Common Data Structures - AP Computer Science A
Card 0 of 308
int foo[] = {1, 2, 3, 4, 5};
number = 100 + foo[4];
What is the value of number?
int foo[] = {1, 2, 3, 4, 5};
number = 100 + foo[4];
What is the value of number?
The answer is 105 because arrays are zero indexed. This means that the first position has the subscript 0, the second subscript 1, and so on. 5 is in the fifth space, located at subscript 4. We would access it by saying foo\[4\]. As another example, if we wanted to access one, we would say foo\[0\].
The answer is 105 because arrays are zero indexed. This means that the first position has the subscript 0, the second subscript 1, and so on. 5 is in the fifth space, located at subscript 4. We would access it by saying foo\[4\]. As another example, if we wanted to access one, we would say foo\[0\].
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Which of these instantiate a matrix called matrx with 5 columns and 4 rows that takes in integers?
Which of these instantiate a matrix called matrx with 5 columns and 4 rows that takes in integers?
You create a matrix also known as a 2 dimensional array the same way you'd instantiate a normal array except the first array space remains blank and you'd insert the number for the amount of rows. Due to the fact that you want 5 columns and 4 rows, you'd only input the 4 into the second array.
You create a matrix also known as a 2 dimensional array the same way you'd instantiate a normal array except the first array space remains blank and you'd insert the number for the amount of rows. Due to the fact that you want 5 columns and 4 rows, you'd only input the 4 into the second array.
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Write a program that iterates through this data structure and prints the data (choose the best answer):
List<List<String>> listOflistOfStrings = new ArrayList<ArrayList<String>>();
Write a program that iterates through this data structure and prints the data (choose the best answer):
List<List<String>> listOflistOfStrings = new ArrayList<ArrayList<String>>();
The correct answer uses a ForEach loop. A ForEach loop is recommended for iterating through Lists because Lists contain iterators. ForEach loops use the iterator to iterate through the List. One of the answers used a regular For loop, while the answer was correct, it was not the best choice.
The correct answer uses a ForEach loop. A ForEach loop is recommended for iterating through Lists because Lists contain iterators. ForEach loops use the iterator to iterate through the List. One of the answers used a regular For loop, while the answer was correct, it was not the best choice.
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Which of the following blocks of code converts an array of characters into a string?
Which of the following blocks of code converts an array of characters into a string?
The easiest way to consider this is by commenting on the correct answer. You must begin by defining the character array:
char\[\] vals = {'A','t', ' ', '6',' ','a','m','!'};
Next, you must initialize the string value s to be an empty string. This is critical. Otherwise, you can't build your string!
String s = "";
Next, you have the loop. This goes through the characters and concatenates the values to the variable s. The operation to concatenate the characters is the "+=". This will give you the string value of the array of characters.
The easiest way to consider this is by commenting on the correct answer. You must begin by defining the character array:
char\[\] vals = {'A','t', ' ', '6',' ','a','m','!'};
Next, you must initialize the string value s to be an empty string. This is critical. Otherwise, you can't build your string!
String s = "";
Next, you have the loop. This goes through the characters and concatenates the values to the variable s. The operation to concatenate the characters is the "+=". This will give you the string value of the array of characters.
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String greet = "Hello ";
String sub;
int len = greet.length();
sub = greet.substring(0, (len/2));
System.out.println (sub);
What is printed?
String greet = "Hello ";
String sub;
int len = greet.length();
sub = greet.substring(0, (len/2));
System.out.println (sub);
What is printed?
The length of greet is 6 characters including the space at the end.
greet.substring(0, (len/2)) is equal to greet.substring(0, 3)
The substring of greet from the zeroth position to second position, not to the third position.
The length of greet is 6 characters including the space at the end.
greet.substring(0, (len/2)) is equal to greet.substring(0, 3)
The substring of greet from the zeroth position to second position, not to the third position.
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int foo[] = {1, 2, 3, 4, 5};
number = 100 + foo[4];
What is the value of number?
int foo[] = {1, 2, 3, 4, 5};
number = 100 + foo[4];
What is the value of number?
The answer is 105 because arrays are zero indexed. This means that the first position has the subscript 0, the second subscript 1, and so on. 5 is in the fifth space, located at subscript 4. We would access it by saying foo\[4\]. As another example, if we wanted to access one, we would say foo\[0\].
The answer is 105 because arrays are zero indexed. This means that the first position has the subscript 0, the second subscript 1, and so on. 5 is in the fifth space, located at subscript 4. We would access it by saying foo\[4\]. As another example, if we wanted to access one, we would say foo\[0\].
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Which of these instantiate a matrix called matrx with 5 columns and 4 rows that takes in integers?
Which of these instantiate a matrix called matrx with 5 columns and 4 rows that takes in integers?
You create a matrix also known as a 2 dimensional array the same way you'd instantiate a normal array except the first array space remains blank and you'd insert the number for the amount of rows. Due to the fact that you want 5 columns and 4 rows, you'd only input the 4 into the second array.
You create a matrix also known as a 2 dimensional array the same way you'd instantiate a normal array except the first array space remains blank and you'd insert the number for the amount of rows. Due to the fact that you want 5 columns and 4 rows, you'd only input the 4 into the second array.
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Write a program that iterates through this data structure and prints the data (choose the best answer):
List<List<String>> listOflistOfStrings = new ArrayList<ArrayList<String>>();
Write a program that iterates through this data structure and prints the data (choose the best answer):
List<List<String>> listOflistOfStrings = new ArrayList<ArrayList<String>>();
The correct answer uses a ForEach loop. A ForEach loop is recommended for iterating through Lists because Lists contain iterators. ForEach loops use the iterator to iterate through the List. One of the answers used a regular For loop, while the answer was correct, it was not the best choice.
The correct answer uses a ForEach loop. A ForEach loop is recommended for iterating through Lists because Lists contain iterators. ForEach loops use the iterator to iterate through the List. One of the answers used a regular For loop, while the answer was correct, it was not the best choice.
Compare your answer with the correct one above
Which of the following blocks of code converts an array of characters into a string?
Which of the following blocks of code converts an array of characters into a string?
The easiest way to consider this is by commenting on the correct answer. You must begin by defining the character array:
char\[\] vals = {'A','t', ' ', '6',' ','a','m','!'};
Next, you must initialize the string value s to be an empty string. This is critical. Otherwise, you can't build your string!
String s = "";
Next, you have the loop. This goes through the characters and concatenates the values to the variable s. The operation to concatenate the characters is the "+=". This will give you the string value of the array of characters.
The easiest way to consider this is by commenting on the correct answer. You must begin by defining the character array:
char\[\] vals = {'A','t', ' ', '6',' ','a','m','!'};
Next, you must initialize the string value s to be an empty string. This is critical. Otherwise, you can't build your string!
String s = "";
Next, you have the loop. This goes through the characters and concatenates the values to the variable s. The operation to concatenate the characters is the "+=". This will give you the string value of the array of characters.
Compare your answer with the correct one above
String greet = "Hello ";
String sub;
int len = greet.length();
sub = greet.substring(0, (len/2));
System.out.println (sub);
What is printed?
String greet = "Hello ";
String sub;
int len = greet.length();
sub = greet.substring(0, (len/2));
System.out.println (sub);
What is printed?
The length of greet is 6 characters including the space at the end.
greet.substring(0, (len/2)) is equal to greet.substring(0, 3)
The substring of greet from the zeroth position to second position, not to the third position.
The length of greet is 6 characters including the space at the end.
greet.substring(0, (len/2)) is equal to greet.substring(0, 3)
The substring of greet from the zeroth position to second position, not to the third position.
Compare your answer with the correct one above
For the following question, consider the following code:
public static void main(String\[\] args)
{
double\[\] vec = {4,8,10,18};
System.out.println(fun(vec));
}
privatestaticdouble fun(double\[\] x)
{
double p = 0;
for(int i = x.length-1; i > -1; i--)
{
p += x\[i\];
}
return p / x.length;
}
Which of the following is a possible output for this program?
For the following question, consider the following code:
public static void main(String\[\] args)
{
double\[\] vec = {4,8,10,18};
System.out.println(fun(vec));
}
privatestaticdouble fun(double\[\] x)
{
double p = 0;
for(int i = x.length-1; i > -1; i--)
{
p += x\[i\];
}
return p / x.length;
}
Which of the following is a possible output for this program?
First of all, you can eliminate the following two options very quickly, as the fun method returns a double value:
\[D@4b71bbc9\]
\[4,9,11,19\]
These are trying to trick you into thinking that it returns an array. (The first value is really what would be the output, since there is no toString defined for the double array by default.)
Now, in fun, the loop simply runs through the values in the array, summing those up. Notice, it runs "backwards" through the array, from the end to the beginning. It then divides by the length of the array, giving you the average of the values in the array.
First of all, you can eliminate the following two options very quickly, as the fun method returns a double value:
\[D@4b71bbc9\]
\[4,9,11,19\]
These are trying to trick you into thinking that it returns an array. (The first value is really what would be the output, since there is no toString defined for the double array by default.)
Now, in fun, the loop simply runs through the values in the array, summing those up. Notice, it runs "backwards" through the array, from the end to the beginning. It then divides by the length of the array, giving you the average of the values in the array.
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Consider the following code:
public static void main(String\[\] args) {
int\[\] vec = {8,-2,4,5,-8};
foo(vec);
}
private static void foo(int\[\] x) {
for(int i = 0; i < x.length; i++) {
int y = Math.abs(x\[i\]);
for(int j = 0; j < y; j++) {
System.out.print(x\[i\] + " ");
}
System.out.println();
}
}
Which of the following represents a possible output for the program above?
Consider the following code:
public static void main(String\[\] args) {
int\[\] vec = {8,-2,4,5,-8};
foo(vec);
}
private static void foo(int\[\] x) {
for(int i = 0; i < x.length; i++) {
int y = Math.abs(x\[i\]);
for(int j = 0; j < y; j++) {
System.out.print(x\[i\] + " ");
}
System.out.println();
}
}
Which of the following represents a possible output for the program above?
In this code's loop, notice that it takes the absolute value of each element. This is done on the line:
int y = Math.abs(x\[i\]);
This value is then used for the second loop, which goes for y times, each time outputting the value of the given member of the original array—but now with its particular sign value. Thus, even numbers like 
will be output multiple times (i.e. 8). This is done line by line for each member of the parameter array.
In this code's loop, notice that it takes the absolute value of each element. This is done on the line:
int y = Math.abs(x\[i\]);
This value is then used for the second loop, which goes for y times, each time outputting the value of the given member of the original array—but now with its particular sign value. Thus, even numbers like
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Consider the following code:
public static void main(String\[\] args) {
double\[\]\[\] matrix = {{1,6,7},{1,4,5}};
graphics(matrix);
}
private static double graphics(double\[\]\[\] x) {
double r = 0;
for(int i = 0; i < x.length; i++) {
for(int j = 0; j < x\[i\].length; j++) {
r += x\[i\]\[j\] * (i + 1);
}
}
return r;
}
What is the return value for graphics in the code below:
double\[\]\[\] matrix = {{1,6,7},{1,4,5}};
graphics(matrix);
Consider the following code:
public static void main(String\[\] args) {
double\[\]\[\] matrix = {{1,6,7},{1,4,5}};
graphics(matrix);
}
private static double graphics(double\[\]\[\] x) {
double r = 0;
for(int i = 0; i < x.length; i++) {
for(int j = 0; j < x\[i\].length; j++) {
r += x\[i\]\[j\] * (i + 1);
}
}
return r;
}
What is the return value for graphics in the code below:
double\[\]\[\] matrix = {{1,6,7},{1,4,5}};
graphics(matrix);
The graphics method takes the 2D array matrix and then runs through each element. This is the point of the double for loop in the method itself. Notice that for each element it does several things. First of all, it is clearly accumulating a value into the variable r. Now, for each element, you are adding:
x\[i\]\[j\] (the current value in the 2D array iteration)
TIMES
(i + 1), or, the current row number. Thus, for the data given you are doing the following:
1* 1 + 1 * 6 + 1 * 7 + 2 * 1 + 2 * 4 + 2 * 5 = 34
The graphics method takes the 2D array matrix and then runs through each element. This is the point of the double for loop in the method itself. Notice that for each element it does several things. First of all, it is clearly accumulating a value into the variable r. Now, for each element, you are adding:
x\[i\]\[j\] (the current value in the 2D array iteration)
TIMES
(i + 1), or, the current row number. Thus, for the data given you are doing the following:
1* 1 + 1 * 6 + 1 * 7 + 2 * 1 + 2 * 4 + 2 * 5 = 34
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Consider the following code:
int\[\] vals = {6,1,41,5,1};
int\[\]\[\] newVals = new int\[vals.length\]\[\];
for(int i = 0; i < vals.length; i++) {
newVals\[i\] = new int\[vals\[i\]\];
for(int j = 0; j < vals\[i\];j++) {
newVals\[i\]\[j\] = vals\[i\] * (j+1);
}
}
What does the code above do?
Consider the following code:
int\[\] vals = {6,1,41,5,1};
int\[\]\[\] newVals = new int\[vals.length\]\[\];
for(int i = 0; i < vals.length; i++) {
newVals\[i\] = new int\[vals\[i\]\];
for(int j = 0; j < vals\[i\];j++) {
newVals\[i\]\[j\] = vals\[i\] * (j+1);
}
}
What does the code above do?
Let's look at the main loop in this program:
for(int i = 0; i < vals.length; i++) {
newVals\[i\] = new int\[vals\[i\]\];
for(int j = 0; j < vals\[i\];j++) {
newVals\[i\]\[j\] = vals\[i\] * (j+1);
}
}
The first loop clearly runs through the vals array for the number of items in that array. After this, it creates in the newVals 2D array a second dimension that is as long as the value in the input array vals. Thus, for 41, it creates a 3rd row in newVals that is 41 columns wide.
Then, we go to the second loop. This one iterates from 0 to the value stored in the current location in vals. It places in the given 2D array location the multiple of the value. Think of j + 1. This will go from 1 to 41 for the case of 41 (and likewise for all the values). Thus, you create a 2D array with all the multiples of the initial vals array.
Let's look at the main loop in this program:
for(int i = 0; i < vals.length; i++) {
newVals\[i\] = new int\[vals\[i\]\];
for(int j = 0; j < vals\[i\];j++) {
newVals\[i\]\[j\] = vals\[i\] * (j+1);
}
}
The first loop clearly runs through the vals array for the number of items in that array. After this, it creates in the newVals 2D array a second dimension that is as long as the value in the input array vals. Thus, for 41, it creates a 3rd row in newVals that is 41 columns wide.
Then, we go to the second loop. This one iterates from 0 to the value stored in the current location in vals. It places in the given 2D array location the multiple of the value. Think of j + 1. This will go from 1 to 41 for the case of 41 (and likewise for all the values). Thus, you create a 2D array with all the multiples of the initial vals array.
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Consider the following code:
String s = "I read logic for fun!";
boolean\[\] b = {true,false,true,false,false,true,true,false,true,false,true,false,false,true,false,true,true,false,true,false,false};
for(int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if(!b\[i\]) {
c = Character.toUpperCase(c);
} else {
c = Character.toLowerCase(c);
}
System.out.print(c);
}
What is the output for this function code?
Consider the following code:
String s = "I read logic for fun!";
boolean\[\] b = {true,false,true,false,false,true,true,false,true,false,true,false,false,true,false,true,true,false,true,false,false};
for(int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if(!b\[i\]) {
c = Character.toUpperCase(c);
} else {
c = Character.toLowerCase(c);
}
System.out.print(c);
}
What is the output for this function code?
This code is using what are called parallel arrays. The boolean array has one value for each character in the String s. Notice the logic in the loop. There is a condition:
if(!b\[i\]) {
c = Character.toUpperCase(c);
} else {
c = Character.toLowerCase(c);
}
This means that if the boolean is false, then you will make the given letter upper case. (This is because of the ! in the condition. Be careful!) Otherwise, you make it lower case.
Carefully walking through the code, you will get:
i rEAd LoGiC fOr FuN!
This code is using what are called parallel arrays. The boolean array has one value for each character in the String s. Notice the logic in the loop. There is a condition:
if(!b\[i\]) {
c = Character.toUpperCase(c);
} else {
c = Character.toLowerCase(c);
}
This means that if the boolean is false, then you will make the given letter upper case. (This is because of the ! in the condition. Be careful!) Otherwise, you make it lower case.
Carefully walking through the code, you will get:
i rEAd LoGiC fOr FuN!
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TWO DIMENSIONAL ARRAYS
Given the following initialized array:
int fourth;
int\[\]\[\] myArray = { {1, 2, 3},
{4, 5, 6},
{7, 8, 9} };
Using myArray, how can I store in variable "fourth", the number 4?
TWO DIMENSIONAL ARRAYS
Given the following initialized array:
int fourth;
int\[\]\[\] myArray = { {1, 2, 3},
{4, 5, 6},
{7, 8, 9} };
Using myArray, how can I store in variable "fourth", the number 4?
When a two dimensional array is created and initialized, the way to access the items inside the matrix is by calling the array with the row and column (i.e. myArray\[ROW\]\[COLUMN\]). Keeping in mind that arrays start at 0, the number four would be in row 1, column 0. Therefore to save that number into the variable "fourth" we'll do the following:
fourth = myArray\[1\]\[0\];
*Note: myArray\[1\]\[0\] is not the same as myArray\[0\]\[1\].
myArray\[1\]\[0\] =4 because it is the item located at row=1 and column = 0.
myArray\[0\]\[1\] =12 because it is the item located at row=0 and column = 1.
When a two dimensional array is created and initialized, the way to access the items inside the matrix is by calling the array with the row and column (i.e. myArray\[ROW\]\[COLUMN\]). Keeping in mind that arrays start at 0, the number four would be in row 1, column 0. Therefore to save that number into the variable "fourth" we'll do the following:
fourth = myArray\[1\]\[0\];
*Note: myArray\[1\]\[0\] is not the same as myArray\[0\]\[1\].
myArray\[1\]\[0\] =4 because it is the item located at row=1 and column = 0.
myArray\[0\]\[1\] =12 because it is the item located at row=0 and column = 1.
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Given the following in C++:
int * data = new int\[12\];
Pick an expression that is equivalent to : data\[5\];
Given the following in C++:
int * data = new int\[12\];
Pick an expression that is equivalent to : data\[5\];
Let's look at this line of code:
int * data = new int\[12\]
An int pointer is created and an array of size 12 is assigned to it.
When data\[5\] is called, the "data" is dereferenced and the value of the 6th position is returned.
The only one of the choices that does this is
*(data + 5)
In line above, the pointer is incremented ot the 6th position and is then dereferenced to get the value.
If the code was
*data
The first item at the first position will be returned.
*(data+1)
This will return the item and the second position and so on.
Let's look at this line of code:
int * data = new int\[12\]
An int pointer is created and an array of size 12 is assigned to it.
When data\[5\] is called, the "data" is dereferenced and the value of the 6th position is returned.
The only one of the choices that does this is
*(data + 5)
In line above, the pointer is incremented ot the 6th position and is then dereferenced to get the value.
If the code was
*data
The first item at the first position will be returned.
*(data+1)
This will return the item and the second position and so on.
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True or False.
The best data structure to represent a set of keys and values is an array.
True or False.
The best data structure to represent a set of keys and values is an array.
Arrays can be two-dimensional. However, when trying to keep track of keys and values it can become complicated when using an array. HashMaps are the best way to represent data containing keys and values.
Arrays can be two-dimensional. However, when trying to keep track of keys and values it can become complicated when using an array. HashMaps are the best way to represent data containing keys and values.
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Define an unwrapped integer array in Swift (iOS).
Define an unwrapped integer array in Swift (iOS).
In Swift, the variable must be declared first with var then given a name. So now we have var arr then to unwrap, we must add a type var arr: \[Int\] and then initialize. Therefore, we have var arr: \[Int\] = \[\]. var arr = \[\] is technically correct, but the prompt asks you to unwrap the variable.
In Swift, the variable must be declared first with var then given a name. So now we have var arr then to unwrap, we must add a type var arr: \[Int\] and then initialize. Therefore, we have var arr: \[Int\] = \[\]. var arr = \[\] is technically correct, but the prompt asks you to unwrap the variable.
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Define an unwrapped string array in Swift (iOS).
Define an unwrapped string array in Swift (iOS).
In Swift, the variable must be declared first with var then given a name. So now we have var arr then to unwrap, we must add a type var arr: \[String\] and then initialize. Therefore, we have var arr: \[String\] = \[\]. var arr = \[\] is technically correct, but the prompt asks you to unwrap the variable.
In Swift, the variable must be declared first with var then given a name. So now we have var arr then to unwrap, we must add a type var arr: \[String\] and then initialize. Therefore, we have var arr: \[String\] = \[\]. var arr = \[\] is technically correct, but the prompt asks you to unwrap the variable.
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