Collegeboard 2015 Practice Exam FRQ 1 • 5 min read

1D & 2D Arrays

• This FRQ was helpful in a lot of ways because it made me go back to the basics and also made me consider new ways to use what I know. The Key Algorithm was 2D Arrays because we used them in part b and c. The 2D arrays were converted into 1D arrays by summing the rows and this conversion was very interesting to me. This is why I think the Key Algorithm is 2D Arrays due to them being used in this FRQ.

Declaration of Data

• I had to remove the declaration of data because it had some issues with liquid. Need to find a way to declare data without having issues.

Part A

// Returns the sum of the entries in the one-dimensional array arr.

public static int arraySum(int[] arr) {
    
    int sum = 0;

    for(int n : arr) {
        sum = sum + n;
    }

    return sum;
}

System.out.print(arraySum(arr1)); // Should return 16

16

Part B

// Returns a one-dimensional array in which the entry at index k is the sum of the entries of row k of the two-dimensional array arr2D.

public static int [] rowSums(int[][] arr2D) {

    int[] newArray = new int[arr2D.length];

    for (int k = 0; k < arr2D.length; k++) {
        newArray[k] = arraySum(arr2D[k]);
    }

    return newArray;
}

System.out.println(Arrays.toString(rowSums(arr2D1))); // Should return [16, 32, 28, 20]
System.out.print(Arrays.toString(rowSums(arr2D2))); // Should return [16, 28, 28, 20]

[16, 32, 28, 20]
[16, 28, 28, 20]

Part C

// Returns true if all rows in arr2D have different row sums; false otherwise.

public static boolean isDiverse(int[][] arr2D) {

    int [] diverseArray =  rowSums(arr2D);

    for (int i = 0; i < diverseArray.length; i++) {
        for (int j = i+1; j < diverseArray.length; j++) {
            if (diverseArray[i] == diverseArray[j]) {
                return false;
            }
        }
    }
    
    return true;
}

System.out.println(isDiverse(arr2D1)); // Should return true
System.out.print(isDiverse(arr2D2)); // Should return false

true
false