Data Collections: Unit Diagnostic (Computer Science A Unit 4)

Answer: (B). Collecting and storing personal data puts personal privacy at risk, especially when it is kept indefinitely. Programmers should try to safeguard the privacy of the user.

• (A) and (C) are engineering costs, not privacy risks.
• (D) is an accuracy detail, not a privacy risk.

If you missed this: Topic 4.1 study guide.

Answer: (B). Algorithmic bias is a systemic and repeated error in a program that creates unfair outcomes for a specific group of users. Here the data set’s collection method made it unrepresentative.

• (A) The code compiles and runs; the issue is the unfair results.
• (C) No exception is involved.
• (D) Autoboxing is a conversion between primitives and wrapper objects.

If you missed this: Topic 4.1 study guide.

Answer: (A). When analysing a data set, values are accessed one at a time and processed according to the desired outcome; here each is tested and counted.

• (B) computes an unrelated quantity.
• (C) gives a median, not a count.
• (D) destroys the data.

If you missed this: Topic 4.2 study guide.

Answer: (B). A new int array is all 0s, so a[0] is 0. a[3] is 5 + 2 = 7 and a.length is 4: 0 + 7 + 4 = 11.

• (A) leaves out the length.
• (C) treats a[0] as 1.
• (D) The indices 0, 1 and 3 are valid for length 4.

If you missed this: Topic 4.3 study guide.

Answer: (A). The enhanced for loop variable v is a copy of each element, so assigning to it does not change the array. n[0] + n[2] is 1 + 3 = 4.

• (B) assumes the elements were doubled (2 + 6).
• (C) and (D) do not follow from the array contents.

If you missed this: Topic 4.4 study guide.

Answer: (B). The consecutive pairs are (4, 9) no, (9, 2) yes, (2, 9) no, (9, 7) yes: two decreases.

• (A) misses a decrease.
• (C) counts a non-decrease.
• (D) is the number of pairs.

If you missed this: Topic 4.5 study guide.

public static int countAboveMean(int[] a)
{
    int sum = 0;
    for (int v : a)
    {
        sum += v;
    }
    double mean = (double) sum / a.length;
    int count = 0;
    for (int v : a)
    {
        if (v > mean)
        {
            count++;
        }
    }
    return count;
}

For {70, 85, 60, 90} the mean is 76.25 and the method returns 2; for {5, 5, 5} it returns 0 (both run).

Suggested Marlbridge rubric (4 points)

If you missed this: Topic 4.5 study guide.

Answer: (C). The loop reads one token per pass across the line break: 3, 8, 1 give 12 (run and confirmed). The enclosing method needs throws IOException.

• (A) reads the first token only.
• (B) stops before the last line.
• (D) joins the digits as text.

If you missed this: Topic 4.6 study guide.

Answer: (B). a is unboxed to an int, so a / 2 is integer division (3), which is then widened to the double 3.0.

• (A) would need a cast such as (double) a / 2.
• (C) A double prints with a decimal point.
• (D) Unboxing and widening are automatic.

If you missed this: Topic 4.7 study guide.

Answer: (A). The list goes [x], [x, y], [z, x, y]. set(1, "w") returns "x" and gives [z, w, y]. remove(0) leaves [w, y]. The size is 2 and get(0) is "w", so 2 + "w" is "2w".

• (B) returns the wrong old element.
• (C) mixes up the final list and adds a String as a number.
• (D) uses size 3.

If you missed this: Topic 4.8 study guide.

Answer: (B). At i = 0 the first 2 is removed ([2, 3, 2]) and the next 2 slides into index 0, but i moves on to 1, skipping it. 3 is kept at i = 1; at i = 2 the last 2 is removed. Result [2, 3] (run and confirmed).

• (A) is what a correct deletion loop would give.
• (C) and (D) do not follow from the passes.

If you missed this: Topic 4.9 study guide.

Answer: (A). sum is 5 and av.size() is 3. Both are int, so the division is integer division (1), widened to 1.0 on assignment. A cast (double) sum / av.size() would give 1.6666666666666667.

• (B) and (C) need the cast.
• (D) is the sum.

If you missed this: Topic 4.10 study guide.

public static void moveMaxToEnd(ArrayList<Integer> list)
{
    int maxIndex = 0;
    for (int i = 1; i < list.size(); i++)
    {
        if (list.get(i) > list.get(maxIndex))
        {
            maxIndex = i;
        }
    }
    list.add(list.remove(maxIndex));
}

Strict > keeps the first occurrence. The run on [3, 9, 2, 9, 5] gave [3, 2, 9, 5, 9].

Suggested Marlbridge rubric (4 points)

If you missed this: Topic 4.10 study guide.

(a) g.length is 3 and g[0].length is 2, so 3 * 2 + 4 = 10 (run and confirmed).

(b) g.length is the number of rows (the length of the outer array). g[0].length is the number of columns (the length of one row array).

Suggested Marlbridge rubric (3 points): 1 for 10; 1 for rows; 1 for columns.

If you missed this: Topic 4.11 study guide.

Answer: (B). The outer loop is over the 2 columns and the inner over the 3 rows: column 0 gives 1, 3, 5 and column 1 gives 2, 4, 6 (column-major order).

• (A) is row-major order.
• (C) visits the columns in reverse.
• (D) stops early.

If you missed this: Topic 4.12 study guide.

Answer: (B). The column sums are 1 + 4 + 2 = 7 and 5 + 3 + 9 = 17. The larger is column 1, and the method returns the column index.

• (A) is the first column.
• (C) is the largest sum, not the index.
• (D) is the sum of column 0.

If you missed this: Topic 4.13 study guide.

public static void rotateRowRight(int[][] m, int r)
{
    int last = m[r][m[r].length - 1];
    for (int c = m[r].length - 1; c > 0; c--)
    {
        m[r][c] = m[r][c - 1];
    }
    m[r][0] = last;
}

The run on { {1, 2, 3, 4}, {5, 6, 7, 8} } with r = 0 gave 4 1 2 3 and left row 1 unchanged.

Suggested Marlbridge rubric (4 points)

A loop that shifts from the front without a saved copy of each element loses values; do not award point 2 for it.

If you missed this: Topic 4.13 study guide.

Answer: (B). The target is absent, so every element is checked (4 comparisons) and -1 is returned after the loop.

• (A) stops after the first mismatch, which is wrong.
• (C) returns the length, not -1.
• (D) misses the last element.

If you missed this: Topic 4.14 study guide.

Answer: (B). Pass 1 gives 4, 7, 9, 2, 5; pass 2 leaves it unchanged; pass 3 inserts 2 at the front by shifting 9, 7 and 4: 2, 4, 7, 9, 5 (run and confirmed).

• (A) is the array after three passes of selection sort (and the fully sorted array).
• (C) is the state after one or two passes.
• (D) is the state after one pass of selection sort.

If you missed this: Topic 4.15 study guide.

Answer: (B). The calls are f(20), f(10), f(5), f(2) and f(1). The base case returns 0 and each of the four earlier calls adds 1: 4 (run and confirmed). The method counts how many times n can be halved (integer division) before it reaches 1.

• (A) is 20 / 2, only the first step.
• (C) counts the five calls rather than the four additions.
• (D) is the base-case value, not the result of the first call.

If you missed this: Topic 4.16 study guide.

Answer: (A). mid is 4 (21 < 27, so low = 5); then mid = (5 + 8) / 2 = 6 (30 > 27, so high = 5); then mid = 5 (27, found). Three comparisons (run and confirmed).

• (B) assumes the second mid lands on 5 directly; it is 6.
• (C) uses 7, the wrong rounding of the midpoint of 5 and 8.
• (D) is the order of a linear search.

If you missed this: Topic 4.17 study guide.

1. split [6, 2, 9, 4] into [6, 2] and [9, 4]
2. split [6, 2] into [6] and [2]
3. merge [6] and [2] into [2, 6]
4. split [9, 4] into [9] and [4]
5. merge [9] and [4] into [4, 9]
6. merge [2, 6] and [4, 9] into [2, 4, 6, 9]

The last merge compares the fronts: 2 and 4 (copy 2), 6 and 4 (copy 4), 6 and 9 (copy 6), then copy the remaining 9.

Suggested Marlbridge rubric (5 points): 1 for the first split into halves; 1 for splitting both halves down to single elements; 1 for the two small merges ([2, 6] and [4, 9]); 1 for the final merge giving [2, 4, 6, 9]; 1 for doing the left half completely before the right half.

If you missed this: Topic 4.17 study guide.