AP® Computer Science A Sample Student Responses and Scoring Commentary

2019

AP

®

Computer Science A

Sample Student Responses

and Scoring Commentary

Inside:

Free Response Question 3

• Scoring Guideline

• Student Samples

• Scoring Commentary

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AP

®

COMPUTER SCIENCE A

2019 SCORING GUIDELINES

Apply the question assessment rubric first, which always takes precedence. Penalty points can only be

deducted in a part of the question that has earned credit via the question rubric. No part of a question (a, b, c)

may have a negative point total. A given penalty can be assessed only once for a question, even if it occurs

multiple times or in multiple parts of that question. A maximum of 3 penalty points may be assessed per

question.

1-Point Penalty

v) Array/collection access confusion

(

[] get)

w) Extraneous code that causes side-effect (

e.g., printing to output, incorrect precondition check)

x) Local variables used but none declared

y) Destruction of persistent data (e.g., changing value referenced by parameter)

z) Void method or constructor that returns a value

No Penalty

o Extraneous code with no side-effec

t (e.g., valid precondition check, no-op)

o Spelling/case discrepancies where there is no ambiguity*

o Local variable not declared provided other variables are declared in some part

o

private or public qualifier on a local variable

o Missing

public qualifier on class or constructor header

o Keyword used as an identifier

o Common mathematical symbols used for operators (× • ÷ < > <> ≠)

o []

vs. () vs. <>

o = instead of == and vice versa

o length / size confusion for array, String, List, or ArrayList; with or without ()

o Extraneous [] when referencing entire array

o [i,j] instead of [i][j]

o

Extraneous size in array declaration, e.g., int[size] nums = new int[size];

o Missing ; where structure clearly conveys intent

o Missing

{ } where indentation clearly conveys intent

o Missing

( ) on parameter-less method or constructor invocations

o Missing

( ) around if or while conditions

*Spelling and case discrepancies for identifiers fall under the “No Penalty” category only if the correction can be

unambiguou sly inferred from context, for example, “

ArayList” instead of “ArrayList”. As a counterexample, note

that if the code declares “

int G=99, g=0;”, then uses “while (G < 10)” instead of

“

while (g < 10)”, the context does not allow for the reader to assume the use of the lower-case variable.

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AP

®

COMPUTER SCIENCE A

2019 SCORING GUIDELINES

Question 3: Delimiters

Part (a) getDelimitersList

4 points

Intent: Store delimiters from an array in an

ArrayList

+1 Creates ArrayList<String>

+1

Accesses all elements in array

tokens

(no bounds errors)

+1 Compares strings in tokens with both instance variables (must be in t he context of a loop)

+1

Adds delimiters into

ArrayList

in original order

Part (b) isBalanced

5 points

Intent: Determine whether open and close delimiters in an

ArrayList are balanced

+1 Initializes accumu

lator(s)

+1

Accesses all elements in

ArrayList delimiters

(no bounds errors)

+1 Compares strings in delimiters with instance variables and updates accumulator(s)

accordingly

+1

Identifies and returns appropriate

boolean

value to implement one rule

+1 Identifies and returns appropriate boolean values for all cases

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AP

®

COMPUTER SCIENCE A

2019 SCORING GUIDELINES

Question 3: Scoring Notes

Part (a)

getDelimitersList

4 points

Points Rubric Criteria Responses earn the point even

if they...

Responses will not earn the point if they...

+1

Creates

ArrayList<String>

•

omit

<String> •

omit the keyword new

+1

Accesses all elements in

array tokens

(no bounds errors)

•

return incorrectly inside the

loop

•

treat

tokens

as a single string

• access elements of tokens as if

from an

ArrayList

(e.g.,

tokens.get(i))

+1

Compares strings in

tokens with both

instance variables (must

be in the context of a loop)

•

access elements of

tokens as if from an

ArrayList

(e.g., tokens.get(i))

•

use == for string comparison

• treat tokens as a single string

+1

Adds delimiters into

ArrayList in original

order

•

add a delimiter by

accessing tokens

incorrectly

(e.g.,

tokens.get(i))

•

add a token that is not a delimiter

• do not maintain the original delimiter

order

Part (b)

isBalanced

5 points

Points Rubric Criteria Responses earn the point even

if they...

Responses will not earn the point if they...

+1

Initializes accumulator(s)

•

initialize inside the loop

•

initialize an accumulator variable but

don't update it

+1

Accesses all elements in

ArrayList

delimiters

(no bounds errors)

• return incorrectly inside the

loop

• access elements of delimiters as

if from an array

(e.g.,

delimiters[i])

+1

Compares strings in

delimiters with

instance variables and

updates accumulator(s)

accordingly

•

access elements of

delimiters as if from

an array

(e.g.,

delimiters[i])

•

use == for string comparison

• adjust an accumulator without a

guarding condition

+1

Identifies and returns

appropriate

boolean

value to implement one

rule

• check for more closing

delimite

rs (inside a loop)

and return

false

• return true if the number

of open and close delimiters

is the same, and

false

otherwise (after a loop)

+1

Identifies and returns

appropriate

boolean

values for all cases

•

have correct logic with the

exception of a loop bounds

error, accessing elements as

if from an array, or using

==

for string comparison

• initialize accumulator inside a loop

• fail to check for more closing

delim

iters inside a loop

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®

COMPUTER SCIENCE A

2019 SCORING GUIDELINES

Question 3: Delimiters

Part (a)

public ArrayList<String> getDelimitersList(String[] tokens)

{

ArrayList<String> d = new ArrayList<String>();

for (String str : tokens)

{

if (str.equals(openDel) || str.equals(closeDel))

{

d.add(str);

}

return d;

}

Part (b)

public boolean isBalanced(ArrayList<String> delimiters)

{

int openCount = 0;

int closeCount = 0;

for (String str : delimiters)

{

if (str.equals(openDel))

{

openCount++;

}

else

{

closeCount++;

}

if (closeCount > openCount)

{

return false;

}

if (openCount == closeCount)

{

return true;

}

else

{

return false;

}

These canonical solutions serve an expository role, depicting general approaches to solution. Each reflects only one

instance from the infinite set of valid solutions. The solutions are presented in a coding style chosen to enhance readability

and facilitate understanding.

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®

COMPUTER SCIENCE A

2019 SCORING COMMENTARY

Question 3

Overview

This question tested the student's ability to:

• Write program code to satisfy methods using expressions, conditional statements, and iterative

stateme

nts; and

• Write program code to create, traverse, and manipulate elements in 1D array or

ArrayList

objects.

This question involved manipulation of both a one-dimensional array and an

ArrayList, both containing

String

values. Students were expected to write two methods in the enclosing Delimiters class, making

use of two instance variables of type

String.

In part (a) students were expected to create an

ArrayList of String objects, then add to it all the values

from the given array that matched either of two instance variables. Students had to construct a new

ArrayList

and write a loop to access each element of an array parameter. Inside the loop, students were

expected to compare each

String value in the array with each of the two instance variables and add each

matching

String value to the constructed ArrayList.

In part (b) students were given an

ArrayList containing String objects representing open and close

delimiters. Students were asked to develop an algorithm to determine whether the given

ArrayList

represents a balanced sequence of open and close delimiters. A sequence is balanced when two conditions are

met: (1) When traversing the

ArrayList from the first element to the last element, there is no point at which

there are more close delimiters than open delimiters. (2) The total number of open delimiters is equal to the total

number of close delimiters. Students had to write a loop to access each element of the given

ArrayList.

Inside the loop, students had to compare each

String value in the ArrayList to the instance variables

and then update accumulator(s) appropriately.

Sample: 3A

Sc

ore: 9

In

part (a) the response constructs a new

ArrayList and assigns it to a local variable. Therefore the response

earned point 1. All elements of

tokens are accessed with an enhanced for loop, so point 2 was earned.

Each token from the array is compared to both instance variables using the

equals method, and the response

earned point 3. Finally, each identified delimiter is added to the

ArrayList, and point 4 was earned. Part (a)

earned 4 points.

In part (b) the response initializes a variable as an accumulator and updates it la

ter; therefore, it earned point 5.

All necessary elements of

delimiters are accessed with a for loop using ArrayList notation

correctly, so point 6 was earned. Each delimiter from the

ArrayList is compared to an instance variable

using the

equals method, and the accumulator is updated. Therefore the response earned point 7. The

response returns

false if there are more close delimiters than open delimiters within the loop by determining

if the number of open delimiters that have not yet matched a close delimiter ever becomes negative. It also

returns

true after the loop if and only if there are no more open delimiters that have not matched a close

delimiter. Because the response returns the correct

boolean value for both conditions, points 8 and 9 were

both earned. Part (b) earned 5 points.

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AP

®

COMPUTER SCIENCE A

2019 SCORING COMMENTARY

Question 3 (continued)

Sample: 3B

Score: 6

In part (a) the response constructs a new

ArrayList and assigns it to a local variable. Therefore the response

earned point 1. The response was not penalized for missing parentheses when constructing the

ArrayList.

Elements of

tokens are accessed with a for loop using array notation correctly. However, the last element

is not accessed, so point 2 was not earned. Each token from the array is compared to both instance variables

using the

equals method; therefore, the response earned point 3. Finally, each identified delimiter is added to

the

ArrayList, and point 4 was earned. Part (a) earned 3 points.

In part (b) the response initializes two variables as accumulators and updates them later; therefore, it earned

point 5. Elements of

delimiters are accessed with a for loop. However, the last element is not accessed,

and elements are accessed as if from an array, so point 6 was not earned. Each delimiter from the

ArrayList

is compared to an instance variable using the equals method, and the accumulators are updated. Therefore

the response earned point 7. After the loop terminates, the response tests that the number of open and close

delimiters is the same, and the correct

boolean value is returned in both cases. Because the response

correctly implements one of the two conditions specified in the question, it earned point 8. However, it did not

earn point 9 because it omits the test for more close delimiters than open delimiters inside the loop. Part (b)

earned 3 points.

Sample: 3C

S

core: 3

In part (a) the response declares but does not construct a new

ArrayList, so point 1 was not earned. The

loop goes out of bounds by accessing an element after the last valid index of the

tokens array, so point 2 was

not earned. Point 3 was not earned because the response uses the

== operator to compare the strings in

tokens

with instance variables. Finally, each identified delimiter is added to the ArrayList, and so point 4

was earned. Part (a) earned 1 point.

In part (b) the response initializes a variable as an accumulator and updates it later; therefore, it earned point 5.

The

loop goes out of bounds by accessing an element after the last valid index of

delimiters, and so point 6

was not earned. Point 7 was not earned because the response uses the

== operator to compare the strings in

delimiters

with instance variables. After the loop terminates, the response tests that the number of open and

close delimiters is the same, and the correct

boolean value is returned in both cases. Because the response

correctly implements one of the two conditions specified in the question, it earned point 8. However, it did not

earn point 9 because it omits the test for more close delimiters than open delimiters inside the loop. Part (b)

earned 2 points.

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