### All Computer Science Resources

## Example Questions

### Example Question #1 : Implementation Techniques

**Consider the following code:**

int[] vals = {6,1,41,5,1};

int[][] newVals = new int[vals.length][];

for(int i = 0; i < vals.length; i++) {

newVals[i] = new int[vals[i]];

for(int j = 0; j < vals[i];j++) {

newVals[i][j] = vals[i] * (j+1);

}

}

What does the code above do?

**Possible Answers:**

It creates a 2D matrix with the *vals* array for every row.

It sums the values in *vals*, placing the outcome in *newVals*.

The code creates a 2D array that contains the multiples for the members of *vals,* using each of those values to determine the number of multiples to be computed.

The code creates a 2D array that contains all multiples from 1 to 5 for the members of *vals.*

It performs a matrix multiplication on *vals* and *newVals*, storing the final result in *newVals*.

**Correct answer:**

The code creates a 2D array that contains the multiples for the members of *vals,* using each of those values to determine the number of multiples to be computed.

Let's look at the main loop in this program:

for(int i = 0; i < vals.length; i++) {

newVals[i] = new int[vals[i]];

for(int j = 0; j < vals[i];j++) {

newVals[i][j] = vals[i] * (j+1);

}

}

The first loop clearly runs through the *vals* array for the number of items in that array. After this, it creates in the *newVals* 2D array a second dimension that is as long as the value in the input array *vals*. Thus, for 41, it creates a 3rd row in *newVals* that is 41 columns wide.

Then, we go to the second loop. This one iterates from 0 to the value stored in the current location in *vals*. It places in the given 2D array location the multiple of the value. Think of *j + 1*. This will go from 1 to 41 for the case of 41 (and likewise for all the values). Thus, you create a 2D array with all the multiples of the initial *vals* array.