scrambleWord and scrambleOrRemove free response answer 2014

scrambleWord and scrambleOrRemove is #1 from the from the 2014 AP Computer Science A Free Response problems.

2014 AP CS A Free Response (Internet Archive)

Part (a) scrambleWord method

public static String scrambleWord(String word)
{
    String scrambledWord = word;

    int i = 1;
    while(i < scrambledWord.length())
    {
        if("A".equals(scrambledWord.substring(i - 1, i)) &&
                ! "A".equals(scrambledWord.substring(i, i + 1)))
        {
            scrambledWord =
                    scrambledWord.substring(0, i - 1) +
                    scrambledWord.substring(i, i + 1) +
                    "A" +
                    scrambledWord.substring(i + 1);

            i += 2;
        }
        else
            i++;
    }

    return scrambledWord;
}

Although String objects are immutable, it is possible to pretend that they are mutable by making a new String object and setting the existing variable to point to it. This approach makes this problem much simpler.

See Strings on the AP CS A Exam for an explanation of String methods and String concatenation.

An alternative approach is to build a new String by concatenating parts of the existing String. This is difficult here, especially with respect to the end of the new String.

A recursive approach is also reasonable here (see below).

Part (a) scrambleWord (recursive approach)

public static String scrambleWord(String word)
{
    if(word.length() <= 1)
        return word;

    if(word.substring(0, 1).equals("A") && ! word.substring(1, 2).equals("A"))
        return word.substring(1, 2) + "A" + scrambleWord(word.substring(2));
    else
        return word.substring(0, 1) + scrambleWord(word.substring(1));
}

Part (a) scrambleWord (alternate approach)

public static String scrambleWord(String word)
{
    String scrambled = "";

    int i = 0;
    while(i < word.length() - 1)
    {
        if(word.substring(i, i + 1).equals("A") && ! word.substring(i + 1, i + 2).equals("A"))
        {
            scrambled += word.substring(i + 1, i + 2);
            scrambled += "A";
            i += 2;
        }
        else
        {
            scrambled += word.substring(i, i + 1);
            i++;
        }
    }

    if(scrambled.length() < word.length())
        scrambled += word.substring(word.length() - 1);

    return scrambled;
}

This approach builds a new string from scratch rather than pretending to mutate the existing string.

Part (b) scrambleOrRemove method

public static void scrambleOrRemove(List<String> wordList)
{
    for(int i = wordList.size() - 1; i >= 0; i--)
    {
        String scrambled = scrambleWord(wordList.get(i));
        if(wordList.get(i).equals(scrambled))
            wordList.remove(i);
        else
            wordList.set(i, scrambled);
    }
}

See ArrayList practice for details on adding to and removing from an ArrayList within a loop.

2014 AP CS Exam Free Response Solutions

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