Given the code fragment:

What is the result?
Given the code fragment:
What is the result?
The code defines a Comparator that sorts the elements in reverse order using b.compareTo(a). When the array 'towns' is sorted using this Comparator, the array order will be ['paris', 'oman', 'boston', 'bangkok']. The binary search operation, also using this Comparator, will then find 'oman' at index 1 in the sorted array. Therefore, the output will be 1.
C please not that Comparator<String> ms = (a,b)->b.compareTo(a) is not going to give you the same order as Comparator<String> ms = (a,b)->a.compareTo(b) one is descending and the other is ascending
Arrays.sort method is then called with the towns array and the ms comparator, sorting the array in reverse order. The resulting sorted array is ["paris", "oman", "boston", "bangkok"]. The Arrays.binarySearch method is then called with the sorted towns array, the search key "oman", and the ms comparator. Since "oman" is found at index 1 in the sorted array, the method returns 1.
However, there is a compilation error in the code because the Comparator interface is generic and requires type arguments. The correct way to define the ms comparator would be: Comparator<String> ms = (a, b) -> b.compareTo(a); which will print 1 after compilation
However, there is a compilation error in the code because the Comparator interface is generic and requires type arguments. The correct way to define the ms comparator would be:Comparator<String> ms = (a, b) -> b.compareTo(a); which will print 1 after compilation
sort by comparator, and then the same comparator indiques to binary search how element is ordered. so print 1=> paris, oman, boston, bangkok
---- reverse order : Comparator<String> ms = (a,b) -> b.compareTo(a); b.compareTo(a) -1 b < a 1 b > a Comparator int compare (T o1, T o2) -1 01<02 Arrays.sort(towns, ms); returns [paris, oman, boston, bankok] System.out.println(Arrays.binarySearch(towns,"oman",ms)); returns 1 because ms comparator sorting is the same as before (Arrays.sort(towns, ms))