WordMatch free response 2

8 May 2021   | 2021 AP Computer Science Exam Answers

WordMatch free response problem from the 2021 AP Computer Science A Exam.

WordMatch is #1 from the from the 2021 AP Computer Science A Free Response problems.

https://apcentral.collegeboard.org/pdf/ap21-frq-computer-science-a.pdf?course=ap-computer-science-a

Part (a) – scoreGuess method

public int scoreGuess(String guess)
{
  int count = 0;

  for(int i = 0; i < secret.length(); i++)
  {
    int j = i + guess.length();

    if(j <= secret.length() && secret.substring(i, j).equals(guess))
      count++;
  }

  return count * (guess.length() * guess.length());
}

I would normally have used indexOf for this problem; however, the description mentioned that the occurrences of guess might overlap (and the second example showed such a situation). This approach seemed more reliable to me.

Part (b) – findBetterGuess method

public String findBetterGuess(String guess1, String guess2)
{
  int score1 = scoreGuess(guess1);
  int score2 = scoreGuess(guess2);
  
  if(score1 > score2)
    return guess1;
  else if(score2 > score1)
    return guess2;
  else
  {
    if(guess1.compareTo(guess2) > 0)
      return guess1;
    else
      return guess2;
  }
}

2021 AP CS Exam Free Response Solutions

2 thoughts on “WordMatch free response”

  1. Dhruv Kaushik May 12,2021 10:25 am

    For part a, it is not required to check if guess’s length is equal to secret’s length?

    • Brandon Horn May 12,2021 10:40 am

      The precondition specified that guess.length() was at least 1 and no longer than secret.length(). You are still required to ensure that you don’t go out of bounds.

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