If you notice, in our for loop, the integer j is used as the iteration variable. However, no where in the code is that variable defined. To fix this issue this could have been done.

for (int j=0;j<3,j++)

Note the bold here is inserted. We need to define j here. We could have also defined j as a double.

Type ifstream objects are used to read from files, NOT to write to a file. To write to a file you can use ofstream. You can also you fstream to both read and write to and from a file.

line d's error is that system.out.println (i); is not capitalized.

line e's error is that there is a missing semicolon.

At first glance we might be tempted to pick O( ) because there are 2 for loops. But, upon closer inspection we can see that the first loop will yield a O( n ) running time but the second loop does not. The second loop has only an O( log(n) ) running time because "j" doubles each iteration and does not increase linearly. That will yield O( log(n) ) since O( log(n) ) is a much faster running time. So the final result is O( n log(n) ).

Code sample #1 relies on i in the second loop where int j = i. Since the code relies on i in the second loop, the order goes from O(N) to O(N2)

Code sample #2 has two separate loops that do not rely on each other. The first for loop loops through the array arr and the second for loop loops through the array arr2. Since the two loops are exclusive, the order is O(N)

O(NlogN) is O(N) * O(logN) which is greater than O(N) alone.

O(N2) is O(N*N) which is greater than O(N).

O(N3) is O(N*N*N) which is greater than O(N).

O(N) is the smallest and therefore is the quickest to compile. Therefore, O(N) is the correct answer.

The two code snippets have the same efficiency. Both operate in O(N) time. ArrayLists use arrays as their underlying data structure so access to the data is also the same.

Code snippet A has a running time of O(N2). Code snippet B has a running time of O(N). While the two code snippets may look the same, the second for loop in code snippet A sets j=i. Since j is relying on i, it's multiplying the first for loop's running time by the second for loop's running time. This gives us O(N*N) or just O(N2).

Squaring a real number will always produce a positive number. The result does not have to be stored in a new variable; it could be a value that is only needed for a one-off expression, thus, not worthy to be stored in memory. Lastly, since the input was passed by value and not by reference, its initial value will stay the same.

The user could input a string and the cast to an (Integer) would cause a runtime exception. If there is a runtime exception, the program will stop and open up vulnerabilities to hackers. Once a hacker knows how to halt a program, they can start input bad data to see a database schema to collect data. One way to fix this would be using a utility method such as parseInt(ui).

The .substring() method takes the character at the first number in the arguments, and goes through the String until it reaches the second number in the arguments, without copying the character at the second number.

In the first part of mystery(), the Strings s1, s2, s3, s4, and s5 are made and filled. If n = # of characters in s, s1 gets the first character in s, s2 gets the second character in s, s3 gets the third through n-2 characters, s4 gets the n-1 character, and s5 gets the last character.

String s1 = s.substring(0,1);

String s2 = s.substring(1,2);

String s3 = s.substring(2, s.length() - 2);

String s4 = s.substring(s.length() - 2, s.length() - 1);

String s5 = s.substring(s.length() - 1);

Let's look at the second portion of mystery().

if (s.length() <= 5)
return s5 + s4 + s3 + s2 + s1;
else
return s1 + s2 + mystery(s3) + s4 + s5;

The if statement checks the length of s, and if it's less than or equal to 5, it returns a String made from s5, followed by s4, etc. If s were equal to "abcde", then the if would evaluate to true, and would return "edcba". In recursion, this is known as the "base case".

The else statement is for strings that are greater than 5 characters in length. It returns s1, followed by s2, then the result of mystery(s3), then s4 and s5. The fact that it calls itself makes this recursion.

Let's step through the example. The argument for mystery(), s, is "ABNORMALITIES". After the first part, this is the result:

s1 = "A"

s2 = "B"

s3 = "NORMALITI"

s4 = "E"

s5 = "S"

Because s is longer than 5 characters, we take the else, so it returns the following:

A + B + mystery(NORMALITI) + E + S

Next, we repeat with the new argument. s = NORMALITI, so after the first part, the result is:

s1 = "N"

s2 = "O"

s3 = "RMALI"

s4 = "T"

s5 = "I"

Because s is longer than 5 characters again, we take the else, so it returns the following:

N + O + mystery(RMALI) + T + I

Which gets added to the previous return, making it this:

A + B + N + O + mystery(RMALI) + T + I + E + S

Once again, we repeat with the argument s = RMALI. After the first part, the result is:

s1 = "R"

s2 = "M"

s3 = "A"

s4 = "L"

s5 = "I"

Because s is less than or equal to 5 characters in length, we take the if this time. It returns the following:

I + L + A + M + R

We can replace all instances of mystery(RMALI) with the above, so the original return becomes this:

A + B + N + O + I + L + A + M + R + T + I + E + S

Which gets printed as ABNOILAMRTIES, the answer.

We start with an array, arr, of size 5, containing {1, 2, 3, 4, 5}.

The loop in the code,

for (int i = 0; i < arr.length - 2; i++)

loops through the array up to the second to last cell, given that arr.length - 2 is the index of the second to last cell, and it starts at the first cell.

Let's look at the code inside the loop.

int temp = arr[i];

arr[i] = arr[i+1];

arr[i+1] = temp;

When i = 0,

arr[0] == 1

arr[1] == 2

temp = 1

arr[0] = 2

arr[1] = 1

arr[] == {2, 1, 3, 4, 5}

When i = 1

arr[1] == 1

arr[2] == 3

temp = 1

arr[1] = 3

arr[2] = 1

arr[] == {2, 3, 1, 4, 5}

When i = 2

arr[2] == 1

arr[3] == 4

temp = 1

arr[2] = 4

arr[3] = 1

arr[] == {2, 3, 4, 1, 5}

When i = 3

arr[3] == 1

arr[4] == 3

temp = 1

arr[3] = 3

arr[4] = 1

arr[] == {2, 3, 4, 5, 1}

As the loop progresses, it moves whatever value is in arr[0] all the way to the end of the array.

The first part of fun assigns the value of the last location of the array to the first location. Then, it returns the a[0] + (a[0] % 2). That last portion will be a "1" if a[0] is odd, and "0" if a[0] is even. Since a[0] == 12, which is even, the expression evaluates to 12 + 0, which is 12.

The loop will run 5 times. The values of x, y, and z after each run will be as follows:

• i == 0: x == 7, y == 1, z == 7
• i == 1: x == 8, y == 6, z == 8
• i == 2: x == 14, y == 2, z == 14
• i == 3: x == 16, y == 12, z == 16
• i == 4: x == 28, y == 4, z == 28

At the end, i will equal 5, and the values will no longer change.

The only error among the options given is the fact that this code assigns a new value to the variable val3, which is defined as a constant. (This is indicated by the keyword final before the rest of its declaration.) You cannot alter constants once they have been declared. Thus, the following line will cause a compile-time error:

val3 = val1 * 12;

This code fills up the 6 member array with alternating Rectangle and Square objects. You can do this because the Square class is a subclass of Rectangle. That is, since Squares are Rectangles, you can store Square objects in Rectangle variables. However, even though rects[1] is a square, you CANNOT immediately reassign that to a Square object. The code has now come to consider all of the objects in the array as being Rectangle objects. You would need to explicitly type cast this to get the line to work:

Square s = (Square)(rects[1]);

The problematic line is this one:

int[] y = remove(x,4);

Notice that the variable y is defined as an array. Now, it is tempting to think (without looking too closely) that the remove method returns the array after the removal has been accomplished; however, this is not how the logic works in the remove method. Instead, it returns a boolean indicating whether or not this removal was successful or not (i.e. it tells you whether or not it actually found the value). Therefore, you cannot make an assignment like the one above, for the two types are not the same. That is, y is an integer array, while remove returns a boolean value!

When you implement an interface, all of the methods defined in that interface must be written in the class that is proposing to be such an implementation. (Or, if they are not implemented there, you need to involve abstract classes—but that is not our concern here.) The methods must match the prototypes proposed in the interface. In the example code, ServerInstance has a method writeBytes that returns a boolean value. However, the MyHost class has implemented this method as returning a byte[] value. Since you cannot have different types of return values for methods with the same parameter set, Java interprets this as being the proposed implementation for writeBytes(byte[] b), and this method must return a boolean if MyHost is to implement ServerInstance.

An error in compilation occurs before any code even attempts to execute. Thus, it is primarily a syntactical error in the code. In the selection above, the line

int x = 10; y = 21, z = 30;

has a semicolon right after 10_._ This causes an error in the code directly following on this, for the code

y = 21, z = 30;

is not valid Java code.

There are other errors in this code. arr[i] = z / i; will cause a divide by 0 error when iis 0. Also, arr[i] = z * i; will overrun the bounds of the array arr. However, these are not compile-time errors—i.e. errors that occur before the code is even able to run!

In this problem, Derived1 and Derived2 are children of the Base class. If we take a look at this line:

Base* bp = new Derived();

We are assigning a new Derived class to a base pointer. This will compile. Think of the Base as a larger object because it is the parent, so copying a smaller object into a larger one is acceptable.

Now let's look at this one:

Derived2* d2p = bp;

This line will cause the program to not compile. Since the Base class is considered the "bigger" object, copying a bigger object into a "smaller" one will result in a failure to copy everything over, this is known as a Slicing Problem.

We don't even have to look at the next line because we know that the program wil crash.