8.4.1. Free Response - String Scramble B¶
The following is part b of a free response question from 2014. It was question 1 on the exam. You can see all the free response questions from past exams at https://apstudents.collegeboard.org/courses/ap-computer-science-a/free-response-questions-by-year.
Question 1. This question involves reasoning about strings made up of uppercase letters. You will implement two related methods that appear in the same class (not shown). The first method takes a single string parameter and returns a scrambled version of that string. The second method takes a list of strings and modifies the list by scrambling each entry in the list. Any entry that cannot be scrambled is removed from the list.
Part b. Write the method scrambleOrRemove
, which replaces each word in the parameter wordList
with its scrambled version and removes any words that are unchanged after scrambling. The relative order of the words in wordList
remains the same as before the call to scrambleOrRemove
.
The following example shows how the contents of wordList
would be modified as a result of calling scrambleOrRemove
.
![../_images/stringScrambleB.png](../_images/stringScrambleB.png)
Figure 1: Example call and result¶
Assume that the method scrambleWord
works as intended and is in the same class. It will return the scrambled word or the same word. You will write the scrambleOrRemove
method to replace each original word with the scrambled word or remove the word if it was not scrambled.
import java.util.List;
public class ScrambledStrings
{
/**
* Modifies wordList by replacing each word with its scrambled version,
* removing any words that are unchanged as a result of scrambling.
*
* @param wordList the list of words Precondition: wordList contains only
* non-null objects Postcondition: - all words unchanged by scrambling have
* been removed from wordList - each of the remaining words has been
* replaced by its scrambled version - the relative ordering of the entries
* in wordList is the same as it was before the method was called
*/
public static void scrambleOrRemove(List<String> wordList)
{
/* to be implemented in part b */
}
}
8.4.1.1. How to solve this problem¶
7-4-1-1: Explain in plain English what your code will have to do to answer this question. Use the variable names given above.
This section contains a plain English explanation of one way to solve this problem as well as problems that test your understanding of how to write the code to do those things. Click on the buttons to reveal the questions.
In the example the first word (at index 0) TAN
is scrambled and replaced. The second word ABRACADABRA
(at index 1) is scrambled and replaced. The third word WHOA
(at index 2) is removed. The fourth word APPLE
(at index 3) is scrambled and replaced. The fifth word EGGS
(at index 4) is removed since the scrambled word is the same as the original. What method of List allows you to replace an element in a list? What method of list allows you to remove an element from a list? How can you loop through a list and not always increment the current index?
- String
- Reread the method header, the datatype returned is to the left of the method name.
- ArrayList
- Reread the method header, the datatype returned is to the left of the method name.
- Array
- Reread the method header, the datatype returned is to the left of the method name.
- void
- Correct!
7-4-1-2: What is returned by this method?
- void
- you cannot have an ArrayList of type void
- String
- Correct!
- List
- This is not an ArrayList of more Lists
- int
- This list does not contain integers.
7-4-1-3: This method accepts an ArrayList
, what is the datatype of the objects contained in this ArrayList
?
- (index != wordList.current())
- the .current() method does not exist
- (int index = wordList.size() - 1; index >= 0; index--)
- this form of range control does not work with while loops
- (index < wordList.size())
- Correct!
- (wordList(index) != wordList.size())
- this does not accurately update the list as you iterate through wordList
7-4-1-4: There are many ways to use loops to solve this problem. If we were to use a while loop, what conditional could we write to make sure the loop does not go out of bounds? (Assume an integer index has already been initialized).
- (int i = wordList.size() - 1; i != wordList.size(); i--)
- this will lead to an infinite loop
- (int i = wordList.size() - 1; i >= 0; i--)
- Correct!
- (int i = wordList.size(); i >= 0; i--)
- This loop starts out of bounds since there isn't an element at wordList.size().
- (int i = wordList.size() - 1; i > 0; i--)
- This loop doesn't iterate all the way through the wordList. It misses the 0th element.
7-4-1-5: You can also use a for loop to solve this problem instead of a while loop. what conditional could we write to make sure the loop does not go out of bounds?
- wordList.get(index)
- Correct!
- wordList[index]
- This accessor method doesn't work for arrayLists.
- wordList(index)
- This accessor method doesn't work for arrayLists.
- wordList.at(index)
- This accessor method doesn't work for arrayLists.
7-4-1-6: How would you access each element in wordList assuming you already have an integer index properly initialized.
- !(word != other)
- This checks to make sure that word does not equal a different space in memory that other.
- word.size() == other.size()
- This only checks the size of the strings, it does not check for equality
- word == other
- This checks the actual addresses in memory of the strings, not their contents.
- word.equals(other)
- Correct!
7-4-1-7: How would you check that a string word
is equal to a different string called other
?
8.4.1.2. The Algorithm¶
Loop through the list and scramble the current word. If the scrambled word and original are equal then remove the word from the list and otherwise replace it. We will have to be careful since the size of the list can change in the loop. If we remove an element all the other elements will shift left. We will only want to increment the index if the word was replaced and not removed. There are many ways to solve this problem but we have outlined 2 in the following optional questions. If you feel that you are ready to solve the problem, please skip ahead to the active code block.
public static void scrambleOrRemove(List<String> wordList)
initialize index counter
while (index less than wordlist size)
initialize a string and set it equal to word in wordList at index
initialize another string and set it equal to the scrambled version
of the word in wordlist at index
if (the normal string equals the scrambled string)
remove the word in wordList at the current index
else
reassign the current word in wordList to be the scrambled version
iterate the index
The method test below contains the correct code for one solution to this problem, but it is mixed up. Drag the needed code from the left to the right and put them in order with the correct indention so that the code would work correctly.
Another way to solve this problem is to start at the end of the list and loop towards the front of the list. That way you don’t have to worry about the index being off if you remove an item from the list.
public static void scrambleOrRemove(List<String> wordList)
for( int i = wordList size - 1; i >= 0; i--)
initialize a string and set it equal to word in wordList at index
initialize another string and set it equal to the scrambled version
of the word in wordlist at index
if (the normal string equals the scrambled string)
remove the word in wordList at the current index
else
reassign the current word in wordList to be the scrambled version
iterate the index
The method test below contains the correct code for another solution to this problem, but it is mixed up. Drag the needed code from the left to the right and put them in order with the correct indention so that the code would work correctly.
8.4.1.3. Try and Solve It¶
Write the method scrambleOrRemove
below. The main has code to test the result.