Crossword free response answer 2016

Crossword is #3 from the from the 2016 AP Computer Science A Free Response problems.

https://secure-media.collegeboard.org/digitalServices/pdf/ap/ap16_frq_computer_science_a.pdf

Part (a) toBeLabeled method

private boolean toBeLabeled(int r, int c, boolean[][] blackSquares)
{
    if(blackSquares[r][c])
        return false;

    if(r == 0 || blackSquares[r - 1][c])
        return true;

    if(c == 0 || blackSquares[r][c - 1])
        return true;

    return false;
}

It is not acceptable to write the opposite checks and return false. The problem specifies that either or both of the conditions could be true for the square to be labeled.

Outside of an exam setting I would write this as:

return ! blackSquares[r][c] && (r == 0 || blackSquares[r-1][c] || c == 0 || blackSquares[r][c-1]);

Part (b) CrossWord constructor

public Crossword(boolean[][] blackSquares)
{
    puzzle = new Square[blackSquares.length][blackSquares[0].length];

    int num = 1;

    for(int r = 0; r < puzzle.length; r++)
    {
        for(int c = 0; c < puzzle[0].length; c++)
        {
            if(toBeLabeled(r, c, blackSquares))
            {
                puzzle[r][c] = new Square(false, num);
                num++;
            }
            else
                puzzle[r][c] = new Square(blackSquares[r][c], 0);
        }
    }
}

It is very common for students to miss the first line of this constructor. Setting elements in an array (or adding to a list) is not the same thing as initializing the variable.

2016 AP CS Exam Free Response Solutions

Additional 2D array resources

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