Exam 1z0-809 topic 1 question 196 discussion

Answer is C, tested

Sorting When sorting in alphabetical order, Java starts with the first character of each string: First character: 1 (for "100, Robin, HR") First character: 1 (for "101, Peter, HR") First character: 2 (for "200, Mary, AdminServices") Since "100, Robin, HR" and "101, Peter, HR" start with 1, Java compares them further: "100, Robin, HR" (less, because after 1 it has 0) "101, Peter, HR" (more, because after 1 it has 01) Finally, when sorting elements from the same group, the sort compares the characters one by one, which gives the order: "100, Robin, HR" "101, Peter, HR" "200, Mary, AdminServices" Final Order The final order after sorting is: Copy code 100, Robin, HR 101, Peter, HR 200, Mary, AdminServices

Answer is c tested. import java.util.Arrays; import java.util.List; public class Main { public static void main(String[] args) { List<String> empDetails = Arrays.asList( "100, Robin, HR", "200, Mary, AdminServices", "101, Peter, HR" ); empDetails.stream() .filter(s -> s.contains("r")) .sorted() .forEach(System.out::println); } }

C is correct.

Answer is C.