CS307 Fall 2005 final exam suggested solutions and grading key

1. answer as written or -2. differences in capitalization and spacing okay. on big o missing O( ) okay, but no extra terms

A.			32
		   /  \
		  5    74
		 / \
		4  31
		
B.	52 25 9 5 30 72 70 75

C.	5 9 30 25 52 70 72 75

D.	5 30 9 25 70 75 72 52

E.	52 25 72 9 70 75 5 30

F.	No

G.	front		back
	3	4	5	3

H.	front	back
	2	1	5
	
I.	999 (1000 okay)

J.	any answers 15 - 35

K.	O(N)

L.	interfaces allow for a limited form of multiple inheritance. They also allow design without any implementation details. This allows many implementations. (or words to that effect)
-1 at discretion of grader

M.	4N + 7 (+/- 2)

N.	O(N^2 log N)

O.	O(N^3)


2.
public boolean isPresent(Comparable data)
{	return helper( data, myRoot );	}

private boolean helper( Comparable data, RBTreeNode n)
{	if( n == null )
		return false;
	else
	{	int dir = data.compareTo( n.getData() )
		{	if( dir == 0 )
				return n.isPresent();
			else if( dir < 0 )
				return helper( data, n.getLeft() );
			else
				return helper( data, n.getRight() );
		}
	}
}

A. 	fall off tree case 3 pts
B. 	found and is present case 3 pts
C.	found and is not present case 3 pts
D.  go left case 3 pts
E.  go right case 3 pts

3.
public Comparable getMin()
{	return minHelper(myRoot, null);	}

private Comparable minHelper( RBTreeNode n, Comparable minSoFar )
{	if( n != null )
	{	minSoFar = minHelper( n.getLeft(), minSoFar )
		//if minSoFar == null I fell off tree or all nodes so far are not present
		if( minSoFar == null && n.isPresent() )
			minSoFar = n.getData();
		// if minSoFar == null, still have not found a value that is present
		if( minSoFar == null )
			minSoFar = minHelper( n.getRight(), minSoFar );
	}
	return minSoFar;
}

okay if do complete traversal

A.	base case, off tree, 4 pts (look ahead okay)
B.	go left, 4 pts
C.	go right, 4 pts
D.	find correct min based on node is present and continue until found, attempt 4 pts
E.	find correct min based on node is present and continue until found, correct 4 pts

4.

public void add(Object item)
{	int index = item.hashCode() % myCon.length;
	while( myCon[index] == null )
	{	index = ( index + 1 ) % myCon.length;
	}
	myCon[index] = item;
	iMySize++;
	
	//rehash?
	if( 1.0 * iMySize / myCon.length > dMyLoadFactorLimit )
		rehash();
}

A.	calculate initial index, attempt 1 pts
B. 	calculate initial index, correct 1 pts
C.	linear probe for first open spot, attempt 3 pts
D.	linear probe for first open spot, correct 4 pts 
E.	increment iMySize, 1 pt
F.	check if above load limit and call rehash, attempt 3 pts
G.	check if above load limit and call rehash, attempt 2 pts (double)

5.
public String encrypt(String[] clearMessage, Map encryptKey)
{	String result = "";
	for(int i = 0; i < clearMessage.length; i++)
	{	if( encryptKey.containsKey( clearMessage[i] )
			result += encryptKey.get( clearMessage[i] )
		else
		{	for(int c = 0; c < clearMessage[i].length(); c++)
			{	if( encryptKey.containsKey( clearMessage[i].charAt{c) + "" )
					result += encryptKey.get( clearMessage[i].charAt(c) + "" );
			}
		}
	}
	return result;
}

A.	iterate through clearMessage, attempt 2 pts
B.	iterate through clearMessage, correct 2 pts
C.	handle case when whole element is key, attempt 4 pts
D.	handle case when whole element is key, correct 4 pts
E.	handle case when must iterate through each char in string, attempt 4 pts
E.	handle case when must iterate through each char in string, correct 4 pts