorseCodeTree(Map<Character, String> codes){
	root = new BNode<Character>();
	Iterator<Character> it = codes.keySet().iterator();
	while(it.hasNext()){
		char letter = it.next();
		addHelper(letter, codes.get(letter));
	}
}

private void addHelper(char letter, String code) {
	BNode<Character> temp = root;
	for(int i = 0; i < code.length(); i++){
		// go correct direction. build child if necessary
		if( code.charAt(i) == '.'){
			if(temp.getLeft() == null)
				temp.setLeft(new BNode<Character>());
			temp = temp.getLeft();
		}
		else{
			if(temp.getRight() == null)
				temp.setRight(new BNode<Character>());
			temp = temp.getRight();
		}
	}    
	// temp pointing at node for letter
	temp.setValue(letter);
}


Create root correctly: 2 points

get iterator from keyset: 1 point

use iterator to access each value in map: 
	attempt: 2 points
	correct: 2 points

For each key, value pair:
	traverse to correct location, going right or left
		attempt: 2 points
		correct: 4 points
	build new nodes as necessary:
		attempt: 1 point
		correct: 4 points
	add letter to leaf node: 2 points

If destroy tree via last path: -7


4. 
public String decode(String encodedMessage){
	String result = "";
	boolean good = encodedMessage.length() > 0;
	BNode<Character> temp = root;
	for(int i = 0; i < encodedMessage.length() && good; i++){
		char ch = encodedMessage.charAt(i);
		if(ch == '.')
			// go left, check to ensure exists
			if(temp.getLeft() == null)
				good = false;
			else
				temp = temp.getLeft();
		else if(ch == '-')
			// go right, check to ensure exists
			if(temp.getRight() == null)
				good = false;
			else
				temp = temp.getRight();
		else
			// ch is an asterisk, check on letter
			if(temp.getValue() == null)
				good = false;
			else{
				result += temp.getValue();
				// move back to start
				temp = root;
			}
	}
	return good ? result : null;
}

start at root: 1 point

loop through string: 2 points

go left or right as appropriate:
	attempt: 2 points
	correct: 3 points. (must ensure node exists)
	
at asterisk check data null or not: 2 points

if !null concat letter to message: 2 points

reset to root: 2 points

return correct value: 1 point


5
private class HashIterator implements Iterator<E>{

	private int conPos;
	private int arrayListPos;
	private int numToGet;
	private boolean removeOK;

	private HashIterator(){
		numToGet = size;
		// for other vars default values correct
	}

	public boolean hasNext(){
		return numToGet > 0;
	}

	public E next(){
		assert hasNext();
		removeOK = true;
		// may have nulled out due to remove
		if(container[conPos] == null 
				|| container[conPos].size() == arrayListPos){
			// move to the next valid array list
			conPos++;
			while(container[conPos] == null)
				conPos++;
			arrayListPos = 0;
		}
		numToGet--;
		return container[conPos].get(arrayListPos++);    
	}

	public void remove(){
		assert removeOK;
		container[conPos].remove(--arrayListPos);
		if(container[conPos].size() == 0)
			container[conPos] = null;
		size--;
	}
}

Instance vars: 
	track current position in container: 2 points
	track current position in current arraylist: 2 points
	way to track hasNext: 1 point
	
constructor: 1 point

hasNext: 4 points

next: 6 points

remove: 4 points