scrambleWord and scrambleOrRemove free response answer 11

scrambleWord and scrambleOrRemove free response problem from the 2014 AP Computer Science A Exam.

This problem is #1 from the from the 2014 AP Computer Science A Free Response problems.

Part (a) – scrambleWord

public static String scrambleWord(String word)
  for(int i = 1; i < word.length(); i++)
    if("A".equals(word.substring(i-1, i)) &&
        ! "A".equals(word.substring(i, i+1)))
      word = word.substring(0, i-1) +
             word.substring(i, i+1) +
             word.substring(i-1, i) +
    return word;

Part (b) – scrambleOrRemove

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

11 thoughts on “scrambleWord and scrambleOrRemove free response answer

  1. Henry May 10,2014 11:21 am

    Hi is it okay if I made mine recursive?
    On the AP test I had a couple cases for a but I’m sure it worked

  2. Edward May 10,2014 7:48 pm

    Hello im just wondering if there would be point reductions for using “=” or “!=” instead of the “equals”

    • Brandon Horn May 11,2014 9:59 am

      While I don’t have the scoring guidelines there is typically a penalty for incorrectly using == to compare String objects. If you obtained char types using charAt then == would be correct.

  3. syed May 11,2014 9:49 am

    is the “else wordList.set(i, scrambled);” necessary?

    • Brandon Horn May 11,2014 10:00 am

      Yes. String objects are immutable. Scrambling the word obtained from the list does not change the word in the list. the set call is what actually changes the list.

  4. Ashna May 20,2014 11:31 pm

    For part B, was it necessary to go backwards in the loop?

    • Brandon Horn May 21,2014 6:29 am

      You must avoid skipping the next element each time you remove an element from the list. You can do that by going backwards or by decrementing your loop control variable every time you remove an element.

  5. CDS May 5,2015 12:49 pm

    I believe your solution to part B is incorrect. You must decrement i after you call wordList.remove(i), otherwise, the new element in position i will not be evaluated.

    • Brandon Horn May 5,2015 4:28 pm

      My loop traverses wordList backwards. The elements that are shifted by the call to remove have already been processed. If the loop traversed the list starting at index 0 a decrement would be necessary.

    • Grady Wetherbee Jul 2,2015 1:29 pm

      Whether or not the element at the index is removed, the index decremement will move to the next (lower) index and test that. If an element is removed, the element that was at an index one greater will slide into the index of the element that was just removed. This will not matter, because the index of the for loop will already be ahead (one less numerically because we are decrementing) of the elements that are sliding because of removals. A way that may be easier to comprehend is using for(int i = 0; i < wordList.size(); ) and only incrementing when the tested element is not removed. Then the for loop would only test the 0 index and let new elements slide in as “bad” ones are removed. If the element isn’t removed, increment and change the “test” index to one larger and let more slide in from there.
      Hope this clarifies!

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