The `Skyview`

problem from the 2013 AP Computer Science Exam is typical of free response problems that test 2 dimensional arrays.

`SkyView`

is #4 from the from the 2013 AP Computer Science A Free Response problems.

https://secure-media.collegeboard.org/digitalServices/pdf/ap/apcentral/ap13_frq_comp_sci.pdf

## Part (a) `SkyView`

constructor

```
public SkyView(int numRows, int numCols, double[] scanned)
{
view = new double[numRows][numCols];
int scannedIndex = 0;
for(int row = 0; row < numRows; row++)
{
for(int col = 0; col < numCols; col++)
{
if(row % 2 == 0)
view[row][col] = scanned[scannedIndex];
else
view[row][numCols - col - 1] = scanned[scannedIndex];
scannedIndex++;
}
}
}
```

Transferring elements from one array to another requires traversing both structures. `SkyView`

traverses the 1D array (`scanned`

) in conventional order from front to back. `SkyView`

uses conventional nested loops to traverse the 2D array (`view`

) as well.

Within each row of the 2D array, the position of the next element is determined based on whether `row`

is even or odd. If `row`

is even, the row is traversed in conventional order. If `row`

is odd, the loop proceeds in conventional order but the elements are placed into the row from right to left.

See 2D array exercises for additional traversals and other challenges.

## Part (a) SkyView constructor (alternate solution)

```
public SkyView(int numRows, int numCols, double[] scanned)
{
view = new double[numRows][numCols];
int scannedIndex = 0;
for(int row = 0; row < numRows; row++)
{
if(row % 2 == 0)
{
for(int col = 0; col < numCols; col++)
{
view[row][col] = scanned[scannedIndex];
scannedIndex++;
}
}
else
{
for(int col = numCols - 1; col >= 0; col--)
{
view[row][col] = scanned[scannedIndex];
scannedIndex++;
}
}
}
}
```

It is possible to reverse the order of the loop that traverses each row (the loop with `col`

as the control variable). This is an alternative to calculating the position within the inner loop.

## Part (b) `getAverage`

method

```
public double getAverage(int startRow, int endRow, int startCol, int endCol)
{
double sum = 0;
int count = 0;
for(int row = startRow; row <= endRow; row++)
{
for(int col = startCol; col <= endCol; col++)
{
sum += view[row][col];
count++;
}
}
return sum / count;
}
```

Conventional nested loops are used to traverse the 2D array. The bounds of each loop are the starting and ending indexes specified by the parameters.

As an alternative to maintaining `count`

it is possible to compute the number of elements as `(endRow - startRow + 1) * (endCol - startCol + 1)`

.

## 2013 AP CS Exam Free Response Solutions

- MusicDownloads Free Response Solution
- TokenPass Free Response Solution
- GridWorldUtilities & JumpingCritter Free Response Solution