Given: ``java import java.util.ArrayList; import java.util.Arrays; public class TestMe { public static void main(String[] args) { String[] fruitNames = {"apple", "orange", "grape", "lemon"…

Question

import java.util.ArrayList;
import java.util.Arrays;

public class TestMe {
 public static void main(String[] args) {
 String[] fruitNames = {"apple", "orange",
 "grape", "lemon", "apricot", "watermelon"};
 var fruits = new ArrayList<>(Arrays.asList(fruitNames));
 fruits.sort((var a, var b) -> -a.compareTo(b));
 fruits.forEach(System.out::println);
 }
}

Options

• Awatermelonorangegrapelemonapricotapple
• Bappleapricotgrapefruit
• Cappleapricotgrapelemonorangewatermelon
• Dlemonappleapricotgrapefruit

• A
  70% (44)
• B
  16% (10)
• C
  5% (3)
• D
  10% (6)

Explanation

Option A is correct because the lambda (var a, var b) -> -a.compareTo(b) negates the natural compareTo result, reversing the sort from ascending to descending alphabetical order. The natural order would be apple, apricot, grape, lemon, orange, watermelon, so the reversed output is watermelon, orange, lemon, grape, apricot, apple - exactly what A shows (the choice text omits line breaks, but println prints each word on its own line).

Option C is wrong because it shows ascending alphabetical order - what you'd get without the negation, i.e., a.compareTo(b) instead of -a.compareTo(b). Options B and D are immediately disqualified because they contain "grapefruit," which is not in the original array at all.

Memory tip: Think of the - before compareTo as a "flip switch" - it inverts the comparison result, which inverts the sort direction. If you see -a.compareTo(b), just reverse the alphabetical order you'd normally expect.

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