Tips and Solutions for AP Computer Science Free Response Problems

1. Some Practice Free ResponseProblems (For CS III AP) March – April 2014

2. Some tips for these Problems… • Focus on what you need to return in each method • Don’t neglect to read the class(es) on which the problem relies • Use the Given Appendix when Needed • (A Better solution is to Memorize the methods for Array, String, Actor, etc.) • Practice makes Perfect

3. Ex 1: What We’re Given…

4. What We’re Given…(cont.) public class StringCoder { private String masterString; /** @param master the master string for the StringCoder * Precondition: the master string contains all the letters of the alphabet */ public StringCoder(String master) { masterString = master; } /** @param parts an ArrayList of string parts that are valid in the master string * Precondition: parts.size()> 0 * @return the string obtained by concatenating the parts of the master string */ public String decodeString(ArrayList<StringPart> parts) { /* to be implemented in part (a) */ } This is what you code

5. Part A: One Solution String result = new String (“”); for (intind = 0; ind < parts.size(); ind ++) { StringPartthisUnit = parts.get(ind); result += masterString.substring (thisUnit.getStart(), thisUnit.getStart()+thisUnit.getLength()); } return result;

6. Part B /** @paramstr the string to encode using the master string * Precondition: all of the characters in str appear in the master string; * str.length() > 0 * @return a string part in the master string that matches the beginning of str . * The returned string part has length at least 1. */ private StringPartfindPart(String str) { /* implementation not shown */ } /** @param word the string to be encoded * Precondition: all of the characters in word appear in the master string; * word.length() > 0 * @return an ArrayList of string parts of the master string that can be combined * to create word */ public ArrayList<StringPart> encodeString(String word) { /* to be implemented in part (b) */ } // There may be instance variables, constructors, and methods that are not shown. }

7. Part B: One Solution String remaining = word; int limit = word.length(); intcurrentInd = 0; ArrayList<StringPart> units = new ArrayList<StringPart>(); while (currentInd < limit) { StringPartthisUnit = findPart (remaining); currentInd += thisUnit.getLength(); remaining = remaining.substring (thisUnit.getLength()); units.add (thisUnit); } return units; ← Could just as easily be (!remaining.equals(“”))

8. Another Interesting Free Response Problem… • 2011 AB #4 – Using a 2-D array and Strings 

10. Part A: One Solution private void fillBlock (String str) { if (str.length() > numRows*numCols) { for (int row = 0; row < numRows; row++) { for (intcol = 0; col < numCols; col++) { letterBlock[row][col] = str.substring (numRows*row + col, numRows*row + col+1); } } } else { for (int row = 0; row < numRows; row++) { for (intcol = 0; col < numCols; col++) { if ((str.substring (numRows * row + col, numRows * row + col + 1).equals (“”))) { letterBlock[row][col] = “A”; } else { letterBlock[row][col] = str.substring (numRows * row + col, numRows * row + col + 1); } } } } }

11. Part B

12. Thank you very much • Please come next week with your solution(s) • We will check back then • Let me know if you have any questions • This ppt will be posted (like the last one) • At http://mthcompsci.wordpress.com/ • With some other sources for Free Response Questions • See you next week! Good luck.