Answer Key: ArrayList Review Handouts

Handout 1: Warm-Up (What’s in the List?)

# Answer
1 [4, 10, 19, 20, 4]
2 words = [yellow, blue, green, purple, blue]; old = "red"
3 list = [20, 15, 10, 25, 5]; a = 15
4 items = [E, D, C, B]; x = "B"; y = "D"
5 [2, 10, 6, 16]
6 [3, 6, 9]
7 [2, 1, 0, 2, 4, 6, 8, 10, 12] (each add(0, i) pushes the earlier ones right, so the new values appear in reverse)
8 [8, 10, 12, 3, 5, 7]
9 [2, 4]
10 [2, 4, 8, 12]

Notes

  • Problem 2: set returns the old value, so the inner words.set(2, "purple") returns "blue", and that is what gets stored at index 3.
  • Problem 4: set(2, x) returns "D", which is y. The last line removes "E" from the end and stores it at index 0.
  • Problem 5: data.get(0) - 1 is an int (2 - 1 = 1), so this is remove(1), which takes an index.
  • Problems 9 and 10: Same condition, opposite directions. Going forward, each remove(i) slides the next element into index i, then i++ skips it. In Problem 10, 8 and 12 are never tested, and 10 is only tested after 8 has slid down.

Handout 2: Trace These

Problem 4: Prints

Alex Bob Carl 
Alex Alex Alex

set returns the old value, so the first loop prints the original names while overwriting each with "Alex".

k students before returned by set students after
0 [Alex, Bob, Carl] Alex [Alex, Bob, Carl]
1 [Alex, Bob, Carl] Bob [Alex, Alex, Carl]
2 [Alex, Alex, Carl] Carl [Alex, Alex, Alex]

Problem 7: numList ends as [2, 20, 16] and the method returns [5, 10]. The intended results were [2, 16] and [5, 10, 20].

i size num divisible? numList after returnList after
0 5 5 yes [2, 10, 20, 16] [5]
1 4 10 yes [2, 20, 16] [5, 10]
2 3 16 no [2, 20, 16] [5, 10]

The loop then ends (i = 3, size 3). It first goes wrong at i = 1: after 10 is removed, 20 slides into index 1, but i++ moves on to index 2, so 20 is never checked.

Problem 8: values ends as [0, 4, 2, 5, 3].

k size nums.get(k) removed? nums after
0 8 0 yes [0, 4, 2, 5, 0, 3, 0]
1 7 4 no same
2 7 2 no same
3 7 5 no same
4 7 0 yes [0, 4, 2, 5, 3, 0]
5 6 0 yes [0, 4, 2, 5, 3]

The loop then ends (k = 6, size 5). The method does not remove every 0. After the first removal, the second 0 slid into index 0 and was never examined, so a 0 survives at the front.

Problem 9: myData ends as [3, 4, 4, 8, 7, 7], not [3, 4, 8, 7].

k get(k) get(k - 1) equal? myData after
1 3 3 yes [3, 4, 4, 4, 8, 7, 7, 7]
2 4 4 yes [3, 4, 4, 8, 7, 7, 7]
3 8 4 no same
4 7 8 no same
5 7 7 yes [3, 4, 4, 8, 7, 7]

The loop then ends (k = 6, size 6). After a removal, the next element slides into index k, but k++ skips past it. The fix is to put an else before k++, so k only advances when nothing was removed.

Problem 11: Prints [200, 400]. The code compiles (raw new ArrayList() only produces a warning).

After statement oldList newList
oldList.add(400) [100, 200, 300, 400] []
newList.add(oldList.remove(1)) [100, 300, 400] [200]
newList.add(oldList.get(2)) [100, 300, 400] [200, 400]

remove(1) returns 200 and shifts the rest left, so get(2) is now 400, not 300.

Handout 3: FRQ averageWithinRange

public double averageWithinRange(double lower, double upper)
{
    double sum = 0.0;
    int count = 0;
    for (ItemInfo item : inventory)
    {
        double cost = item.getCost();
        if (item.isAvailable() && cost >= lower && cost <= upper)
        {
            sum += cost;
            count++;
        }
    }
    return sum / count;
}

What to look for

  • Traverses every element of inventory.
  • Tests availability and both bounds, inclusive (>= and <=).
  • Accumulates a sum and a count of only the qualifying items.
  • Returns sum divided by count, as a double.

Common mistakes

  • Using > or <, which excludes the endpoints.
  • Dividing by inventory.size() instead of the count of qualifying items.
  • Declaring sum as an int, which loses the cents.
  • Forgetting the isAvailable() check (the watch is the test case for this).
  • Guarding against count == 0, which the precondition makes unnecessary but is harmless.