AP JAVA Practice Exam

Below are 175 questions to help you prepare for taking = the AP=20 Computer Science A test. Click the "Score Test" button at the bottom of = the page=20 to see how well you did. You do not need to answer all the = questions at=20 one time=97only the questions you answer will be scored. That way, = you can=20 practice taking smaller portions of the test, remembering where you left = off=20 each time.

2005

What is the size of a double variable = in=20 Java?

2 bytes=20
4 bytes=20
8 bytes=20
It depends on the = compiler=20 setting=20
It depends on the = operating=20 system

What is displayed by

System.out.println("1" + new Integer(2) +=20 3);

The statement has a = syntax error=20 and won't compile=20
6=20
15=20
123=20
ClassCastException =

Which of the following best describes the set of all = pairs=20 of values for boolean variables a and = b, such=20 that

(!a && b) =3D=3D !(a || = b)

evaluates to true?

Empty set=20
Only one pair: = a =3D=3D true, b=20 =3D=3D false=20
Two pairs in which = a =3D=3D=20 true=20
Two pairs in which = a !=3D=20 b=20
All four possible = combinations of=20 values

4. (AB)

When you try to compile MyClass, the = java=20 compiler gives an error message

MyClass is not abstract and does not override abstract = method=20 <some method> in = java.util.Comparator

Which of the following is <some method> in = the error=20 message?

= equals(myClass)=20
= compareTo(myClass)=20
= compare(myClass,=20 myClass)=20
=20 compareTo(java.lang.Object)=20
= compare(java.lang.Object,=20 java.lang.Object)

Consider the following classes:

public class Year2005 { = public=20 String toString() { return "2005"; } }

public class Test2005 extends=20 Year2005 { public void=20 print() =20 { = <missing=20 statement> = } }

Which of the following could replace <missing=20 statement> so that Test2005 would compile = with no=20 errors and

Test2005 test =3D new=20 Test2005(); test.print();

would display 2005?

I. System.out.println(new=20 Year2005());
II. System.out.println(new=20 Test2005());
III.=20 System.out.println(this);

I only=20
II only=20
I and II=20
II and III=20
I, II, and III =

What is the value of a[1] after the = following=20 code is executed?

int[] a =3D {0, 2, 4, 1, 3}; for (int i =3D 0; i = < a.length;=20 i++) a[i] =3D a[(a[i] + 3) %=20 a.length];

0=20
1=20
2=20
3=20
4

Consider the method

public String mystery(String=20 s) { String s1 =3D=20 s.substring(0,1); String s2 =3D = s.substring(1,=20 s.length() - 1); String s3 =3D=20 s.substring(s.length() - 1); if = (s.length()=20 <=3D = 3) =20 return s3 + s2 + s1; =20 else = return s1=20 + mystery(s2) + s3; }

What is the output of

=
System.out.println(mystery("DELIVER"));

<= /TD>

DELIVER=20
DEVILER=20
REVILED=20
RELIVED=20
DLEIEVR =

8. (AB)

Consider the following code segment:

List list =3D new = LinkedList(); list.add("["); =20 list.add("A"); =20 list.add("]"); System.out.println(list);

ListIterator it =3D=20 = list.listIterator(); while(it.hasNext() { &nbs= p;=20 if ("[".equals(it.next()) ||=20 = "]".equals(it.next())) = ; =20 it.remove(); =20 else =20 = it.add("*"); } System.out.println(list);

The first output line is

[[, A, ]]

What is the second output line?

[A]=20
[A, B] =
[B, A] =
= ClassCaseException=20
=20 NoSuchElementException

Which of the following is not a method of=20 java.util.ArrayList?

add(Object = x);=20
remove(Object = x);=20
insert(int i, = Object=20 x);=20
= contains(Object x);=20
set(int i, = Object=20 x);

Questions 10-11 refer to

public interface InterfaceA { void methodA(); = }

public interface InterfaceB extends InterfaceA { void=20 methodB(); }

public class ClassA implements=20 InterfaceA { public void = methodA()=20 {} public void methodB() = {} }

public class ClassB extends ClassA implements=20 InterfaceB { public ClassB()=20 {} ... <methods not=20 shown> }

10.

What is the minimum number of methods that must be = defined=20 in classB for it to compile with no errors?

No particular = methods are=20 required=20
= methodA=20
= methodB=20
methodA =and=20 methodB=20
methodA =,=20 methodB, and toString

11.

Which of the following statements causes a syntax=20 error?

InterfaceA = obj =3D new=20 ClassA();=20
InterfaceB = obj =3D new=20 ClassA();=20
InterfaceA = obj =3D new=20 ClassB();=20
InterfaceB = obj =3D new=20 ClassB();=20
ClassA obj = =3D new=20 ClassB();

Questions 12-13 refer to the following method=20 rotate that takes a two-dimensional array=20 a and returns a two-dimensional array b = made by=20 rotating a by 90=B0 counterclockwise.

public int[][]=20 rotate(int[][]a) { int rows =3D=20 <expression 1>; int cols =3D=20 <expression 2>; int[][] b =3D=20 <expression 3>; for (int r = =3D 0; r=20 < rows;=20 r++) = for (int=20 c =3D 0; c < cols;=20 = c++) = ; =20 b[r][c] =3D <expression = 4>; return=20 b; }

For example, if a holds

1 2 3 4 5 6

b =3D rotate(a) will hold

3 6 2 5 1 4

12. (AB)

Which of the following should replace =

<expression=20
      1>

, <expression 2>, and =

<expression=20
      3>

<expression=20 1> =20 <expression=20 2> =20 <expression 3>

=20 = a.cols = a.rows new = int(rows, cols)=20
=20 = a.cols = a.rows new = int[][](rows, cols)=20
=20 a.cols() =20 a.rows() new int(rows, = cols)=20
=20 a[0].length =20 a.length new=20 int[rows][cols]=20
=20 a[0].length =20 a.length new = int[][](rows,=20 cols)

13. (AB)

Which of the following should replace =

<statement=20
      4>

= a[c][r]=20
a[c][rows - = 1 - r]=20
a[cols - 1 - = c][r]=20
a[cols - 1 - = c][rows - 1 -=20 r]=20
a[rows - 1 - = r][cols - 1 -=20 c]

14.

What is the value of n after the = following code=20 is executed?

int n =3D 2005; for (int i =3D 0; i < 50;=20 i++) n =3D (n + 3) /=20 2;

0=20
1=20
2=20
3=20
65

15.

What is the output of the following code segment?

List cities =3D new=20 = ArrayList(); cities.add("Atlanta"); cities.add("Boston");
for (int i =3D 1; i < cities.size();=20 i++) cities.add(i, "+");
=
System.out.println(cities);

[Atlanta, = Boston]=20
[Atlanta, +, = Boston]=20
[Atlanta, = Boston,=20 +]
[Atlanta, +, = Boston,=20 +]
No output because = the program=20 goes into an infinite loop

16.

Which of the following statements is true?

A static variable = cannot be=20 initialized in a constructor=20
A static variable = must be=20 declared final=20
An instance = variable can't be=20 declared final=20
A static method = can't access an=20 instance variable=20
Only a static = method can access=20 a static variable

17. (AB)

a and b are arrays of = integers and=20 each of them has n elements. a is sorted = in=20 ascending order and b is sorted in descending = order. =20 What is the big-O, in terms of n, for the number of=20 comparisons in an optimal algorithm that determines whether there=20 exists i, such that

a[i] =3D=3D =
b[i]

O(log = n)=20
O( (log=20 n)² )=20
O(n) =
O(n = log n)=20
= O(n²)=20

18. (AB)

What is the output of the following code?

String key =3D ""; Map map =3D new TreeMap(); for = (int k =3D 0;=20 k < 3; k++) { key +=3D=20 k; String value =3D=20 "A"; map.put(key,=20 value); value +=3D=20 "B"; map.put9key,=20 = value); } System.out.println(map.size());

1=20
2=20
3=20
4=20
6

19. (AB)

Consider the following method:

private TreeNode mysteryProcess(TreeNode=20 root) { if (root =3D=3D null ||=20 (root.getLeft() =3D=3D null && root.getRight() =3D=3D=20 null)) = return=20 null; = else =20 { =20 = root.setLeft(mysteryProcess(root.getLeft())); = =20 = root.setRight(mysteryProcess(root.getRight())); &nbs= p; =20 return root; = } }

If root refers to the root of the tree

&nbs= p; 1 = =20 = / \ &= nbsp; =20 2 =20 = 3 &n= bsp;=20 / /=20 = \ =20 4 5 6

which of the following trees is returned by=20 mysteryProcess(root)?

a. 1 b. 1 &nb= sp; =20 \ =20 3 c. 1 &nb= sp; =20 / \ 2 = 3 d. 1 &nb= sp; / \ =20 2 3 = / =20 4 e. 1 &nb= sp; =20 \ =20 3 /=20 \ = 5 =20 6

20. (AB)

What is the output of the following code?

Stack stk =3D new=20 = ArrayStack(); stk.push("A"); stk.push("B"); stk.push(stk); wh= ile(!stk.isEmpty()) { =20 Object obj =3D stk.pop(); if (obj = instanceOf=20 Stack) =20 { =20 = while(!((Stack)obj).isEmpty()) &nb= sp; =20 { =20 = System.out.print(((Stack)obj).pop()); =20 } =20 else =20 System.out.print(obj); }

BA=20
ABBA=20
BAAB=20
BABA=20
ClassCastException =

21. (AB)

A priority queue is represented as a minimum heap, = stored in=20 an array. When a node is added to the heap, it is first added at = the=20 leftmost vacant spot in the current level (or, if the current = level is=20 full, at the leftmost spot in the next level), and then the = reheap-up=20 procedure is applied. What is the order of the values in the array = after=20 "M", "A", "N", = "G",=20 "O", "E", "S" are added to = the=20 queue?

= AEGMNOS=20
= AGEMONS=20
= AEGMNOS=20
= AEGSOMN=20
= AGESNOM

22. (AB)

Suppose an array of n elements is stored in = ascending order.=20 Then 5 elements are picked randomly and assigned random values. = Which of=20 the following sorting algorithms guarantees to restore the = ascending order=20 in than array using no more than O(n) = comparisons?

I. Selection Sort II. Insertion=20 Sort III. = Mergesort

I only=20
II only=20
I and II=20
II and III=20
I, II, and III =

23.

For any object obj, a call=20 obj.getClass().getName() returns the name of the=20 obj's class.

Suppose

System.out.println(new A() + "+" + new=20 B());

displays

A+B

Which of the following implementations would produce that = result?

I. Class A has a method

          = public=20 String toString() { return "A"; }

   and class B has a method

          = public=20 String toString() { return "B"; }

II. Both class A and class B extend class X that has a = method

          = public=20 String toString() { return getClass().getname(); }

III. Both class A and class B extend an abstract class X that = has=20 methods

          = public=20 abstract String=20 = getName();          =20 public String toString() { return getname(); }

   Class A has a method

          = public=20 String toString() { return "A"; }

   and class B has a method

          = public=20 String toString() { return "B"; = }

I only=20
II only=20
I and II=20
II and III=20
I, II, and III =

24. (AB)

Consider the following class:

public class Widget implements=20 Comparable {      private int nuts,=20 bolts;

     public Widget(int nuts, int = bolts) {=20 this.nuts =3D nuts; this.bolts =3D bolts; }

     public int compareTo(Object = other) {=20 return nuts - ((Widget)other).nuts; }

     public boolean equals(Object = other) {=20 return bolts =3D=3D ((Widget)other).bolts; }

     public String toString() { = return "" +=20 nuts; } }

Suppose a non-empty List list holds several=20 Widget objects and the following statements have been = executed:

=
Collections.sort(list); System.out.println(list); Set=20 tSet =3D new TreeSet(list); Set hSet =3D new=20 HasSet(list);

Which of the following statements is NOT necessarily = true?

tSet.size() = <=3D=20 list.size()=20
hSet.size() = <=3D=20 list.size()=20
hSet.size() = =3D=3D=20 tSet.size()=20
=20 tSet.contains(list.get(list.size() - 1))returns=20 true=20
The output lists = the values of=20 nuts in Widget objects in = list in=20 ascending order

25.

A multiple-choice test has 25 questions, each with = five=20 answer choices, A - E. On the eve of the test, Casey and other = students=20 had learned that the correct answers on the test were not balanced = properly: 3 correct answers were C, 7 were D, and 15 were E (A and = B were=20 never correct answers). Casey spent the rest of the night devising = an=20 optimal strategy for using this information (neglecting to review = the test=20 material). In the worst case, how many questions is Casey = guaranteed to=20 get right using the optimal strategy?

3=20
5=20
7=20
10=20
15

2004

26.

twist is defined as follows:

public void twist(String[] = w) { =20 String temp =3D w[0].substring(0, = 1); w[0] =3D=20 w[1].substring(0, 1) + = w[0].substring(1); =20 w[1] =3D temp + w[1].substring(1); }

What is the output of the following code segment?

String[] words =3D {"HOW",=20 "NEAT"}; twist(words): System.out.println(words[0] + " " + = words[1]);

NOW = NOW=20
HOW = HOW=20
NOW = HOW=20
HOW = NEAT=20
NOW = HEAT=20

27.

If Crossword extends Puzzle and implements Solvable, = and

Puzzle p =3D new Puzzle(); Crossword cw =3D new = Crossword(10,=20 20);

are declared, which of the following statements will cause a = syntax=20 error?

p =3D = cw;=20
cw =3D new = Puzzle();=20
p =3D new = Crossword(10,=20 20);=20
Solvable x = =3D new=20 Crossword(10, 20);=20
All of the above = will compile=20 with no errors.

28.

What is the output of the following code?

List nums =3D new ArrayList(3); nums.add(New=20 Integer(1)); nums.add(New Integer(2)); nums.add(0,=20 nums.get(1));

Object x =3D nums.get(0); Object y =3D = nums.get(2);

if (x =3D=3D y) = System.out.println(x +=20 " is equal to " + y); else =20 System.out.println(x + " is NOT equal to " +=20 y);

1 is equal = to 2=20
1 is NOT = equal to 2=20
2 is equal = to 2=20
2 is NOT = equal to 2=20
=20 IndexOutOfBoundsException

29. (AB)

Which of the following best describes the loop = invariant in=20 the following code?

double x =3D 3.0, y =3D 1.0; while (x - y >=20 0.01) { x =3D (x + y) /=20 2; y =3D 3.0 /=20 x; }

3=20
x =3D 3=20
xy =3D 3=20
xy =3D 3 = and x - y=20 > 0.01=20
x =3D = 1.7321428571428572=20

30.

What is the output of the following code?

int sum =3D 0, p =3D 1; for (int count =3D 1; count = <=3D 50;=20 count++) { sum +=3D=20 p; p *=3D = 2; }

-1=20
562949953421311 = (=3D=20 2⁴⁹ - 1)=20
1125899906842623 = (=3D=20 2⁵⁰ - 1)=20
= ArithmeticException=20
=20 IllegalArgumentException

31.

What is the output of the following code?

String barb =3D=20 = "BARBARA"; scramble(barb); System.out.println(barb);
The=20 method scramble is defined as follows:

public String scramble(String=20 str) { if (str.length() >=3D=20 2) =20 { int = n =3D=20 str.length() /=20 2; str = =3D=20 scramble(str.substring(n)) + str.substring(0,=20 n); } = return=20 str; }

= BARBARA=20
= ARBABAR=20
AABAR =
= ARBABARB=20
= ARABARBARB=20

32.

What are the values in arr after the = following=20 statements are executed?

int[] arr =3D {1, 1, 0, 0, 0};

for (int i =3D 2; i < arr.length;=20 i++) arr[i] =3D arr[i-1] +=20 arr[i-2];

11011=20
11210=20
11222=20
11235=20
11248 =

33.

Given

double x =3D 5, y =3D 2;

what is = the value of=20 m after the following statement is executed?

int m =3D (int)(x + y + x / y - x * y - x / (10 *=20 y));

-1=20
-0.75=20
-0.5=20
0=20
1

34.

What is the value of sum after the = following=20 code segment is executed?

int p =3D 3, q =3D 1, sum =3D 0; while (p <=3D=20 10) { sum +=3D p %=20 q; p++; =20 q++; }

0=20
10=20
12=20
14=20
52

35. (AB)

What is the output of the following code segment?

List words =3D new LinkedList(); int k;

for (k =3D 1; k <=3D 9; = k++) =20 words.add("word", k);

for (k =3D 1; k <=3D words.size();=20 k++) if (k % 3 =3D=3D=20 0) =20 words.remove(k);
=
System.out.println(words);

[word1, = word2, word4,=20 word5, word7, word8]=20
[word2, = word3, word5,=20 word6, word7, word8]=20
[word2, = word3, word5,=20 word6, word8, word9]=20
[word1, = word2, word3,=20 word5, word6, word7, word8]=20
[word1, = word2, word3,=20 word5, word6, word8, word9]

36.

A class Particle has a private field=20 double velocity and public methods

double=20
      getVelocity()

and void setVelocity(double v). = It also=20 has a method

public void hit(Particle p) { < Missing = statements=20 > }

Which of the following could=20 replace < Missing statements > in=20 hit to make it compile with no errors?

I.     double v =3D=20 getVelocity();      =20 setVelocity(p.getVelocity());      =20 p.setVelocity(v);

II.    double v =3D=20 velocity;       velocity =3D=20 p.getVelocity();      =20 p.setVelocity(v);

III.    double v =3D=20 velocity;       velocity =3D=20 p.velocity();       p.velocity =3D=20 v;

I only=20
II only=20
I and II=20
II and III=20
I, II, and III =

37.

Which of the following conditions must always be = true after=20 the while loop finishes?

while (hours < hours0 || (hours =3D=3D hours0 = && mins=20 < mins0)) { mins +=3D=20 5; if (mins >=3D=20 60) =20 { mins = -=3D=20 60; =20 hours++; = } }

hours > = hours0=20 && mins >=3D mins0=20
hours = >=3D hours0=20 && mins >=3D mins0=20
hours < = hours0 ||=20 (hours =3D=3D hours0 && mins < mins0)=20
hours = >=3D hours0=20 && (hours !=3D hours0 && mins >=3D = mins0)=20
hours = >=3D hours0=20 && (hours !=3D hours0 || mins >=3D mins0) =

38.

Consider the following class:

public class SaleItem implements=20 Comparable {      public SaleItem(int = p) {=20 prcice =3D p; }

     < Comparison method = header=20 > { < Code not shown > }

     public String toString() { = return=20 String.valueOf(price); }

     private int=20 price; }

Which of the following could = replace=20 < Comparison method header > to make = this class=20 compile with no errors?

I.    public int compare(SatleItem = item1,=20 SaleItem item2)

II.    public boolean compareTo(SaleItem = item1)

III.   public int compareTo(Object=20 obj)

I only=20
II only=20
III only=20
I or II=20
II or III =

int money;

     public Gambler(int m) { money = =3D m;=20 }

     public int currentMoney() { = return=20 money; }      public void addMoney(int m) = { money=20 +=3D m; }

     public void work() { money = +=3D 100;=20 }      public void play() { money /=3D 2; = }      public liveAnotherDay() { work(); = play();=20 }

     public String toString() { = return=20 String.valueOf(money); } }

public class CompulsiveGambler extends=20 Gambler {      public = CompulsiveGambler(int=20 m)     =20 {           < = Missing statements >      = }

     public void work() { /* do = nothing */=20 }

     public void=20 play()     =20 {           = while=20 (currentMoney() >=20 1)          =20 = {            &n= bsp;  =20 = super.play();          =20 }      = } }

39.

Given that

System.out.println(new=20 CompulsiveGambler(300)); =20

displays 300, which of the = following=20 could replace < Missing statements > in=20 CompulsiveGambler's constructor?

I.    addMoney(m);

II.    super(m);

III.   super();     =20 addMoney(m);

I only=20
II only=20
I or II=20
II or III=20
I, II, or III =

40.

What is the output of the following code segment?

CompulsiveGambler x =3D new=20 = CompulsiveGambler(300); x.liveAnotherDay(); System.out.println(x);<= /CODE>

200=20
150=20
100=20
1=20
0

41.

Consider the following method:

public int goFigure(int = x) { if=20 (x < = 100) x=20 =3D goFigure(x + 10); return (x - = 1); }=20

What does goFigure(60) return?

59=20
69=20
95=20
99=20
109 =

42. (AB)

Suppose a List list contains strings =

"A",=20
      "*", "B", "*", "C"

(in this order). What is the output of = the=20 following code segment?

ListIterator it =3D list.listIterator(); while=20 (it.hasNext) { if=20 = ("*".equals(it.next())) &nbs= p; =20 it.add(it.next() + "*"); }
=
System.out.println(list);

No output: the = program goes into=20 an infinite loop=20
[A, *, **, B, *, = **, C]=20
[A, **, *, B, **, = *, C]=20
[A, *, B, B*, *, = C, C*]=20
[A, *, **, ***, = ****, B, *, **,=20 ***, ****, C]

43. (AB)

A linked list pointed to by ListNode=20 head contains Comparable objects. Which = sorting=20 algorithm is implemented by the following method sort and its = helper=20 method doSomething?

public ListNode sort(ListNode=20 head) {      if (head !=3D=20 null)           = head =3D=20 doSomething(sort(head.getNext())), head);

     return head; }

private ListNode doSomething(ListNode head, ListNode=20 node) {      if (head =3D=3D null || = ((Comparable)=20 node.getValue()).CompareTo(head.getValue()) <=20 0)     =20 {          =20 = node.setNext(head);         &n= bsp;=20 head =3D node;      = }     =20 else          =20 head.setNext(doSomething(head.getNext(), node));

     return=20 head; }

Selection sort=20
Insertion sort=20
Mergesort=20
Quicksort=20
Heapsort =

44. (AB)

A Map thingsToDo associates a=20 Resort object with a Set of activities = available=20 at that resort. The following code segment is intended to = remove

=20
      "Golf"

from the activity sets in all resorts:

Iterator iter =3D = thingsToDo.keySet().iterator(); while=20 (inter.hasNext()) < Missing=20 statement >

Which of the following should replace < Missing=20 statement >?

(Set)=20 iter.next().remove("Golf");=20
((Set)=20 iter.next()).remove("Golf");=20
= thingsTodo.remove((Set)=20 iter.next(), "Golf");=20
= thingsTodo.remove((Resort)=20 iter.next(), "Golf");=20
((Set)=20 thingsToDo.get(iter.next())).remove("Golf"); =

45. (AB)

What is the output of the following code segment?

TreeNode node6 =3D new TreeNode("6", null,=20 null);
TreeNode node5 =3D new TreeNode("5", = null,=20 null); TreeNode node4 =3D new TreeNode("4", null, = null); TreeNode=20 node3 =3D new TreeNode("3", node5, node6); TreeNode node2 =3D = new=20 TreeNode("2", null, node4); TreeNode node1 =3D new = TreeNode("1", node2,=20 node3); TreeNode root =3D node1;

Object[] arr =3D new Object[8]; toArray(root, 1,=20 arr);

for (int i =3D 0; i < arr.length;=20 i++) System.out.println(arr[i] + "=20 ");

The method toArray is defined as follows:

private void toArray(TreeNode root, itn i, Object[]=20 arr) { if (root !=3D=20 null) =20 { = arr[i] =3D=20 = root.getValue(); = ;=20 toArray(root.getLeft(), 2*i,=20 arr); =20 toArray(root.getRight(), 2*i + 1, = arr); =20 } }

null 1 2 = null 3 4 5=20 6=20
null 1 2 3 = null 3 4=20 5=20
null 1 2 3 4 = null 3=20 4=20
null 1 2 3 4 = 5 null=20 3=20
null 1 2 3 4 = 5 6=20 null

46. (AB)

Suppose all valid five-digit zip codes are = represented as=20 Integer objects and stored in a set containing about = 4000 zip=20 codes. Compare two implementations of this set: one is a=20 HashSet with 400 buckets; another is a = TreeSet.=20 Assuming that various zip codes are matched against the set with = roughly=20 the same frequency, which of the following statements about the = average=20 performance of these implementations is true?

= HashSet works more=20 than 100 times faster than TreeSet=20
= HashSet works about=20 20 times faster than TreeSet=20
= HashSet works 2-4=20 times faster than TreeSet=20
= HashSet works=20 slower than TreeSet=20
= HashSet works=20 roughly as fast as TreeSet, but takes more than = twice as=20 much space

47. (AB)

What is the output of the following code segment?

String[] letters =3D { "A", "B", "C" }; Queue = qLetters =3D new=20 ListQueue(); String sLetters =3D ""; Stack stk =3D new=20 ArrayStack();

for (int i =3D 0; i < letters.length;=20 i++) { =20 qLetters.enqueue(letters[i]); =20 stk.push(qLetters); sLetters +=3D=20 letters[i]; } while=20 (!stk.isEmpty()) { =20 System.out.print(stk.pop() + " "); = Queue q =3D=20 (Queue) stk.pop(); =20 System.out.print("["); while=20 = (!q.isEmpty()) = System.out.print(q.dequeue()); =20 System.out.print("] "); }

ABC [ABC] AB = [] A=20 []=20
ABC [] ABC = [] ABC=20 []=20
ABC [ABC] AB = [AB] A=20 [A]=20
A [A] AB = [AB] ABC=20 [ABC]=20
= ClassCastException=20

48. (AB)

An n-by-n two-dimensional array = contains=20 Comparable values. The values in each row are = increasing. The=20 columns alternate: in the first, third, and other odd columns the = values=20 are increasing and in the second, fourth, and other even columns = the=20 values are decreasing. What is the average big-O for an optimal = algorithm=20 that finds a given value in such an array?

O(log = n)=20
O((log=20 n)²)=20
O(n) =
O(n = log n)=20
= O(n²)=20

49.

Consider the following classes:

public abstract class = PointXY {     =20 private int x, y;

     public PointXY(int x, int y) { = this.x=20 =3D x; this.y =3D y; }      public int = getX() {=20 return x; }      public int getY() { = return y;=20 }      public String toString() { return = "(" +=20 getX + "," + getY() + ")"; }

     public abstract PointXY = moveBy(int dx,=20 int dy);      public abstract double=20 distanceFrom(PointXY p); }

public abstract class CartesianPoint extends=20 PointXY {      public = CartesianPoint(int x,=20 int y) { < Code not shown > = }     =20 public double distanceFrom(PointXY p) { < Code not = shown >=20 } }

public class BoardPosition extends=20 CartesianPoint {      < Code not = shown > }

Which of the following is the minimal set of public = constructors and/or=20 methods required in the BoardPosition class, so that the = statements

BoardPosition pos =3D new BoardPosition(0,=20 0); System.out.println(pos.moveBy(10,=20 10).distanceFrom(pos));

compile with no errors?

public = PointXY moveBy(int=20 dx, int dy)=20
public = boardPosition(int=20 x, int y)andpublic PointXY moveBy(int dx, int = dy)=20
public = double=20 distanceFrom(PointXY pos)andpublic PointXY = moveBy(int dx,=20 int dy)=20
public = double=20 distanceFrom(BoardPosition pos)andpublic = BoardPosition=20 moveBy(int dx, int dy)=20
public=20 BoardPosition()andpublic BoardPosition(int x, int=20 y)and public PointXY moveBy(int dx, int dy)=20

50.

A multiple-choice question deals with the scores = that four=20 students received in a contest. The question offers the following = answer=20 choices:

Tim got more points than Jenny.=20
Tim is the contest winner.=20
Jenny is in last place.=20
Tim's score is above average, and Jenny's score is below = average.=20
While Nina did better than Phil, the boys' combined score = is=20 higher than the girls' combined score.

The question assumes that one option is true and all the rest = are=20 false. But the question is badly designed, making it possible to = guess the=20 correct answer from the given choices without even looking at the=20 question. What is the correct answer?

a=20
b=20
c=20
d=20
e

2003

51.

fun is defined as follows:

public int fun(int[] = v) { =20 v[0]--; return v[0] +=20 2; }

What is the value of v[0] after the = following code segment is executed?

int [] v =3D { 3, 4, 5 }; v[0] =3D fun(v);=20

1=20
2=20
3=20
4=20
5

52.

The method xProperty is defined as = follows:

public boolean xProperty(int=20 a) { return a =3D=3D 2 * (a / 10 = + a %=20 10); }

For which of the following = values of=20 a does xProperty(a) return=20 true?

2=20
8=20
18=20
28=20
128 =

53.

What are the values of m and n after the following = code=20 runs?

int m =3D 20, n =3D 2, temp;

for (int count =3D 1; count <=3D 20;=20 count++) { temp =3D=20 m; m =3D n +=20 count; n =3D temp -=20 count; }

m =3D = 230 n =3D -208=20
m =3D = 30 n =3D -8=20
m =3D = 12 n =3D -10=20
m =3D = -12 n =3D 8=20
m =3D = -190 n =3D 212 =

54.

Consider the method

public int[] copyX(int[] = arr) { =20 int[] result =3D new int[arr.length];

for (int i =3D 0; i < = arr.length;=20 i++) =20 { if = (arr[i]=20 <=3D=20 = 0) &= nbsp; =20 break; = if=20 (arr[i] >=20 = 10) = =20 break; = result[i] =3D arr[i]; =20 } return result; }=20

Suppose it is rewritten as follows:

public int[] copyX(int[]=20 arr) { int[] result =3D new=20 int[arr.length]; int i =3D = 0;

while (< condition = >) =20 { = result[i]=20 =3D = arr[i]; =20 i++; } = return=20 result; }

Which of the following expressions can replace <=20 condition > so that the second version is = equivalent to=20 the first one (i.e., for any int[] parameter=20 arr, it returns the same result as the first=20 version)?

i < = arr.length=20 && (arr[i] <=3D 0 || arr[i] > 10)=20
= (arr[i] <=3D 0 ||=20 arr[i] > 10) || i >=3D arr.length=20
= (arr[i] <=3D 0 ||=20 arr[i] > 10) && i < arr.length=20
i < = arr.length=20 && !(arr[i] <=3D 0 || arr[i] > 10)=20
i < = arr.length=20 && arr[i] > 0 && arr[i] <=3D 10) =

55.

Given

double x =3D < any positive value less than = 2003 >;=20

which of the following code fragments set int y to the smallest = integer=20 that is NOT less than three quarters of x?

I.    int y =3D (int)(3 * x /=20 4);      if (y < 3 * x / 4) = y++;

II.   int y =3D = 1;     =20 while (y < 3 * x / 4) y++;

III.   int y =3D 2010 - (int)(2010 - x * 3 = /=20 4);

I only=20
II only=20
I and II=20
I and III=20
I, II, and III =

56.

Alicia is five years older than her brother Ben. = Three years=20 from now Alicia's age will be twice Ben's age. How old are Alicia = and Ben=20 now? Ha wrote the following program to solve this puzzle:

public class = AliciaAndBen { =20 public static void main(String[] = args) =20 { for = (int a =3D=20 1; a <=3D 100;=20 = a++) = ; =20 for (int b =3D 1; b <=3D 100;=20 = b++) = ; =20 if (< condition=20 = >) &nbs= p; = ; =20 System.out.println("Alicia is " + a + " and Ben is " +=20 b); } } =

Which of=20 the following expressions should replace

< =
condition=20
      >

in Hal's program so that it displays the correct = solution to=20 the puzzle?

(a =3D=3D b = - 5) && (a=20 - 3 =3D=3D 2 * (b - 3))=20
a =3D=3D b + = 5 && a +=20 3 =3D=3D 2 * b + 6=20
(a =3D=3D (b = + 5)) &&=20 ((a + 3) ++ (2 * b + 3))=20
a =3D=3D (b = - 5) && (2=20 * a - 3) =3D=3D (b - 3)=20
None of the above = works=20

57. (AB)

Suppose a two-dimensional array of = chars=20 m has 4 rows and 5 columns and holds the values = represented=20 in the picture below:

=
x.xxx xx..x .x.xx xx...

What = are=20 the values in m after the following code segment is = executed?

int r, c; for (r =3D 1; r < m.length;=20 r++) for (c =3D 1; c < = m[0].length - 1;=20 c++) = if=20 (m[r-1][c-1] =3D=3D 'x' && m[r-1][c+1] =3D=3D=20 = 'x') = ; =20 m[r][c] =3D 'x';

a. x.xxx = ; =20 xx..x =20 .xxxx = xx... b. xxxxx = ; =20 xx..x =20 .xxxx = xx... c. x.xxx = ; =20 xx.xx =20 xxxxx = xxx.. d. x.xxx = ; =20 xx.xx =20 .xxxx = xxxx. e. x.xxx = ; =20 xx.xx =20 .x.xx =20 xxx..

58.

What is the output of the following code?

String s =3D = "ONION"; System.out.println(s.substring(1,=20 5).substring(1, 4).substring(0, = 3));

I=20
IO=20
ION=20
ONI=20
NION =

59.

What is the smallest required number of three = comparisons in=20 an optimal algorithm (based on comparing values) that puts any = THREE=20 distinct values into a list in ascending order? What is the answer = for=20 FOUR distinct values?

2 and 3=20
3 and 4=20
3 and 5=20
3 and 6=20
6 and 12 =

60.

Given

int counts[] =3D {7, 2, 9, 0, 1, 5, 5, 3,=20 9};

What does find3(counts, 9) = return?=20 find3 is defined as follows:

public int find3(int a[], int=20 targetsum) {      int i =3D 0, sum =3D = 0;

     while (i <=20 3)     =20 {           sum = +=3D=20 a[i];          =20 i++;      }

     if (sum =3D=3D=20 = targetsum)          =20 return 1;

     while (i <=20 a.length)     =20 {           sum = +=3D a[i]=20 - = a[i-3];           if=20 (sum =3D=3D=20 = targetsum)           = ;    =20 return i - = 1;          =20 i++;      = } }

-1=20
1=20
2=20
3=20
8

61. (AB)

Let us call an array "oscillating" if its values = alternate=20 going up and down, as follows: a[i-1] < a[i] and=20 a[i] > a[i+1] for any odd i,

0 =
< i=20
      < n-1

, where n is the number of elements in = a. What is the "big-O" for an optimal algorithm that=20 determines the minimum value for an oscillating array of length=20 n? The median value?

= ; =20 = Minimum =20 Median

=20 = O(1) &nb= sp; =20 O(1)=20
=20 = O(n/2) = =20 O(1)=20
=20 = O(n) &n= bsp; =20 O(n)=20
=20 = O(n) &n= bsp; =20 O(n log n)=20
=20 = O(n) &n= bsp; =20 O(n²)

62. (AB)

Suppose an n by n matrix of "black" and "white" = pixels=20 (e.g.,1s and 0s) represents a picture of a black blob that fills = the=20 southeastern corner of the picture. The blob's boundary extends in = a=20 generally southwest-to-northeast direction. All the pixels below = and to=20 the right of any black pixel are black. For example:

=
0000001 0000001 0000011 0001111 0111111 11111111111111

What=20 is the worst case "big-O," in terms of n, for the total number of = integer=20 additions plus pixel comparisons in an optimal algorithm that = determines=20 the area of a blob (the number of black pixels in the = blob)?

O(log = n)=20
O((log=20 n)²)=20
O(n) =
O(n log = n)=20
= O(n²)=20

Questions 63-67 refer to the following interface = and=20 classes, written by Kim, a novice programmer, for modelling a = "Caller ID"=20 device:

public interface Call {      = String=20 getSource(); }

public class IncomingCall implements=20 Call {      private String=20 telephoneNumber;

     public IncomingCall() {=20 telephoneNumber =3D ""; }      public=20 IncomingCall(String tel) { telephoneNumber =3D tel; }

     public void = setTelephoneNumber(String=20 tel)      { telephoneNumber =3D tel; = }

     public String getSource() { = return=20 telephoneNumber; }      public String = toString()=20 { return getSource(); } }

public class IncomingCallWithName extends=20 IncomingCall {      private String=20 callerName;

     public = IncomingCallWithName(String=20 tel, String name)     =20 {           < = missing statement=20 >          =20 callerName =3D name;      }

     public String getName() { = return=20 callerName; }

     public String=20 getSource()      { return = super.getSource() + " "=20 + getName(); }

     public String=20 toString()      { return = super.getSource() + " "=20 + getName(); } }

63.

Kim's teacher specified that

System.out.println(new = IncomingCallWithName("800-749-2000",=20 "Pizza Palace"));

should display

800-749-2000 Pizza Palace

Which of the following statements can replace < = missing=20 statement > in the=20 IncomingCallWithName constructor to achieve = that?

I. =20 = super(tel);II. setTelephoneNumber(tel);=III. telephoneNumber=20 =3D tel;

I only=20
II only=20
I and II=20
II and III=20
I, II, and III =

64.

What is the result of the following code segment?

Call[] calls =3D new Call[3]; calls[0] =3D new=20 = IncomingCall("888-888-8888"); &n= bsp; &nb= sp;=20 // Line 1 calls[1] =3D (IncomingCallWithName)=20 = calls[0]; &nbs= p; =20 // Line 2 System.out.println(calls[0] + " " + calls[1] + " " = +=20 calls[2]); // Line 3

Syntax error on = line 1=20
Syntax error on = line 2=20
= ClassCastException=20 on Line 2=20
=20 NullPointerException on Line 3=20
Compiles and runs = with no=20 errors; the output is:
=20 888-888-8888 888-888-8888 null

65.

After correctly completing the=20 IncomingCallWithName constructor, as requested in = Question=20 63, Kim wrote the following test code:

IncomingCall call =3D new = IncomingCallWithName("800-749-2000",=20 = "PizzaPalace"); System.out.println(call);

= ;The=20 output was

800-749-2000 PizzaPalace

Kim's teacher suggested that Kim try to compile and run the = program=20 again with IncomingCallWithName's = toString=20 method commented out. What would be the result of this=20 experiment?

Syntax error=20 "IncomingCallWithName shoul dbe declared abstract"=20
Infinite = recursion, stack=20 overflow=20
The program = compiles with no=20 errors and displays 800-749-2000
The program = compiles with no=20 errors and displays the same output as before, = 800-749-2000=20 PizzaPalace=20
The program = compiles with no=20 errors and displays IncomingCallWithName@47e553 =

66.

Suppose calls and name are initialized as follows:

Call[] calls =3D { new=20 IncomingCallWithName("800-749-2000", "Pizza=20 Palace"), new=20 IncomingCall("888-237-3757"), new=20 IncomingCallWithName("800-555-1234", "Burger = Heaven") };

String name =3D "Pizza = Palace";

The=20 following code segment is intended to count the number of=20 Call objects in the calls array that came from a = given=20 source:

String name =3D IO.readline(); // Read the name of the=20 source int count =3D 0; for (int i =3D 0; i < = calls.length;=20 i++) if ( < condition > = ) =20 count++;

Which of the following = replacements=20 for ( < condition > ) will compile with = no=20 errors and correctly set count to 1?

=20 calls[i].getSource().indexOf(name) >=3D 0=20
=20 calls[i].toString().indexOf(name) >=3D 0=20
=20 calls[i].getName().equals(name)=20
= ((IncomingCallWithName)=20 calls[i]).getName().equals(name)=20
None of the above =

67. (AB)

Consider the following class:

public class CallerID {      = public=20 List calls;

     public=20 CallerID()     =20 {           = calls =3D new=20 LinkedList();      }

     // precondition: calls hold=20 IncomingCall objects      // = postcondition: all=20 calls that came from target are removed from the calls=20 list      public void removeAll(String = target) {=20 < code not shown > }

     < other methods not = shown=20 > }

If removeAll works as specified in its pre- and=20 postconditions, which of the following code segments can serve as=20 removeAll's body?

I. for (int i =3D 0; i < = calls.size();=20 t++) =20 { if = (((IncomingCall)=20 = calls.get{i}).getSource().equals(target)) &nbs= p; =20 calls.remove(i); =20 }II. int i =3D=20 0; while (i < calls.size();=20 t++) =20 { if = (((IncomingCall)=20 = calls.get{i}).getSource().equals(target)) &nbs= p; =20 = calls.remove(i); = ;=20 = else = ; =20 i++; =20 }III. Iterator iter =3D=20 calls.iterator(); while=20 (iter.hasNext()) =20 { if = (((IncomingCall)=20 = calls.get{i}).getSource().equals(target)) &nbs= p; =20 iter.remove(i); =20 }

I only=20
II only=20
I and II=20
II and III=20
I, II, and III =

68. (AB)

What is the output of the following code segment?

Map m =3D new TreeMap(); m.put("La", = "La"); m.put("La-La",=20 "La"); m.put("La-La-La", "Ye-Ye"); Iterator it =3D=20 m.keySet().iterator(); while=20 (it.hasNext()) =20 System.out.println(m.get(it.next()) + " = ");

La = Ye-Ye=20
La La = Ye-Ye=20
La = La-La-La=20
La La = La-La-La=20 Ye-Ye=20
La La La-La = La La-La-La=20 Ye-Ye

69. (AB)

Suppose ListNode head refers to the = first node=20 of a linked list. Consider the following code fragment:

ListNode node, temp;

for (node =3D head; node !=3D null;=20 node.getNext()) { while = (node.getNext()=20 !=3D null &&=20 = node.getNext().getValue().equals(node.getValue())) &= nbsp;=20 { =20 = node.setNext((node.getNext().getNext()))l =20 } }

If head refers to a = linked list=20 with 11 nodes that hold letters

"M", "I", "S", "S", "I", "S", "S", "I", "P", "P",=20 "I"

(in that order), what letters are = stored in the=20 nodes of this list after the above code is executed?

"M"=20
"M", "I", = "S"=20
"M", "I", = "I", "I"=20
"M", "I", = "S", "P"=20
"M", "I", = "S", "I", "S",=20 "I", "P", "I"

70. (AB)

Consider the following method:

public int mysteryCount(TreeNode=20 root) { int count =3D = 0;

if (root !=3D=20 null) =20 { = count =3D=20 mysteryCount(root.getLeft()) +=20 = mysteryCount(root.getRight()); &nb= sp; =20 if ((root.getLeft() =3D=3D null && root.getRight() = =3D=3D null)=20 = || &= nbsp; =20 (root.getLeft() !=3D null && root.getRight() !=3D null = &&=20 = root.getLeft().getValue().equals(root.getRight().getValue()))) &= nbsp; &n= bsp;=20 count++; =20 } }

If root refers to the = tree

=
&nbs= p; =20 = A &n= bsp; =20 / =20 = \ &n= bsp; =20 / =20 = \ &n= bsp; =20 B =20 = C &n= bsp;=20 / \ / =20 = \ =20 D D D =20 = D &n= bsp; =20 / \ /=20 = \ &n= bsp; =20 E E E = E

which value is=20 returned by mysteryCount(root)?

1=20
3=20
4=20
5=20
10

71. (AB)

Consider the following method that creates a binary = tree=20 from a linked list:

public TreeNode listToTree(ListNode=20 head) { if (head =3D=3D null ||=20 head.getNext() =3D=3D=20 null) = return=20 null; =20 else = return=20 new=20 = TreeNode(head.getValue(), &n= bsp; &nb= sp; =20 = lisToTree(head.getNext()), &= nbsp; &n= bsp; =20 = listToTree(head.getNext().getNext())); }

If=20 head refers to a linked list with five nodes=97

"A", "B", "C", "D", "E"

=97how = many nodes in=20 the tree returned by listToTree(head) hold = "E"?

0=20
3=20
4=20
5=20
8

72. (AB)

Consider the following method that builds a linked = list from=20 a binary tree:

public ListNode treeToList(TreeNode=20 root) { if (root =3D=3D=20 null) = return=20 null; else if (Math.random() <=20 0.5) = return=20 new ListNode(root.getValue(),=20 treeToList(root.getLeft())); =20 else = return=20 new ListNode(root.getValue(),=20 = treeToList(root.getRight())); }

 root=20

refers to the tree

=
&nbs= p; =20 = A &n= bsp; =20 / =20 = \ &n= bsp; =20 / =20 = \ &n= bsp; =20 B =20 = C &n= bsp;=20 / \ =20 = / =20 D E F

which = of the=20 following lists could possibly result from = treeToList(root)?

I. A, B,=20 DII. A, B, C,=20 DIII. A, C,=20 F

I only=20
II only=20
I and II=20
I and III=20
I, II, and III =

73. (AB)

Consider the following method that builds a linked = list from=20 a queue:

// precondition: q holds Comparable objects public = TreeNode=20 queueToTree(ListQueue q) {      if=20 = (q.isEmpty())          =20 return null;

     ListQueue q1 =3D new=20 ListQueue();      ListQueue q2 =3D new=20 ListQueue();      Comparable x, = y;

     x =3D (Comparable)=20 q.dequeue();      while=20 (!q.isEmpty())     =20 {           y = =3D=20 (Comparable)=20 = q.dequeue();          =20 if (y.compareTo(x) <=20 = 0)            &= nbsp;  =20 = q1.enqueue(y);           = = else            = ;   =20 q2.enqueue(y);      }

     return new TreeNode(x,=20 queueToTree(q1), = queueToTree(q2)); }

 q=20

contains letters A, P,

=20
      R

, I, C, O,

=
T

=20 (represented by Character or String = objects, in=20 this order, from front to rear), what is the result of inorder = transversal=20 (left-to-right) of the tree returned by=20 queueToTree(q)?

= A,=20 C, I, O, P,=20 R, T=20
= A,=20 P, R, T, I,=20 C, O=20
= A,=20 P, R, C, I,=20 O, T=20
= A,=20 C, O, I, T,=20 R, P=20
= A,=20 P, I, C, O,=20 R, T

74. (AB)

Consider the following method eval:

String eval(Queue q) {      = Stack stk=20 =3D new ArrayStack();      String s =3D=20 "";

     while=20 (!q.isEmpty())     =20 {           s = =3D (String)=20 q.dequeue();

          = if=20 = (!s.equals("+"))          = ;     =20 = stk.push(s);          =20 else          =20 = {            &n= bsp;  =20 String s1 =3D "", s2 =3D=20 = "";            =    =20 if=20 = (!stk.isEmpty())          = ;          =20 s2 =3D (String)=20 = stk.pop();           = ;    =20 if=20 = (!stk.isEmpty())          = ;          =20 s1 =3D (String)=20 = stk.pop();           = ;    =20 stk.push(s1 +=20 s2);          =20 }      }      if=20 = (!stk.isEmpty())          = ;=20 s =3D (String) stk.pop();      return=20 s; }

The program reads strings (which = may hold=20 single characters), separated by spaces, one by one from the input = line=20 and puts them into a queue q. Then it displays the = string=20 returned by eval(q). For which of the following input = lines=20 is the output HELLO?

= ; =20 = front
&= nbsp; =20 ↓=20

H + E + L + = L + O=20
H E L + + L = O + +=20
H E + + L L = + O=20
+ + + + O L = L E H=20
None of the above =

75.

A multiple-choice test contains 25 questions. The = correct=20 answers include five A's, five B's, five C's, five D's, and five = E's. Both=20 Constance and Randy solve the first 15 questions correctly but = then run=20 out of time. Constance picks the letter (or one of the letters) = that is=20 least frequent among the first 15 answers and fills the remaining = 10=20 answers with this letter. Randy picks random answers for the = remaining 10=20 questions, but he makes sure that at the end each letter A-E = appears=20 exactly five times among all 25 answers. Which of the following = statements=20 is FALSE?

Randy and = Constance can get the=20 same score=20
Randy's score can = be any number=20 from 15 to 25=20
Constance scores = not more than=20 17=20
Constance scores = not more than=20 20=20
Neither Randy nor = Constance can=20 score exactly 24

2002

76.

If a is and int array of = length 2=20 and a[0] and a[1] holds values 7 and 13=20 respectively, what are their values after fun(a) is = called?=20 The method fun is defined as follows:

public void fun(int[] = x) { x[0]=20 =3D (int)(100.0 * x[0] / (x[0] + = x[1])); x[1]=20 =3D (int)(100.0 * x[1] / (x[0] + x[1])); } =

7 and 13=20
35 and 27=20
34 and 64=20
35 and 65=20
34 and 66 =

77.

Consider a method

public boolean isProcessedX(int n, int[]=20 v) { if (n >=3D 2 &&=20 isProcessedX(n-1, v)) =20 { = v[n-1] =3D=20 = v[n-2]; return = true; } =20 else = return=20 false; }

What happens if an int array s holds = values=20 1, 2, 3, 4, 5 and isProcessedX(5, s) is = called?

s = holds 1, 2, 3, 4,=20 5 and isProcessedX returns false=20
s = holds 1, 2, 3, 4,=20 4 and isProcessedX returns true=20
s = holds 1, 1, 1, 1,=20 1 and isProcessedX returns true=20
=20 indexOutOfBoundsException=20
Stack overflow = error

78.

An array mix holds seven elements. For = which=20 seven values in mix will the value of the variable=20 property be true after the following = code=20 segments is executed?

int i; boolean property =3D true;

for (i =3D 1; i < 6; = i++) { if=20 (mix[i] >=3D mix[i-1] || mix[i] >=3D=20 = mix[i+1]) =20 property =3D false; }

1, 2, 3, 4, 5, 6, = 7=20
7, 6, 5, 4, 3, 2, = 1=20
7, 1, 6, 2, 5, 3, = 4=20
1, 2, 3, 4, 3, 2, = 1=20
1, 5, 2, 6, 3, 7, = 4

79. (AB)

The class Game has a data member

 =
char[][]=20
      board

and a constructor defined as follows:

public Game() {      board = =3D new=20 char[4][4];      int r, c;

     for (r =3D 0; r < 4;=20 r++)           = for (c =3D=20 0; c < 4;=20 = c++)            = ;   =20 board[r][c] =3D '.';

     for (r =3D 0; r < 4;=20 r++)          =20 board[r][(r + 1) % 4] =3D 'x'; }

What = values are=20 stored in board when a Game object is=20 constructed with the above constructor?

a. ...x = =20 ..x. =20 .x.. = x... b. ..x. = =20 .x.. =20 x... = ...x c. ...x = =20 x... =20 .x.. = ..x. d. .x.. = =20 ..x. =20 ...x = x... e. .x.. = =20 x... =20 ...x =20 ..x.

80.

Consider the following method prepare:

public String prepare(String=20 s) { int k =3D s.length() /=20 2; if (k <=3D=20 1) = return=20 s; return s.charAt(k - 1) +=20 prepare(s.substring(0, k - 1) + s.substring(k + 1, 2*k)) +=20 s.charAt(k); }

what doesprepare("LEMONADE") =return?

= ONMAEDLE=20
= OLEMADEN=20
= OMELEDAN=20
= OOOONNNN=20
= LEMONADE=20

81.

What are the values of a and = b=20 after the following code fragment is executed?

int a =3D 3, b =3D 5, s; for (int i =3D 0; i < = 10;=20 i++) { s =3D a +=20 b; b =3D a - = b; a=20 =3D s; }

0 and 96=20
96 and 0=20
96 and 96=20
96 and 160=20
160 and 96 =

82.

If int weekDay contains = the code=20 for the day of the week on November 1 (0 for Sunday, 1 for Monday, = ..., 6=20 for Saturday), which of the following expressions gives the date = for=20 Thanksgiving (the fourth Thursday in November)?

weekDay + = 26=20
26 - = weekDay=20
(4 - = weekDay) % 7 +=20 22=20
(11 - = weekDay) % 7 +=20 22=20
(weekDay + = 3) % 7 +=20 22

83.

The method average3 below "simultaneously" replaces = all=20 elements in an array. Each element is replaced with the average of = that=20 element's current value and its left and right neighbors. If a = "neighbor"=20 is outside the array, its value is assumed to be 0:

public void average3 (double=20 a[]) {      int i, len =3D=20 a.length;      double tempSaved[] =3D new = double[len];      for (i =3D 0; i < = len;=20 i++)          =20 tempSaved[i] =3D a[i];

     double leftNeighbor,=20 rightNeighbor;

     for (i =3D 0; i < len;=20 i++)     =20 {           if = (i >=20 = 0)            &= nbsp;  =20 leftNeighbor =3D=20 = tempSaved[i-1];          = =20 = else            = ;   =20 leftNeighbor =3D=20 0;           if = (i <=20 len -=20 = 1)            &= nbsp;  =20 rightNeighbor =3D=20 = tempSaved[i+1];          = =20 = else            = ;   =20 rightNeighbor =3D=20 0;           = a[i] =3D=20 (leftNeighbor =3D tempSaved[i] =3D rightNeighbor) /=20 3;      } }

Suppose the first and lest elements in v are = zeroes. Which=20 of the following statements is FALSE after = average3(v) is=20 called? (Disregard all inaccuracies that may be introduced due to=20 floating-point arithmetic.)

If v = holds the=20 values 0, 3, 6, 9, 12, 9, 6, 3, 0, the resulting array will hold = 1, 3,=20 6, 9, 10, 9, 6, 3, 1.=20
The sum of all the = elements in=20 v remain the same.=20
If in the original = array the sum=20 of all the values in even positions, v[0] + v[2] + = ..., is=20 the same as the sum of all the values in the odd positions, = v[1] +=20 v[3] + ..., then the same will remain true for the = resulting=20 array.=20
The maximum value = in the=20 resulting array does not exceed the maximum value in the = original array.=20
The position of = the maximum=20 element in the resulting array remains the same as in the = original array=20 or shifts to one of the two neighboring positions. =

84. (AB)

An array arr contains n = integer=20 elements whose values form an arithmetic sequence (i.e., =

arr[i+1] -=20
      arr[i] =3D=3D arr[i] - arr[i-1]

for any 0 <

i =

<=20 n-1). Whatr is the "big-O" for an optimal algorithm = that=20 determines whether such an array contains two given = values?

O(1)=20
O(log = n)=20
O(2 log = n)=20
O(n) =
= O(n²)=20

85.

Given

Random generator =3D new Random(); int bigNum =3D = 10000; int=20 r =3D generator.nextInt(bignum);

which of the following expressions is the best way to = initialize=20 x to the value of a randomly chosen element from an = array=20 arr of 3 values? (The odds for choosing any element = must be=20 the same or almost the same.)

x =3D arr[r = / bignum *=20 3];=20
x =3D = arr[(int)(3.0 * r /=20 bignum)];=20
x =3D = arr[(int)(2.9 * r /=20 bignum)];=20
x =3D = arr[(int)(3.0 * (r -=20 1) / (bignum -1)];=20
x =3D arr[3 = * (int)((double)=20 r / bigNum)];

86.

Given three integer variables, a,=20 b, and c, with small non-negative = values, which=20 of the following code fragments tests the condition that any two = of the=20 values are zeroes while the third one is positive? The variable=20 ok should be set to true if and only if the above = condition=20 is true.

I. boolean ok=20 =3D a = =3D=3D 0=20 && b =3D=3D 0 && c > 0=20 || b = =3D=3D 0=20 && c =3D=3D 0 && a > 0=20 || c = =3D=3D 0=20 && a =3D=3D 0 && b >=20 0;II. boolean ok =3D = a + b + c=20 > 0 && a*b + b*c + c*a =3D=3D=20 0;III. boolean ok =3D a = > 0 || b=20 > 0 || c > 0; if (ok) = ok =3D a +=20 b =3D=3D 0 || b + c =3D=3D 0 || c + a =3D=3D=20 0;

I only=20
II only=20
I and II=20
I and III=20
I, II, and III =

Questions 87-91 are based on the following = classes:

public class Person implements=20 Comparable {      private String=20 name;

     public Person(String name) { = this.name=20 =3D name; }      public String getName() = { return=20 name; }

     public boolean equals(Object=20 other)     =20 {           = return other=20 !=3D null && name.equals(((Person)=20 other).name);      }

     public int compareTo(Object=20 other)     =20 {           = return=20 name.compareTo((Person other).name);      = }

     public int hashCode() { return = name.hashCode(); } }

public class SoccerPlayer extends=20 person {      private int = numGoals;

     public SoccerPlayer(String = name, int=20 n)     =20 {          =20 = super(name);          =20 numGoals =3D n;      }

     public int getNumGoals() { = return=20 numGoals; }      public void score() [=20 numGoals++; }

     public int = compareTo(SoccerPlayer=20 other)     =20 = {            return = getNumGoals() - other.getNumGoals();      = }

     public String=20 toString()     =20 {           = return=20 gatName() + "/" + getNumGoals();     =20 } }

87.

Which of the following declarations is = invalid?

Person p =3D = new Person("Mia=20 Hamm");=20
SoccerPlayer = p =3D new=20 Person("Mia Hamm");=20
Comparable p = =3D new=20 Person("Mia Hamm");=20
Person p =3D = new=20 SoccerPlayer("Kristine Lilly", 0);=20
Comparable p = =3D new=20 SoccerPlayer("Kristine Lilly", 0);

88.

What is the result of the following code?

Person players[] =3D { new SoccerPlayer("Mia Hamm", 7), = new=20 SoccerPlayer("Kristine Lilly", 6)=20 }; System.out.println(players[0].compareTo((SoccerPlayer)=20 players[1])); // Line=20 ***

Syntax error in = the class=20 Person: other.name is not accessible=20
Syntax error in = the class=20 SoccerPlayer: compareTo is redefined=20
= ClassCaseException=20 on Line ***=20
Compiles with no = errors,=20 displays 1=20
Compiles with no = errors,=20 displays 2

89. (AB)

Suppose the Person and=20 SoccerPlayer classes are changed as follows:

other.name is replaced with = other.getame()=20 and compareTo in SoccerPlayer is renamed = into=20 compareGoals. What will be the result of running the=20 following code segment?

SoccerPlayer mia =3D new SoccerPlayer("Mia Hamm",=20 6); SoccerPlayer kristine =3D new SoccerPlayer("Kristine = Lilly",=20 5); Set team =3D new=20 = HashSet(); team.add(mia); team.add(kristine); kristine.score();<= BR>team.add(kristine); Iterator=20 iter =3D team.Iterator(); // Line=20 *** while(iter.hasNext()) =20 System.out.print(iter.next() + " = ");

Kristine = Lilly/5 Mia=20 Hamm/6=20
Kristine = Lilly/6 Mia=20 Hamm/6=20
Mia Hamm/6 = Kristine=20 Lilly/5 Kristine Lilly/6=20
Kristine = Lilly/5 Kristine=20 Lilly/6 Mia Hamm/6=20
Syntax error on = Line ***=20

Questions 90-91 are concerned with a class = SoccerTeam=20 that represents a team of soccer players:

public class = SoccerTeam {     =20 private SoccerPlayer[] players;      = private=20 ArrayList mvps;     //holds all the players = who=20 scored the same highest number of goals

     public void score(int=20 k)     =20 {          =20 players[k].score();     // Line=20 *1*           = int goals=20 =3D=20 = players[k].getNumGoals();        &n= bsp; =20 int maxGoals =3D ((SoccerPlayer)=20 mvps.get(0)).getNumGoals();     // Line=20 *2*           if = (goals=20 >=3D=20 = maxGoals)          =20 = {            &n= bsp;  =20 if (goals =3D=3D maxGoals)     // Line=20 = *3*            =         =20 = mvps.add(players[k]);         =       =20 = else            = ;   =20 = {            &n= bsp;       =20 // mvps is left with only one player in it,=20 = players[k]:          &nbs= p;         =20 < missing statements=20 = >            = ;   =20 }          =20 }      }

     < constructors and other = methods=20 not shown > }

90.

SoccerTeam's score method = is=20 intended to update the number of scored goals for a given player = on the=20 team and update the list of "most valuable players" (all of whom = have the=20 same score, the highest on the team). If the player's new score is = higher=20 than the old best, the mvps list is updated to = contain only=20 that one player. However, the score method has an = error.=20 Which of the following actions would correct that error?

I.      Move Line *2* before Line=20 *1*
II.     Replace Line *3*=20 with
        = if (goals=20 =3D=3D maxGoals && players[k] !=3D=20 mvps.get(0))
III.     Replace Line = *3*=20 with
         = if=20 (goals =3D=3D maxGoals && !=3D=20 mvps.contains(players[k]))

I only=20
II only=20
I and II=20
II and III=20
I, II, and III =

91.

Which of the following would be an appropriate = replacement=20 for < missing statements > in=20 SoccerTeam's score method?

mvps.set(0,=20 players[k]);=20
=20 mvps.reize(0); = mvps.add(players[k]);=20
mvps =3D new = ArrayList(); mvps.add(players[k]);=20
mvps =3D new = ArrayList(1); mvps.set(0, players[k]);=20
delete=20 mvps; mvps =3D new ArrayList(): =20 mvps.add(players[k]);

92. (AB)

The elements in an array of size n are first = increasing=20 (a[i] < a[i+1]), until they reach a maximum value, then = decreasing=20 (a[i] > a[i+1]). What are the respective "big-O" estimates for = the=20 number of comparisons in two optimal algorithms, one that finds a = maximum=20 value in such an array and the other that sorts the array in = ascending=20 order?

O(log=20 ²) and O(n)=20
O(log = n) and=20 O(n log n)=20
O(n) = and=20 O(n)=20
O(n) = and=20 O(n log n)=20
O(n) = and=20 O(n²)

93. (AB)

What is the output from the following code segment?

Set set =3D new TreeSet(); String str =3D=20 "A"; set.add(str); str +=3D "B"; set.add(str); str = +=3D=20 "C"; set.add(str); Iterator iter =3D = set.Iterator(); while=20 (iter.hasNext()) =20 System.out.print(iter.next() + = "-");

A-=20
ABC-=20
A-B-C-=20
A-AB-ABC-=20
None of the above =

94. (AB)

Suppse ListNode p1, p2 = initially=20 refer to two nodes in the same circular linked list. Under what = conditions=20 does the following loop terminate?

do { p1 =3D=20 p1.getNext(); p2 =3D=20 ps.getNext().getNext(); } while (p1 !=3D=20 p2);

Always=20
If and only if = p1 =3D=3D=20 p2.getNext()=20
If and only if the = total number=20 of nodes in the list is even=20
If and only if the = number of=20 nodes from p2 to p1 (excluding both = ends of=20 this segment of the list) is even=20
If and only if the = list contains=20 two nodes with the same info

95. (AB)

What does the following code display?

String expr =3D "(a + b) / (2 * (a - b)"; Stack stk = =3D new=20 ArrayStack(); int i, k;

for (k =3D 0; k < expr.length();=20 k++) { if (expr.charAt(k) =3D=3D=20 '(') =20 { = stk.push(new=20 Integer(k + 1)); =20 } else =20 { i = =3D=20 ((Integer)=20 = stk.pop()).intValue(); = ; =20 System.out.println(expr.substring(i, k)); =20 } }

(2 * (a -=20 b)) (a - b) (a + b)=20
(a - = b) (2=20 * (a - b)) (a + b)=20
a -=20 b a + b (2 * (a - = b))=20
a +=20 b a - b 2 * a - = b=20
a +=20 b) a - b) 2 * (a -=20 b))

96.

The following method packed analyzes a string passed = to it=20 and returns a new string:

String packed(String = msg) { =20 String packedMsg =3D "";

for (int i =3D 0; i < = msg.length();=20 i+) =20 { if=20 (msg.charAt(i) !-=20 '.') =20 = { &n= bsp; =20 int len =3D=20 = packedMsg.length(); &n= bsp; =20 if (len =3D=3D 0 || msg.charAt(i) !=3D=20 = packedMsg.charAt(len-1)) &nb= sp; =20 packedMsg +=3D msg.substring(i,=20 i+1); =20 } } = return=20 packedMsg; }

For which of the following values of msg, is= packed(msg) NOT equal to=20 packed("xxo.ooo.xx.x")?

xxo..ooo..xx..x=20
..xxooooooxxxx..=20
xxoooxxx=20
xxooooooxxox=20
xox =

97. (AB)

What is the value of sum after the = following=20 code is executed?

int sum =3D 0; int r =3D 0, c =3D 0; int temp, d1 = =3D 1; d2 =3D=20 2; int table[][] =3D new int[8][8];

for (int k =3D 0; k < 64;=20 k++) { table [r][c] =3D=20 1; r =3D (r + d1) %=20 8; c =3D (c + d2) %=20 8; temp =3D d1; d1 =3D d2; d2 =3D=20 temp; }

for (r =3D 0; r < 8; = r++) for (c =3D=20 0; c < 8;=20 c++) = sum +=3D=20 table[r][c];

8=20
15=20
16=20
22=20
64

Questions 98-99 assume that TreeNode root points = to the=20 following binary tree: =

&nbs= p; =20 = A &n= bsp; =20 /=20 = \ &n= bsp; =20 B =20 = C &n= bsp;=20 / /=20 = \ =20 D E =20 = F &n= bsp; =20 = \ &n= bsp; =20 G

98. (AB)

Consider the following method traverse:

void traverse(TreeNode = root) { =20 if (root !=3D null) =20 { =20 = traverse(root.getLeft()); &n= bsp; =20 = traverse(root.getRight()); &= nbsp; =20 = System.out.print(root.getValue()); = ; =20 = traverse(root(getRight()); &= nbsp; =20 traverse(root.getLeft()); =20 } }

How many letters will be displayed = when=20 traverse(root) is called?

1=20
7=20
13=20
14=20
25

99. (AB)

Consider the following method:

public void grow(TreeNode=20 root) { if (root !-=20 null) =20 { if=20 (root.getLeft() !=3D null && root.getRight !=3D=20 = null) &nbs= p; =20 root.setRight(new TreeNode("X", null,=20 = null)); else=20 if (root.getLeft() =3D=3D null && root.getRight !=3D=20 = null) &nbs= p; =20 root.setLeft("X", null,=20 = null)); =20 = grow(root.getLeft()); = =20 grow(root.getRight()); =20 } }

Which of the following represents the resulting tree after=20 grow(root) is called?

a. = A /=20 \ B =20 C / /=20 \ D E =20 = F =20 = \ &n= bsp;=20 = G &n= bsp; =20 = \ &n= bsp; =20 X b. =20 A /=20 \ / =20 \ B =20 C / \ /=20 \ D X E =20 = F =20 /=20 = \ X&= nbsp; =20 G c. &n= bsp; =20 = A &n= bsp;=20 / =20 = \ =20 / =20 = \ =20 B =20 C = / =20 \ / =20 \ =20 D X =20 E =20 = F / \ =20 / \ / \ /=20 \ X =20 X X X X X X = X d. =20 = A =20 / =20 = \ =20 / =20 \ =20 B =20 C =20 / \ / =20 \ = D =20 X E =20 = F / \ = =20 / \ /=20 \ X = X =20 X X X =20 = G &n= bsp; =20 /=20 = \ &n= bsp; =20 X X e. None=20 of the above

100.

An exam contains 24 questions for 40 minutes. Some = questions=20 are easy, while other questions are really hard. Constance and = Skip always=20 solve any easy question in 20 seconds, but a hard question always = takes=20 each of them 3 minutes. Constance's strategy is to take each = question in=20 turn and take whatever time it takes to solve it. Skip tries a = question=20 for 20 seconds and if he can't solve it, he moves on to the next = one. If=20 he has looked at all the questions, Skip returns to the unsolved = hard=20 questions, but he has to start solving them from scratch because = by then=20 he has forgotten what they were about. If the first half of the = test=20 contains at least 8 easy questions and the second half contains at = least 8=20 hard questions, which of the following statements is = FALSE?

Skip will solve = at least 18=20 questions=20
Constance will = solve at least=20 20 questions=20
If at most 10 of = the questions=20 are hard, both Skip and Constance will solve all 24 questions=20
Skip will solve = at most 10 hard=20 questions=20
Constance will = always solve at=20 least as many questions as Skip

2001

101.

What is the output of the following code?

int a =3D 1, b =3D 2, c =3D 3; a +=3D b + c; b = +=3D a + c; c +=3D=20 a + b; System.out.println(a + " " + b + " " +=20 c);

3 3 = 4=20
3 5 = 6=20
5 4 = 3=20
5 8 = 13=20
6 11 = 20=20

102.

Consider two methods:

public int f(int x) { = return x +=20 2; }

and

public int g(int x) { = return x *=20 2; }

What is the value of x after the following code=20 executes?

int x =3D 1; x +=3D f(g(x)) - g(f(x));=20

-9=20
-2=20
-1=20
3=20
7

103.

What is the value of y after the = following code=20 is executed?

int x =3D 123; while (x >=20 0) { y *=3D = 10; =20 y +=3D x % 10; x /=3D=20 10; }

1=20
3=20
6=20
12=20
321 =

104. (AB)

The method xWon below is supposed to = return=20 true is a tic-tac-toe board (represented by a 3-by-3 = array of=20 characters) has three x's in any line, = false=20 otherwise:

public boolean xWon (char[][]=20 b) { if (b[1][1] =3D=3D=20 'x') =20 { if = (b[0][0]=20 =3D=3D 'x' && b[2][2] =3D=3D 'x') return=20 true; = if=20 (b[0][1] =3D=3D 'x' && b[2][1] =3D=3D 'x') return=20 true; = if=20 (b[1][0] =3D=3D 'x' && b[1][2] =3D=3D 'x') return=20 true; = if=20 (b[0][2] =3D=3D 'x' && b[2][0] =3D=3D 'x') return=20 true; } = else if=20 (b[0][0] =3D=3D 'x') =20 { if = (b[0][1]=20 =3D=3D 'x' && b[0][2] =3D=3D 'x') return=20 true; = if=20 (b[1][0] =3D=3D 'x' && b[2][0] =3D=3D 'x') return=20 true; } = else if=20 (b[2][2] =3D=3D 'x') =20 { if = (b[2][0]=20 =3D=3D 'x' && b[2][1] =3D=3D 'x') return=20 true; = if=20 (b[0][2] =3D=3D 'x' && b[1][2] =3D=3D 'x') return=20 true; } = return=20 false; }

a. x.. = =20 xxo = xoo b. o.x = =20 ox. = xox c. xxo = =20 o.o = oxx d. x.o = =20 oxo = x.x e. oox = =20 o.. =20 xxx

105.

Consider the following method:

public void divide5 (int[] a, int[] q, int[]=20 r) { q[0] =3D a[0] /=20 5; r[0] =3D a[0] % 5; }=20

What is the value of y[0] after the following = statements=20 are executed?

int[] x =3D {21, 22, 23], y =3D new = int[3]; divide5(x, y, z);=20

0=20
1=20
2=20
4=20
21 =

106. (AB)

Suppose the method fun is defined as = follows:

public void fun (int m[][], String=20 s) { for (int i =3D 1; i < = s.length();=20 i++) =20 { int = r =3D=20 Character.digit(s.charAt(i),=20 10); = int c =3D=20 Character.digit(s.charAt(i - 1),=20 10); =20 m[r][c]++; } } =

What is the output when the following code fragment is = executed?

int m[][] =3D new int[10][10]; String s =3D=20 "20012002";

fun(m, s); int sum =3D 0; for (int k =3D 0; k = < 10;=20 k++) sum +=3D=20 m[k][k]; System.out.println(sum); =

1=20
2=20
4=20
7=20
8

107.

Which of the following conditions is always true = after the=20 while loop in the following code fragment has = finished=20 (assuming it executes without errors)?

int k =3D 0; n =3D 10; while (k < n && = a[k] >=3D=20 0) k++; =

k >=3D n = && a[k]=20 < 0=20
k < n = && a[k]=20 < 0=20
k < n || = a[k] <=20 0=20
k >=3D n = || a[k] <=20 0=20
None of the above =

108.

The method

public boolean xyz (String = s) { =20 return s.length() >=3D 3=20 = && =20 ((s.charAt(0) =3D=3D s.charAt(1)=20 = && = ; =20 s.charAt(1) =3D=3D s.charAt(2)) || xyz(s.substring(1))); }=20

returns true if and only if

s = contains three=20 or more of the same characters in a row=20
s = starts with=20 three or more of the same characters=20
s = ends with three=20 or more of the same characters=20
s = contains three=20 or more of the same characters=20
s[0] = is the same=20 as two other characters in s

109.

The following version of Selection Sort is supposed = to sort=20 an array in ascending order. For better performance it tries to = tackle the=20 array from both ends simultaneously:

public void sort (int = a[]) { int=20 left =3D 0, right =3D a.length() - = 1; int=20 k;

while (left <=20 right) =20 { for = (k =3D=20 left + 1; k < right;=20 k+) =20 = { &n= bsp; =20 if (a[k] <=20 = a[left]) &= nbsp; =20 swap(a, k,=20 = left); &nb= sp; =20 else if (a[k] >=20 = a[right]) = =20 swap(a, k,=20 = right); =20 } =20 = left++; =20 right--; } } =

swap(a, i, j) correctly swaps a[i] =and=20 a[j]. This code has a bug, though. Which one of the = following=20 changes would assure that the method sorts the array = correctly?

I.      Remove else=20 in
         = else if=20 (a[k] > a[right]) = ...II.    =20 = Replace
         =20 for (k =3D left + 1; k < right;=20 k+)        =20 = with
          for = (k =3D left; k <=3D right; = k+)III.    =20 Add
        =20       if (a[left] >=20 = a[right])           =      =20 swap(a, left,=20 = right);         at=20 the beginning of the while loop (before the=20 for loop).

I only=20
II only=20
III only=20
I or II=20
II or III =

110. (AB)

Insertion Sort is a sorting algorithm that works as = follows:=20 keep the first k elements of the array sorted; find the right = place and=20 insert the next element among the first k. These steps are = repeated for k=20 - 1, ..., n. Sequential Search is usually used to find the right = spot in=20 which to insert the next element. Suppose we use Binary Search = instead of=20 Sequential Search in this algorithm. How would Big-O for the = number of=20 comparisons among the elements (not counting the number of moves = or swaps)=20 change?

No change=20
From = O(n) to=20 O(log n)=20
From=20 O(n²) to = O(n²/2)=20
From=20 O(n²) to O(n log n)=20
From=20 O(n²) to O(n) =

Questions 111-112 are based on the following = class that=20 represents a moving ball on a rectangular pool table:

public class = MovingBall {     =20 private int mLength, mWidth;      private = int=20 mPosX, mPosY;      private int mDirX,=20 mDirY;

     public MovingBall(int length, = int=20 width, int dx, int dy)     =20 {           = mLength =3D=20 = length;           mWidth = =3D = width;           mPosX=20 =3D length / = 2;          =20 mPosY =3D width /=20 2;           = mDirX =3D=20 dx;           = mDirY =3D=20 dy;      }

     public void=20 move()     =20 {           = mPosX +=3D=20 mDirX;           = mPosY=20 +=3D = mDirY;           if=20 (mPosX =3D=3D 0 || mPosX =3D=3D mLength) mDirX =3D=20 = -mDirX;           id=20 (mPosY =3D=3D 0 || mPosY =3D=3D mWidth) mDirY =3D=20 -mDirY;      = } }

111.

Giiven

MovingBall b =3D new MovingBall(8, 4, 1,=20 -1);

what are the values of mPosX and = mPosY after=20 70 moves (i.e., 70 calls to b.move())?

74 and -68=20
6 and 4=20
4 and 2=20
3 and 1=20
1 and -1 =

112.

MovingBall b =3D new MovingBall(9, 4, 1,=20 1);

is defined, how many moves (i.e., calls to = b.move()) will=20 ball b hot position mPosX =3D 6, mPosY =3D=20 1?

Never=20
2=20
3=20
7=20
30 =

113.

Suppose all the elements in an array have different = values.=20 Let us say the array is "nearly sorted" (in ascending order) if = each=20 element's position differs from its appropriate position in the = sorted=20 arrangement of the same array by at most 2 (in either direction). = The=20 following method takes an array where the first n elements = are=20 "nearly sorted" and properly sorts the array.

public void sortNearlySorted(int[] arr, int=20 n) { int i =3D 0;

while (i < n =3D=20 1) =20 { if = (arr[i+1]=20 <=20 = arr[i]) &n= bsp; =20 swap(arr, i, i +=20 1); if = (i+2=20 < n && arr[i+2] <=20 = arr[i]) &n= bsp; =20 swap(arr, i,=20 i+2); =20 i++; } }

Which of the following is a loop invariant for the = while=20 loop in the above method?

arr[0] ... = arr[n-1] are sorted=20 and 0 <=3D i < n=20
arr[0] ... = arr[n-1] are nearly=20 sorted and 0< n-1=20
arr[0] ... = arr[i-1] are placed=20 where they belong in the sorted array and arr[i] ... arr[n-1] = are=20 "nearly sorted"=20
arr[i] < = arr[i+1] and arr[i]=20 < arr[i+2]=20
arr[0] ... = arr[n-3] are sorted=20

114.

In the ABBAB language the alphabet has only two = letters. A=20 string of letters (including and one-letter strings) is a valid = word, if=20 and only if the isValid method returns = true for=20 that string. isValid is defined as follows:

public boolean isValid(String=20 word) { int n =3D = word.length();

return n < 1=20 || =20 (isValid(word.substring(0, n-1) &&=20 = &nb= sp; =20 word.charAt(n-1) =3D=3D 'B')=20 || =20 (isValid(word.substring(0, n-2)=20 = && = ; =20 "BA".equals(word.substring(n-2))); }

How many valid words of length 7 are there in the ABBABA=20 language?

2=20
3=20
15=20
23=20
34 =

115. (AB)

The method below takes a linked list pointed to by=20 head, removes the first node, appends it at the end = of the=20 list, and returns a reference to the head of the new list.

public ListNode firstToLast (ListNode=20 head) {      if (head =3D=3D null) ||=20 head.getNext() =3D=3D=20 null)           = return=20 head;

     ListNode p =3D=20 head;      while (p.getNext() !=3D=20 null)           = p =3D=20 p.getNext();

     < missing statements = >

     return head; }=20

Which of the following code fragments correctly completes this=20 method?

I. = ListNode=20 temp =3D head; = head =3D=20 head.getNext(); =20 p.setNext(temp); =20 temp.setNext(null);II. =20 ListNode temp =3D=20 head.getNext(); =20 = head.setNext(null); =20 p.setNext(head); = head =3D=20 temp;III. = =20 p.setNext(head); = head =3D=20 head.getNext(); =20 =20 p.getNext().setNext(null);

I only=20
II only=20
I and II=20
II and III=20
I, II, and III =

116. (AB)

The following method goodHand checks = whether s=20 stack of Integers (that represent ranks of playing = cards in a=20 small deck) has a certain property:

public boolean goodHand (Stack=20 hand) { int[] count =3D new=20 int[2]; Object x, y;

for (int k =3D 0; k <=3D 1; = k++) =20 { if=20 = (hand.isEmpty()) = ; =20 return = false; =20 x =3D=20 = hand.pop(); =20 Stack stk =3D new=20 = ArrayStack(); =20 while=20 = (!hand.isEmpty()) &nbs= p;=20 = { &n= bsp; =20 y =3D=20 = hand.pop(); &nbs= p; =20 if=20 = (x.equals(y)) &n= bsp; =20 = count[k]++; &nbs= p; =20 = else = ; =20 = stk.push(y); =20 } hand = =3D=20 stk; } = return=20 Math.abs(count[0] - count[1]) <=3D 1 && = hand.isEmpty(); }=20

For which of the following stacks does goodHand = return=20 true?

top = =3D=3D>

a. 2 &n= bsp; =20 3 =20 4 =20 5 = 6 b. 2 &n= bsp; =20 2 =20 3 =20 3 = 4 c. 2 &n= bsp; =20 3 =20 2 =20 2 = 3 d. 3 &n= bsp; =20 3 =20 3 =20 3 = 2 e. 2 &n= bsp; =20 2 =20 2 =20 2 =20 2

117. (AB)

An n by n square image contains black = and=20 white pixels. The first several columns contain one or more black = pixels=20 at the top with only white pixels below; the remaining columns are = all=20 white. For example:

=
xxxx.. xxxx.. x.xX.. x.x... ..x... ......<= /P>

A program is allowed to examine individual pixels in the image; = its=20 task is to find the position of the lowest black pixel in the = rightmost=20 column that has at least one black pixel (the uppercase pixel in = the above=20 example). What is the worst case big-O for the number of examined = pixels=20 in the best possible algorithm.

O(log n)=20
O((log=20 n)²)=20
= O(n)=20
O(n log = n)=20
= O(n²)=20

Questions 118-122 use the following classes:

public class Point {      = private int=20 x;      private int y;

     public Point(int x, int y) { = this.x =3D=20 x; this.y =3D y; }

     public int getX() { return x;=20 }      public int getY() { return y; = }

     public boolean equals(Point=20 other)     =20 {           = return=20 getX() =3D=3D other.getX() && getY() =3D=3D=20 other.getY();      }

     public int=20 hashCode()     =20 {           = return (new=20 Integer(getX() + = getY())).getHashCode();     =20 }

     public void setX(int x) { = this.x =3D x;=20 }      public void setY(int Y) { this.y = =3D y;=20 }

     public String=20 toString()     =20 {           = return "(" +=20 getX() + ", " + getY() + ")";     =20 } }

public class MovingPoint extends=20 Point {      private Point = myPoint;

     public MovingPoint(Point=20 p)     =20 {           < = missing statements=20 >           = myPoint =3D=20 p;      }

     public int getX() { return=20 myPoint.getX(); }      public int getY() = { return=20 myPoint.getY(); }

     public void move(int x, int=20 y)     =20 {          =20 = myPoint(setX(x));         &nbs= p;=20 myPoint(setY(y));      }

     public String=20 toString()     =20 {           = return=20 myPoint.toString();     =20 } }

118.

Which of the following can replace < missing=20 statements > in MovingPoint's = constructor?

= super();=20
super(0, = 0);=20
setX(0); = setY(0);=20
super(); = setX(p.getX());=20 seyY(p.getY());=20
// set = myPoint to=20 p:

119.

Which line in the following code segment causes = a =20 syntax error?

Point p1 =3D new Point(100,=20 100); // = Line=20 1 MovingPoint p2 =3D new MovingPoint(p1); // Line = 2 Point p3 =3D=20 new MovingPoint(p2); = // Line=20 3 MovingPoint p4 =3D new MovingPoint(p2); // Line 4=20

No syntax errors = on these lines=20
Line 1=20
Line 2=20
Line 3=20
Line 4 =

120.

What is the output of the following code?

Point p =3D new Point(0, 0); MovingPoint mp =3D new=20 MovingPoint(p); mp.move(1, 1); System.out.println(p + " " = +=20 p.equals(mp));

=20 NullPointerException=20
= ClassCastException=20
(0, 0) = true=20
(0, 0) = false=20
(1, 1) = true=20

121. (AB)

Consider the following two code segments: <= /TR>

Set points =3D new HashSet(); MovingPoint = p;

p =3D new MovingPoint(new Point(0,=20 0)); points.add(p);

p =3D new MovingPoint(new Point(0,=20 2)); points.add(p);

p =3D new MovingPoint(new Point(2,=20 2)); points.add(p);

p =3D new MovingPoint(new Point(2,=20 0)); points.add(p);

p =3D new MovingPoint(new Point(1,=20 1)); points.add(p);
=
System.out.println(points.size());

Set points =3D new HashSet(); MovingPoint = p;

p =3D new MovingPoint(new Point(0,=20 0)); points.add(p);

p.move(0, 2); points.add(p);

p.move(2, 2); points.add(p);

p.move(2, 0); points.add(p);

p.move(1, 1); points.add(p);
=
System.out.println(points.size());

What are their respective outputs?

3 = and=20 1=20
3 = and=20 3=20
5 = and 1=20
5 = and=20 3=20
5 = and=20 5

122. (AB)

What is the output of the following code segment?

Point p =3D new Point(0, 0); MovingPoint p1 =3D new=20 MovingPoint(p); movingPoint p2 =3D new = MovingPoint(p); Queue q =3D new=20 ListQueue(); p1.move(1, 0); q.enqueue(p1); p2.move(0,=20 1); q.enqueue(p2); System.out.println(q.dequeue() + " " +=20 q.dequeue());

(0, 0) (0, = 0)=20
(1, 0) (1, = 0)=20
(0, 1) (0, = 1)=20
(1, 0) (0, = 1)=20
(0, 1) (1, = 0)=20

123. (AB)

Consider the following method:

public int magic (TreeNode=20 root) { if (root !=3D=20 null) =20 { if=20 (root.getLeft() =3D=3D null && root.getRight() =3D=3D=20 = null) &nbs= p; =20 return = 0; if=20 (magic(root.getLeft()) + magic(root.getRight()) =3D=3D=20 0) =20 = { &n= bsp; =20 TreeNode temp =3D=20 = root.getLeft(); = =20 = root.setLeft(root.getRight()); &nb= sp; =20 = root.setRight(temp); &= nbsp;=20 } } = return=20 -1; }

What does this method do to a = tree=20 referred to by root?

Nothing, leaves = the tree=20 unchanged=20
Swaps the left = and right=20 branches of the tree at the root=20
Replaces the tree = with its=20 mirror image=20
Swaps any two = leaves that have=20 the same parent=20
Swaps the = leftmost and=20 rightmost leaves

124. (AB)

Suppose traversePreOrder and=20 traversePostOrder are defined as follows:

public void traversePreOrder(TreeNode root, Stack=20 s) { if (root !=3D=20 null) =20 { =20 = s.push()(root.getValue()); &= nbsp; =20 traversePreOrder(root.getLeft(),=20 s); =20 traversePreOrder(root.getRight(), = s); } }=20

public void traversePostOrder(TreeNode root, Stack=20 s) { if (root !=3D=20 null) =20 { =20 traversePostOrder(root.getLeft(),=20 s); =20 traversePostOrder(root.getRight(),=20 s); =20 root.setValue(s.pop()); } }=20

If root initially refers to

=
&nbs= p; =20 = A &n= bsp; =20 /=20 = \ &n= bsp; =20 / =20 = \ &n= bsp; =20 B =20 = C &n= bsp;=20 / \ /=20 = \ =20 D = E F G

what is=20 the resulting tree after the following statements are executed?

Stack s =3D new ArrayStack = (); traversePreOrder(root,=20 s); traversePostOrder(root, = s);

a. `= A /=20 \ / = \ =20 B C = / \ /=20 \ D = E F G`	b. `= A /=20 \ / = \ =20 C B = / \ /=20 \ G = F E D`	c. `= G /=20 \ / = \ =20 D F = / \ /=20 \ A = B E C`
d. `= A /=20 \ / = \ =20 C B = / \ /=20 \ F = G D E`	e. `= G /=20 \ / = \ =20 F D = / \ /=20 \ E = C B A`

125.

A multiple-choice question offers three options, I, = II, and=20 III, and asks which ones fit into a given situation. The offered = answers=20 are a) I only; b) II only; c) III only; d) I and II; and e) II and = III.=20 Assuming that it takes the same amount of time to examine any one = of the=20 three options, that the odds that any option fits are 50-50, and = that s=20 student always gets all answers right and doesn't waste time = considering=20 unnecessary options, which option should she consider = first?

Doesn't matter=20
Either I or II=20
Either I or III=20
Definitely I=20
Definitely II =

2000

126.

Which of the following statements does NOT display=20 2/3?

=20 System.out.println("2/3");=20
= System.out.println("2" +=20 "/" + "3");=20
=20 System.out.println(2/3);=20
= System.out.println(2 +=20 "/" + 3);=20
= System.out.println((int)2=20 + "/" + (int)3);

127.

If array arr has five elements with = values=20 0 1 2 3 4, what are the values in arr = after the=20 following code is executed?

for (int k =3D 0; k < 5;=20 k++) arr[k] =3D=20 arr[arr[k]];

0 1 2 3 = 4=20
1 2 3 4 = 0=20
1 2 2 2 = 2=20
0 0 0 0 = 0=20
1 2 3 4 = 4=20

128.

If c and d are=20 boolean variables, which of the answer choices is NOT = equivalent to the following expression?

(c && d) !=3D (c || = d)

(c = && !d) || (!c=20 && d)=20
(c || d) = && (!c=20 && !d)=20
(c || d) = && (!c=20 || !d)=20
(c || d) = && !(c=20 && d)=20
c !=3D = d

129.

The value of (1 + (1/n))n for large enough n (e.g., = n >=3D=20 50) approximates the base of the natural logarithm e =3D = 2.71828... A=20 student decided to test this property and wrote the following = method:

public double = approxE() { double=20 e =3D 1.0 + 1.0 / 64; for (int count = =3D 1;=20 count <=3D 6;=20 = count++) e = *=3D=20 e; return = e; }

What value is returned by approxE()? =

No value is = returned: the=20 method throws an ArithmeticException=20
1.0=20
2.6973449525651=20
1.642359568597906 =
7.275669793128421 =

130.

What is the output from the following code?

int[] counts =3D new int[3]; int i, j; for (int i = =3D 0; i=20 < 100; i++) for (j =3D 0; j < = 10;=20 j++) = counts[j=20 % 3]++; System.out.println((counts[1] + counts[2]) /=20 counts[0]);

1=20
1.5=20
2=20
2.5=20
3

131.

Suppose an array A has n=20 elements. Let's call it periodic with a period of = p if=20 0 < p < n and A[i] = =3D=3D=20 A[i+p] for all 0 <=3D i <

 =
n-p

=20 and p is the smallest such number. What is the period = of=20 array v after the following code is executed?

int v[] =3D new=20 int[100]; v[0] =3D 0; v[1] =3D = 1;

for (int i =3D 2; i < 100;=20 i++) = v[i] =3D=20 v[i-1] - v[i-2];

2=20
3=20
4=20
6=20
No period =

132.

What is the output from the following code?

StringBuffer s =3D new=20 StringBuffer("WYOMING"); int k, i, n = =3D=20 s.length(); char temp;

for (k =3D 0; k < 3;=20 k++) =20 { temp = =3D=20 = s.charAt(n-1); = for (i =3D k+1; i < n;=20 = i++) = ; =20 s.setCharAt(i,=20 = s.charAt(i-1)); = =20 s.setCharAt(k, temp); =20 } =20 System.out.println(s);

= YOMINGW=20
= MINGWYO=20
= GWWWWWW=20
= MINGOYW=20
= GNIMOYW=20

133.

A recursive method upNdown is defined = as=20 follows:

public void upNdown(int = n) { if=20 (n > 1) =20 { if = (n % 2 !=3D=20 0)=20 = upNdown(n+1); =20 else=20 = upNdown(n/2); =20 System.out.println("*"); =20 } }

How many stars are displayed when upNdown(5) is=20 called?

1=20
2=20
3=20
4=20
5

134.

Given

String a =3D "a", b =3D "b", zero =3D ""; Integer c = =3D new=20 Integer(0);

what is the output of

System.out.println( =20 ((a+b)+c).equals(a+(b+c)) + " " + (c = +=20 zero).equals(zero + c) + " " + (a +=20 null).equals(a + "null"));

false false = false=20
false true = true=20
true false = true
true true = false=20
true true = true=20

135. (AB)

The method below uses a stack to check whether = parentheses=20 and brackets match in a string of characters:

public boolean parensAndBracketsMatch(String=20 expr) { Stack s =3D new=20 ArrayStack(); char ch0, = ch;

for (int k =3D 0; k < = expr.length();=20 k++) =20 { ch = =3D=20 = expr.charAt(k); = =20 if (ch =3D=3D '(' || ch =3D=3D=20 = '[') = ; =20 s.push(new=20 = Character(ch)); = =20 else if (ch =3D=3D ')' || ch =3D=3D=20 ']') =20 = { &n= bsp; =20 if=20 = (s.isEmpty()) &n= bsp; =20 return=20 = false; &nb= sp; =20 ch0 =3D=20 = ((Character)s.pop().charValue(); &= nbsp; =20 if ((ch0 =3D=3D '(' && ch !=3D ')')=20 = || &= nbsp; =20 (ch =3D=3D '[' && ch !=3D=20 = ']')) &nbs= p; =20 return = false; =20 } } = return=20 true; }

However, it has a bug. For which of the following strings does = this=20 method return a result that is DIFFERENT from what is = expected?

= ; = &= nbsp; &n= bsp; =20 Expected result:

"[(a+b) *=20 = (c+d)]/2" &nbs= p; =20 true=20
"[(a+b) *=20 = (c+d)/2]" &nbs= p; =20 true=20
"[(a+b) *=20 = (c+d)]/2[" &nb= sp; =20 false=20
"](a+b) *=20 = (c+d)/2[" &nbs= p; =20 false=20
"[(a+b] *=20 = [c+d)/2" = ; =20 false

136.

Consider the following two versions of the method=20 mixUp:

public void mixUp(int[] x, int[]=20 y) { x[0] =3D x[0] - 2 * = x[0] *=20 y[1]; y[1] =3D y[1] - 2 * x[0] = *=20 y[0]; }

public void mixUp(int[] x, int[]=20 y) { int d =3D 2 * x[0] *=20 y[1]; x[0] -=3D=20 d; y[1] -=3D=20 = d; }

Suppose the=20 following arrays are declared and initialized:

int[] a =3D {1, 1}, b =3D {0, 0}, c =3D {1, 1};=20

Which of the following calls to mixUp result in = the same=20 values in a, b, and c for = bother=20 versions of the code?

I. mixUp(a,=20 a)II. mixUp(a,=20 b)III. mixUp(a,=20 c)

I only=20
II only=20
I and II=20
II and III=20
I, II, and III =

137.

The Binary Search algorithm normally finds a value = in an=20 array sorted in ascending order. Suppose that by mistake the = algorithm is=20 used on an unsorted array with the following seven elements:

1 3 2 5 13 8 21

Which of the following target values will NOT be found? =

1=20
3=20
5=20
13=20
21 =

138.

An array has 4095 =3D 2¹²-1 elements, = arranged in=20 ascending order. Binary Search is used to find the position of a = target=20 value. This Binary Search is implemented iteratively, in such a = way that=20 in each iteration the target is compared to one of = the=20 elements of the array. Suppose we know that the = target is=20 somewhere in the array. What number of iterations guarantees that = a target=20 value is found.

10=20
11=20
12=20
2047=20
4095 =

139.

Which of the following arithmetic expressions maps = the=20 values x =3D 0, 1, 2, 3, 4, 5, 6 onto

y =

=3D 4, 3, 9,=20 8, 7, 6, 5, respectively?

y =3D 11 - = x + (x + 4) %=20 7;=20
y =3D (4 - = x) % 7 + 2 *=20 x;=20
y =3D 3 + = (8 - x) %=20 7;=20
y =3D 9 - = (x - 2) %=20 7;=20
y =3D 4 + x = % 7 - 2 *=20 x;

140.

The method below implements a simplified square = cipher:

public char[][] encrypt(char[][] key, String=20 msg) {      int i, j, n =3D = key.length, k =3D=20 0;      char[][] result =3D new=20 char[n][n];   // fills with spaces

     for (i =3D 0; i < n;=20 i++)           = for (j =3D=20 0; j < n;=20 = j++)            = ;   =20 if (key[i][j] =3D=3D=20 = 'x')            = ;        =20 result[i][j] =3D msg.charAt(k++);

     for (i =3D 0; i < n;=20 i++)           = for (j =3D=20 0; j < n;=20 = j++)            = ;   =20 if (key[i][j] =3D=3D=20 = 'x')            = ;        =20 result[n-i-1][n-j-1] =3D msg.charAt(k++);

     return=20 result; }

If key is a 4 by 4 matrix

.x.. x..x .xx. xx.x

and msg is "ransom received ", which = of the=20 following matrices is returned from encrypt(key,=20 msg)?

a. rean = ; =20 csoe =20 mivr d = e b. =20 rde =20 avin =20 esoc m = er c. =20 rvc =20 aein =20 dsoe m = er d. rdsr = ; =20 no =20 aeve = iemc e. rcan = ; =20 esoi =20 mvre d=20 ee

Questions 141-142 use the following class:

public class Circle {      = private=20 int xCenter, yCenter, radius;

     public Circle(int x, int y, = int=20 r)     =20 {           = xCenter =3D=20 x;           = yCenter =3D=20 y;           = radius =3D=20 r;      }

     public void moveTo(int x, int=20 y)     =20 {           = xCenter =3D=20 x;           = yCenter =3D=20 y;      }

     // Draw this circle in the = graphics=20 context g      public void draw(Graphics = g)=20 {     < code not shown=20 >     = } }

141.

Suppose the following code is added to a method that = repaints a window within a graphics context g:

for (int x =3D 10; x <=3D 30; x +=3D=20 10) { Circle circle =3D new = Circle(x + 100,=20 100, x); =20 circle.draw(g); }

The origin of the coordinate system in in the upper left corner = of the=20 window with the y-axis pointing down. Which of the following = pictures will=20 be displayed?

a.	b.	c.
d.	e.

142.

The method drawOrnament draws several circles, as = follows:

public void drawOrnament(Graphics g, int=20 k) { if (k <=20 8) =20 return; Circle c =3D new Circle(100 = - k, 100 -=20 k, k); =20 c.draw(g); c.moveTo(100 + k, 100 +=20 k); = c.draw(g); =20 drawOrnament(g, k/2); }

Which of the following pictures is produced by = drawOrnament(g,=20 32)?

a.	b.	c.
d.	e.

143.

What is displayed by

=
System.out.println(expand(expand("1001")));
where=20 expand is defined as follows?

public String expand(String = s) { =20 String d =3D "";

for (int i =3D 0; i < = s.length();=20 i++) = switch=20 = (s.charAt(i)) =20 = { &n= bsp; =20 case '0': d +=3D "01";=20 = break; &nb= sp; =20 case '1': d +=3D "10";=20 break; = } return = d; }

1001 =
= 1*0*0*1=20
= 10010110=20
= 10*01*01*10*=20
= 1001011001101001=20

Questions 144-146 use the following interface and = class:

public interface = Toggleable { =20 void toggle(); boolean=20 isOn(); }

public class toggleSwitch = implements=20 Toggleable =20 { = private=20 boolean on;

= public=20 ToggleSwitch() { on =3D false;=20 } public = ToggleSwitch(boolean state) { on =3D state; }

= public=20 void toggle() { on !=3D on;=20 } public = boolean=20 isOn() { return on; } = }

144.

Suppose we have found the=20 ToggleSwitch.class file but have no access to its = source=20 code. Which of the following code segments, if it compiles with no = errors,=20 will convince us that ToggleSwitch implements=20 Toggleable?

I. Toggleable x =3D new=20 ToggleSwitch();II. = ToggleSwitch x=20 =3D new ToggleSwitch(); =20 x.toggle();III. ToggleSwitch x = =3D new=20 ToggleSwitch(); if (!x.isOn())=20 x.toggle();

I only=20
II only=20
III only=20
II and III=20
I, II, and III =

145.

Consider the following class that represents a set = of=20 checkboxes

public class = CheckBoxSet {     =20 private Togleable[] buttons;

     public CheckBoxSet(int=20 nButtons)     =20 {           < = missing statements >      = }

     public int numButtons() { = return=20 buttons.length; }      public void = push(int k) {=20 buttons[k].toggle(); }      public = boolean=20 isOn(int k) { return buttons[k].isOn(); }

     public void=20 clear()     =20 {           for = (int k =3D=20 0; k < buttons.length;=20 = k++)            = ;   =20 if=20 = (buttons[k].isOn())         &n= bsp;          =20 buttons[k].toggle();     =20 } }

Which of the following can replace < missing = statements>=20 in the CheckBoxSet constructor?

I. buttons =3D new=20 Toggleable[nButtons]; for (int k =3D 0; k = <=20 buttons.length; = k++) =20 buttons[k] =3D new = ToggleSwitch();II. =20 buttons =3D new ToggleSwitch[nButtons]; =20 clear();III. buttons =3D new=20 ToggleSwitch[nButtons]; for (int i =3D 0; = i <=20 buttons.length; = i++) =20 if=20 = (buttons[i].isOn()) &n= bsp; =20 buttons[i].toggle();

I only=20
II only=20
I and II=20
II and III=20
I, II, and III =

146. (AB)

A set of "radio buttons" is a user interface control = used to=20 choose one of several options. For example:=20

small medium =20 large

Consider the = following class=20 RadioSet that implements a set of radio buttons:

public class RadioSet extends=20 CheckBoxSet {      // = precondition: =20 nButtons > k >=3D 0      // = postcondition:=20 creates an array of nButtons and     =20 = //            = ;   =20 sets the k-th button to "on"      public=20 RadioSet(int nButtons, intk)     =20 {          =20 = super(nButtons);          = ;=20 push(k);      }

     // postcondition: sets the = k-th button=20 to "on", all other     =20 = //            = ;   =20 buttons to "off"      public void = push(int=20 k)     =20 {           < = missing code >      = }

     // postcondition: returns the = number=20 of the button that is "on";     =20 = //            = ;   =20 throws an exception if none are "on"      = {           for = (int k =3D=20 0; k < numButtons();=20 = k++)            = ;   =20 if=20 = (isOn(k))           =          =20 return = k;          =20 throw IllegalStateException();     =20 } }

Which of the following can replace < missing code> = in this=20 class?

I. = clear(); =20 = push(k);II. clear(); = ;=20 super.push(k);III. for (int j = =3D 0; j=20 < numButtons();=20 j++) if=20 = (buttons[j].isOn()) &n= bsp; =20 buttons[j].toggle(); =20 buttons[k].toggle();

I only=20
II only=20
I and II=20
II and III=20
I, II, and III =

147. (AB)

char[][] m holds the following picture:

=
........ ..xxxxx. ..xx.... ..xxx... ....xxx. ....= ..x. .x.xxx.. ..xxx... ........=20

What picture is stored in m = after=20 process(m) is called? The method process = is=20 defined as follows:

public void process(char[][]=20 m) {      int rows =3D m.length, cols = =3D=20 m[0].length, r, c;      char[][] t =3D = new=20 char[rows][cols];

     for (r =3D 0; r < rows;=20 r++)           = for (c =3D=20 0; c < cols;=20 = c++)            = ;   =20 t[r][c] =3D m[r][c];

     for (r =3D 1; r < rows;=20 r++)           = for (c =3D=20 1; c < cols;=20 = c++)            = ;   =20 if (t[r][c] =3D=3D 'x' && (t[r-1][c] !=3D=20 = 'x'            =             &= nbsp;           &n= bsp; =20 || t[r][c-1] !=3D=20 = 'x'))           &nbs= p;        =20 m[r][c] =3D=20 = 'x';            = ;   =20 = else            = ;        =20 m[r][c] =3D '.'; }

a. ........ &= nbsp; =20 ..xxxxxx =20 ..xxxxx. =20 ..xxxx.. =20 ..xxxxxx =20 ....xxxx =20 .xxxxxx. =20 .xxxxx.. =20 ..x..... b. ........ &= nbsp; =20 ..xxxxx. =20 ..x..... =20 ..x.x... =20 ....xxx. =20 ......x. =20 .x.xxx.. =20 ..x..... =20 ........ c. ........ &= nbsp; =20 ..xxxxxx =20 ........ =20 ..xxx... =20 .....xx. =20 ........ =20 .x.xxx.. =20 ..x..... =20 ........ d. ........ &= nbsp; =20 ...xxxx. =20 ...x.... =20 ...xx... =20 .....xx. =20 ........ =20 ....xx.. =20 ...xx... =20 ........ e. ........ &= nbsp; =20 ..xxxxx. =20 ..xxx... =20 ..xxx... =20 ....xxx. =20 ......x. =20 .x.xxxx. =20 ..xxxx.. =20 ........

148.

Suppose Mergesort is implemented as follows:

public void sort(int[] a, int n1, int=20 n2) { if (n1 =3D=3D=20 n2) =20 return; else if (n2 =3D=3D n1 +=20 1) =20 { if = (a[n2]=20 <=20 = a[n1]) &nb= sp; =20 swap(a, n1, n2); // swaps a[n1] and=20 a[n2] =20 return; = } int m=20 =3D (n1 + n2) / 2; sort(a, n1,=20 m); sort(a, m+1,=20 n2); if (a[m] >=20 = a[m+1]) =20 merge(a, n1, m, n2); // merges a[n1]...a[m] = and=20 a[m+1]...a[n2] }

How many times will = the method=20 merge be called if an array a contains = the=20 values

2 1 4 3 6 5 8 7

and =

sort(a, 0,=20
      7)

is called?

7=20
4=20
3=20
1=20
0

149. (AB)

Consider the following method:

public void xQ(Queue = q) {      if=20 = (q.isEmpty())          =20 return;

     Queue q1 =3D new ListQueue(), = q2 =3D new=20 ListQueue();      Object x =3D=20 q.dequeue();

     while=20 (!q.isEmpty())     =20 {           = Object y =3D=20 = q.dequeue();          =20 if (((Comparable)y).compareTo(x) <=3D=20 = 0)            &= nbsp;  =20 = q1.enqueue(y);           = = else            = ;   =20 q2.enqueue(y);      }

     = xQ(q1);     =20 xQ(q2);

     while=20 = (!q1.isEmpty())          = =20 q.enqueue(q1.dequeue());

     q.enqueue(x);

     while=20 = (!q2.isEmpty())          = =20 q.enqueue(q2.dequeue()); }

What are the values in q after xQ(q) = is=20 called, if q initially holds Integer = objects=20 with the values 7, 11, 3,=20 5, 6, 0 (listed starting = from the=20 front)?

0, = 3,=20 5, 6, 7, 11=20
0, = 6,=20 5, 3, 11, 7=20
3, = 5,=20 6, 0, 7, 11=20
11,=20 3, 5, 6, 0,=20 7=20
11,=20 7, 6, 5, 3,=20 0

150. (AB)

Consider the following method:

public int xSum(TreeNode = root) { =20 if (root =3D=3D=20 null) = return=20 0; =20 else = return=20 ((Integer)root.getValue()).intValue()=20 = + &n= bsp; =20 xSum(root.getRight()) - xSum(root.getLeft()); }=20

What value is returned by xSum(root) if = root=20 points to the following tree?

=
&nbs= p; =20 = 5 &n= bsp; =20 /=20 = \ &n= bsp; =20 4 =20 = 2 &n= bsp; =20 / /=20 = \ &n= bsp;=20 3 =20 = 7 1 &= nbsp; &n= bsp; =20 = \ &n= bsp; =20 12

-9=20
-10=20
5=20
12=20
20 =

1999

151.

What is the output from the following code?

int x =3D 5, y =3D 2; System.out.println(x/y - = (double)(x/y));=20

0=20
0.5=20
-0.5 =
-2.5 =
None of the above =

152.

What are the values of u and = v=20 after the following code is executed?

int u =3D 3, v =3D 5;

u +=3D v; v +=3D u; u -=3D v; v -=3D u;=20

0, 0=20
3, 5=20
5, 3=20
-5, -3=20
-5, 18 =

153.

Which of the following does NOT display=20 9.95?

= System.out.println(9 +=20 0.95);=20
=20 System.out.println(995/100.0);=20
= System.out.println(9. +=20 95/100);=20
= System.out.println(9 +=20 95.0/100);=20
= System.out.println(9 +=20 "." + 95);

154.

What is the output from the following code?

int count1 =3D 0, count2 =3D 0, inc =3D 1; for (int = i =3D 0; i <=20 11; i++) { count1 +=3D=20 inc; inc =3D = -inc; =20 count2 +=3D inc; } System.out.println(count1 -=20 count2);

0=20
2=20
-2=20
22=20
-22 =

155.

What is the result of the following code segment?

Integer i =3D new=20 = Integer(0); &n= bsp; &nb= sp; =20 // Line 1 if=20 = (!i.equals(0)) = ; = &= nbsp; =20 // Line 2 System.out.println(i + " = does not=20 equal to " + 0); // Line = 3 else =20 System.out.println("Done");

Syntax error on = Line 2=20
Syntax error on = Line 3=20
= ClassCastException=20
0 is not = equal to=20 0=20
Done =

156.

What is the output from the following code?

int a =3D 0, b =3D 0; while (a <=20 3) { switch(a +=20 b) =20 { case = 0:=20 a++; = case=20 1: = case 2:=20 b++; = break; =20 case 3: a++;=20 break; = default: b =3D 0; break; =20 } =20 System.out.print(b); }

0123 =
= 122011=20
= 0122011=20
= 0122123=20
= 112300

157.

What is the value of n after the = following code=20 is executed?

int i, n =3D 0; while (n <=20 90) { for (i =3D 0; i < 10;=20 i++) =20 { n = +=3D=20 3; if = (n >=20 = 50) = =20 break; } = n++; }

51=20
61=20
91=20
93=20
104 =

158.

What is the set of all possible outputs of the = following=20 code when the user correctly follows the input instructions and = the=20 program terminates without an error?

System.out.print( "Enter an = integer=20 from -1000 to 1000 (inclusive): ");

int n =3D console.readInt(); // Read an = integer

while (n > 0 && Math.sqrt(n) <=20 10.0) n++;
=
System.out.println(n);

{0, ..., 99, 100} =
{100, ..., 999, = 1000}=20
{-1000, -999, = ..., 0} union=20 {100, ..., 999, 1000}=20
{-1000, -999, = ..., 0, ..., 999,=20 1000}=20
{-1000, -999, = ..., 0, ..., 99}=20

159.

An integer array a has 19 elements. What is the = value of the=20 middle element after the following code is executed?

int i, j, n =3D 19;

for (i =3D 0; i < n; = i++) a[i] =3D=20 i+1;

for (i =3D 0, j =3D n-1; i <=3D j; i++,=20 j--) a[(i+j)/2] -=3D (a[i] + a[j]) / = 2;

0=20
10=20
-41=20
-62=20
-71 =

160.

Consider the following method:

public void mysteryMix(Integer a, Integer=20 b) { a =3D new Integer(2 *=20 a.intValue()); b =3D a; }=20

What is the output of the following code?

Integer a =3D new Integer(1); Integer b =3D new=20 Integer(2); mysteryMix(a, a); mysteryMix(a,=20 b); System.out.println(a + " " + b); =

1 1=20
1 2=20
2 2=20
1 4=20
4 4 =

161.

The following interface Index describes = the=20 location of a document:

public interface Index { = String=20 getKey(); Document = getAddress(); }=20

Document is a class that has a public method=20 getSize(). An ArrayList folder contains=20 Index objects and describes a collection of = documents. Which=20 of the following expressions refers to size of the=20 k-th document in folder?

=20 folder[k-1].getAddress().getSize()=20
(Index)=20 folder.get(k-1).getAddress().getSize()=20
((Index)=20 folder.get(k-1)).getAddress().getSize()=20
((Document) = (folder[k-1].getAddress())).getSize()=20
((Document) = (folder.get(k-1).getAddress())).getSize()

162.

Which of the following could safely appear and make sense in = place of=20 < condition > in some suitable context?

if ( < condition = >=20 = ) msg =3D = new message("o");

I. msg =3D=3D null ||=20 = msg.getStatus().equals("x")II. msg.= getStatus().equals("x")=20 || msg =3D=3D nullIII. =20 "x".equals(msg.getStatus()) || msg =3D=3D=20 null

I only=20
II only=20
I and II=20
II and III=20
I, II, and III =

Questions 163-167 use the following interface and = classes=20 used in a picture drawing application:

public interface = Drawable { void=20 draw(Graphics g); }

public class Line implements=20 Drawable =20 { = private int=20 xBeg, yBeg, xEnd, yEnd;

= public=20 Line(int x0, int y0, int x1, int=20 y1) =20 = { &n= bsp; =20 xBeg =3D x0; yBeg =3D y0; xEnd =3D x1; yEnd =3D=20 y1; = }

= public=20 void draw(Graphics=20 g) =20 = { &n= bsp; =20 g.drawLine(xBeg, yBeg, xEnd,=20 yEnd); =20 } }

public class Picture implements=20 Drawable =20 { = private List=20 pictures;

= public=20 = Picture() =20 = { &n= bsp; =20 pictures =3D new=20 = LinkedList(); =20 }

= public=20 void add(Drawable=20 obj) =20 = { &n= bsp; =20 = pictures.add(obj); &nb= sp;=20 }

= public=20 void draw(Graphics=20 g) =20 = { &n= bsp; =20 Iterator it =3D=20 = pictures.iterator(); &= nbsp; =20 while=20 = (it.hasNext()) &= nbsp; =20 ((Drawable)=20 = it.next()).draw(g); &n= bsp;=20 } }

163.

Which of the following code segments compiles with = no=20 errors?

Drawable = picture =3D new=20 Picture(new Line(0, 0, 100, 100));
=20
Drawable = picture =3D new=20 Picture();
=20 picture.add(new Line(0, 0, 100, 100));
=20
Picture = picture1 =3D new=20 Picture();
= Picture=20 picture2 =3D new Picture(picture1);
=20
Picture = picture =3D new=20 Picture();
=20 picture.pictures.add(new Line(0, 0, 100, = 100));
=20
Drawable = picture =3D new=20 Picture();
((Picture)=20 picture).add(new Picture());

164.

Consider the following code segment:

Picture picture =3D new Picture(); picture.add(new = Line(100,=20 100, 200, 50)); picture.add(new Line(200, 50, 300, = 100)); Picture=20 box =3D new Picture(); box.add(new Line(100, 100, 100,=20 300)); box.add(new Line(100, 300, 300, 300)); box.add(new=20 Line(300, 300, 300, 100)); box.add(new Line(300, 100, 100,=20 100));

Which of the following pictures will be displayed when=20 pictures.draw(g) is called from an appropriate=20 paint method within the graphics context=20 g?

a. Empty picture

165.

Suppose the class Box extends=20 Picture and has the following constructor:

public Box(int x, int y, int width, int=20 height) { =20 super(); add(new Line(x, y, x + = width,=20 y)); add(new Line(x + width, y, x + = width, y=20 + height)); add(new Line(x + width, = y +=20 height, x, y + height)); add(new = Line(y, y +=20 height, x, y)); }

Suppose the statements

Box box =3D new Box(100, 100, 300, = 200); box.draw(g);=20

=97when executed from an appropriate paint method = within the=20 graphics context g=97draw a box with the upper left corner at = (100, 100),=20 width 300 and height 200. Besides the above constructor, which = methods=20 must be supplied in the Box class for this to = happen?

No methods are = needed=20
= add(Drawable)=20
= draw(Graphics)=20
= add(Drawable) and=20 add(Graphics)=20
= add(Drawable), add(Line),=20 and draw(Graphics)

166.

Suppose the following statements have been added to the Line = class:

public String getName() { return "Line"; }

public String = toString() { =20 return "[" + getName()=20 = + =20 /* " " + xBeg + " " + yBeg=20 = + &n= bsp; =20 " " + xEnd + " " + yEnd + =20 = */ = "]"; }

Also, the following methods have been added to the = Picture=20 class:

public String getName() { return "Picture"; = }

public String = toString() { =20 String str =3D "[" + getName() + " = "; Iterator=20 it =3D pictures.iterator(); while=20 = (it.hasNext()) = str +=3D it.next(); return str + = "]"; }=20

Finally, the following method has been added to the = Box=20 class mentioned in the previous question:

public String getName() { return "Box"; }=20

What is the output of the following code?

Box box =3D new Box(100, 100, 100, 100); box.add(new = Line(100,=20 100, 200, 200)); box.add(new Line(100, 200, 200, = 100)); Picture=20 picture =3D new=20 = Picture(); picture.add(box); System.out.println(picture);

[Picture = [Line] [Line]=20 [Line] [Line] [Line] [Line]=20
[Picture = [Box =20 [Line] [Line] [Line] [Line] ] ]=20
[Picture = [Box =20 [Line] [Line] [Line] [Line] [Line] [Line] ] ]=20
[Picture = [Picture [Line]=20 [Line] [Line] [Line] [Line] [Line] ]=20
[Picture = [Picture [Line]=20 [Line] [Line] [Line] [Line] [Line] ] ]

167.

The class DrawingBoard extends=20 JFrame and represents a window on the screen:

public class DrawingBoard extends=20 javax,Swing,JFrame {      private = Drawable=20 picture;

     public DrawingBoard(Drawable=20 picture)      {    =20 this.picture =3D picture;     }

     public void paint(Graphics g) = {=20 picture.draw(g); } }

Consider the following code segment:

Picture picture =3D new=20 = Picture(); picture.add(picture); &n= bsp; =20 // Line *** DrawingBoard w =3D new=20 DrawingBoard(picture); w.setBounds(50, 50, 400, = 400); w.show();=20

What happens when we try to compile and execute this = code?

Syntax error on = Line ***=20
= ClassCastException=20
The code compiles = and runs, but=20 goes into an infinite loop=20
= StackOverflowError=20
The code compiles = and runs; a=20 blank window is displayed

Questions 168-169 use the following class:

public class PetDog implements=20 Comparable =20 { = private String=20 myName, myBreed;

= public=20 petDog(String name, String=20 breed) { = myName=20 =3D name; myBreed =3D breed; }

= public=20 String getName() { return myName;=20 } public = String=20 getBreed() { return myBreed; }

= public=20 boolean equals(Object=20 other) { = return=20 other !=3D null && getName.equals(other.toString()); = }

= public int=20 compareTo(Object=20 other) =20 = { &n= bsp; =20 return=20 = getName().compareTo(other.toSstring()); = =20 }

= public int=20 hashCode() { return getName().hashCode(); }

= public=20 String=20 = toString() {=20 return getName() + "--" + + getBreed(); = } =20 }

Suppose the following variables are declared and = initialized:

PetDog honey =3D new PetDog("Honey", "Cocker = Spaniel"); PetDog=20 lucie =3D new PetDog("Lucie", "Springer Spaniel"); PetDog = murray =3D new=20 PetDog("Murray", "Golden = Retriever");

168. (AB)

The (AB)

a Cocker Spaniel

Map map =3D new=20 HashMap(); = map.put(honey.getName(),=20 honey); map.put(lucie.getName(),=20 lucie); map.put(murray.getName(),=20 murray);

= System.out.println(honey.getNamer()=20 + " is a " = + =20 < missing expression >=20 );

Which of the following can replace < missing = expression=20 >?

I. =20 = PetDog(map.get(honey.getName())II. = ((PetDog)map.get(honey)).getBreed()III. = = ((PetDog)map.get(honey.getName())).getBreed()

I only=20
II only=20
I and II=20
II and III=20
I, II, and III =

169. (AB)

Which of the following code segments will compile = with no=20 errors and produce an alphabetical list of all three pets:

Honey -- Cocker Spaniel Lucie -- Springer Spaniel = Murray=20 -- Golden Retriever

I.     Set set =3D new=20 TreeSet();     = set.add(honey);    =20 set.add(lucie);    =20 set.add(murray);     Iterator iter =3D=20 set.iterator();     while=20 (iter.hasNext())        =20 = System.out.println(iter.next());II.   =20 Map map =3D new TreeMap();
    = map.put(homey.getName(),=20 honey);
    map.put(lucie.getName(),=20 lucie);
    map.put(murray.getName(),=20 murray);
    Iterator iter =3D=20 map.values().iterator();
    while=20 (iter.hasNext())
       =20 System.out.println(iter.next());

III.  =20 Map map =3D new HashMap();    =20 map.put(homey.getName(), honey);    =20 map.put(lucie.getName(), lucie);    =20 map.put(murray.getName(), murray);     = Iterator iter =3D=20 map.keySet().iterator();     while=20 (iter.hasNext())        =20 = System.out.println(map.get(iter.next()));

I only=20
II only=20
I and II=20
II and III=20
I, II, and III =

170. (AB)

head1 points to the first node of a = linked list=20 that has three nodes with Integer values = 0,=20 1, 2; head2 points to the = first=20 node of a list that has three nodes with Integer = values=20 3, 4, 5. What is the output = from=20 the following code?

ListNode node; if (head1 !=3D null && head2 = !=3D=20 null) { node =3D=20 head1.getNext(); =20 head1.setNext(head2.getNext()); =20 head2.setNext(node); }

for (node =3D head1; node !=3D null; node =3D=20 node.getNext()) =20 System.out.println(node.getValue() + " "); for (node =3D = head2; node !=3D=20 null; node =3D node.getNext()) =20 System.out.println(node.getValue() + " = ");

0 1 2 3 4 = 5=20
0 4 5 3 1 = 2=20
3 4 5 0 1 = 2=20
3 0 2 0 4 = 5=20
3 0 1 2 4 = 5=20

171. (AB)

What is the contents of the stack s1 = (its=20 elements listed starting from the top) after the following code is = executed?

Stack stk =3D new ArrayStack(); Stack stk1 =3D new=20 ArrayStack(); Stack stk2 =3D new ArrayStack();

int n; Integer obj; for (n =3D 1; n <=3D 6;=20 n++) stk.push(new = Integer(n));

while (!stk.isEmpty()) { = obj =3D=20 (Integer stk.pop()); n =3D=20 obj.intValue(); if (N % 2 !=3D=20 0) =20 stk1.push(obj); =20 else =20 stk2.push(obj); } while=20 (!stk1.isEmpty()) =20 stk.push(stk1.pop()); while=20 (!stk2.isEmpty()) =20 stk.push(stk2.pop());

1, 2, 3, 4, 5, 6=20
6, 5, 4, 3, 2, 1=20
1, 3, 5, 2, 4, 6=20
2, 4, 6, 1, 3, 5=20
None of the above =

172. (AB)

Consider the following method:

public boolean someProperty(TreeNode=20 root) { return root !=3D null=20 = && = ; =20 (root.getLeft() !=3D null && root.getRight() !=3D null=20 = || &= nbsp; =20 someProperty(root.getLeft())=20 = || &= nbsp; =20 someProperty(root.getRight())); }

This method returns true if and only if the tree = pointed=20 to by root

is not empty.=20
is not empty and = the root is=20 not a leaf.=20
is not empty and = the root is=20 either a leaf or has two children.=20
has at least one = node with two=20 children.=20
is a full tree. =

173. (AB)

The method max(TreeNode root) assumes a = precondition that root points to a non-empty binary = search=20 tree containing Comparable objects. max = returns=20 the value from the tree's largest node. Which of the following = three=20 versions of max return the correct answer when the=20 precondition is met?

I.     public Object = max(TreeNOde=20 root)     =20 {           = while=20 (root.getRight() !=3D=20 = null)           &nbs= p;   =20 root =3D=20 = root.getRight();          = ;=20 return root.getValue();     =20 }II.   public Object = max(TreeNOde=20 root)     =20 {           = Object=20 maxValue =3D=20 = root.getValue();          = ;=20 if (root.getRight() !=3D=20 null)          =20 = {            &n= bsp;  =20 Object temp =3D=20 = max(root.getRight());         =       =20 if (((Comparable)temp).compareTo(maxValue) >=20 = 0)            &= nbsp;       =20 maxValue =3D=20 temp;          =20 }           = return=20 maxValue;      = }III.  =20 public Object max(TreeNOde = root)     =20 {           = Object=20 maxValue =3D root.getValue();

          = if=20 (root.getLeft() !=3D null=20 = &&           = ;    =20 = ((Comparable)max(root.getLeft())).       = ;            = =20 compareTo(maxValue) >=20 = 0)            &= nbsp;  =20 maxValue =3D max(root.getLeft());

          = if=20 (root.getRight() !=3D null=20 = &&           = ;    =20 = ((Comparable)max(root.getRight())).      &nbs= p;            = ; =20 compareTo(maxValue) >=20 = 0)            &= nbsp;  =20 maxValue =3D max(root.getRight());

          = return=20 maxValue;      = }

I only=20
II only=20
I and II=20
II and III=20
I, II, and III =

174.

Given

int[] a =3D {1, 3, 5, 7, 9, 11, 13}; =

what are the values in a after = disarray(int[] a, int=20 n) is called? The method disarray is defined = as=20 follows:

public void disarray(int[] a, int=20 n) { if (n >=20 1) =20 { = disarray(a,=20 n-1); = a[n-1]=20 +=3D a[n-2]; } }=20

1, 4, 9, 16, 25, = 36, 49=20
1, 4, 8, 12, 16, = 20, 24=20
1, 8, 12, 16, 20, = 24, 13=20
1, 24, 20, 16, = 12, 8, 4=20
None of the above =

175.

The method mixup is defined as follows:

String mixup(String = word) { if=20 (word.length() =3D=3D=20 1) = return=20 ""; =20 else = return=20 mixup(word.substring(0, word.length() -=20 = 1)) = &= nbsp;=20 + word.charAt(word.length() - 1); }

What is the value of the string returned by=20 mixup("IDEAL")?

IDEAL=20
IDEA=20
LEAD=20
LEDA=20
DEAL =

=20