Sorting second array accordingly to the indexs of first sorted array

I have two arrays with same size.

ArrayList<ProfileClass> array1= new ArrayList<>()
ArrayList<Double> array2= new ArrayList<>()

array1 = {b,d,a,c}
array2 = {2.0,4.0,1.0,3.0}

Now I need to sort array2 in descending order.
Collections.sort(array2, Collections.reverseOrder());
Expected result: {4.0,3.0,2.0,1.0}

Therefore, indexes of these values before sorting against each value is not same as it is after sorting. However the index are changed against each value in array2, I want array1 to change indexes against each value accordingly to array2.

Final Expected result of array1: {d,c,b,a}

You can get the result by sorting the indexes (0, 1, 2, 3) by the value of array2 and mapping the result to the value of array1.

System.out.println("array1=" + array1);
System.out.println("array2=" + array2);
ArrayList<ProfileClass> sortedArray1 = IntStream.range(0, array2.size())
    .boxed()
    .sorted(Collections.reverseOrder(Comparator.comparingDouble(i -> array2.get(i))))
    .map(i -> array1.get(i))
    .collect(Collectors.toCollection(ArrayList::new));
System.out.println("sorted array1=" + sortedArray1);

output:

array1=[b, d, a, c]
array2=[2.0, 4.0, 1.0, 3.0]
sorted array1=[d, c, b, a]

Or you can also do like this.

ArrayList<ProfileClass> sortedArray1 = IntStream.range(0, array2.size())
    .mapToObj(i -> new Object() {
        double sortKey = array2.get(i);
        ProfileClass value = array1.get(i);
    })
    .sorted(Collections.reverseOrder(Comparator.comparingDouble(obj -> obj.sortKey)))
    .map(obj -> obj.value)
    .collect(Collectors.toCollection(ArrayList::new));

You should create a temp Double arrayList. And you should keep the indexes of array2 before sorting.

expected tmpArrayList = {1, 3, 0, 2}

then, with these indexes, you can recreate array1.

If you don't have to use array1 as linkedlist, it is better array1 = new ProfileList[]; for this solution.

Here, i want to say, why is this necessary, you can just put another field to your profileclass object such as array2Place?? and update it. I don't get what are you trying to do here actually. Such as, is array2 index of something about array1?

Create a Map using both arrays and get the first array (map value array 1) from the map using the key (Array 2).

//Map it
Map<Double, ProfileClass> map = IntStream.range(0, array2.size()).boxed()
	    .collect(Collectors.toMap(i -> array2.get(i), i -> array1.get(i)));
Collections.sort(array2, Collections.reverseOrder());

//Create new array for array1
List<ProfileClass> array11 = new ArrayList<>();
for (Double d : array2) {
	array11.add(map.get(d));
}

System.out.println(array11);

On the assumption that both arrays (especially the first) can contain duplicates I would suggest the following solution

ProfileClass[] arr1 = new String[] { b,d,a,c };
double[] arr2 = new double[] { 2.0,4.0,1.0,3.0 };

List<Tuple<ProfileClass, Double>> list 
  = IntStream.range(0, arr2.length) // create a stream of all possible indexes
  .mapToObj(i -> new Tuple<ProfileClass, Double>(arr1[i], arr2[i])) // zip the two arrays by the index
  .sorted((a,b) -> Double.compare(a.getItem2(), b.getItem2())) // sort the stream according to the second array
  .collect(Collectors.toList());

// write the result back to the array
for (int i = 0; i < arr1.length; i++) {
  arr1[i] = list.get(i).getItem1();
  arr2[i] = list.get(i).getItem2();
}

here the class Tuple is defined as:

public class Tuple<T1, T2> {
  private T1 item1;
  private T2 item2;

  public Tuple(T1 item1, T2 item2) {
      this.item1 = item1;
      this.item2 = item2;
  }

  public T1 getItem1() { return item1; }
  public T2 getItem2() { return item2; }
}

Simple to use TreeMap for sorting orders quickly.

TreeMap<Double, ProfileClass> map = new TreeMap<>();
    for (int i = 0; i < array1.size(); i++) {
        map.put(array2.get(i), array1.get(i));
    }
    final Collection< ProfileClass> sortedArray = map.descendingMap().values();
    System.out.println(sortedArray);

If you are able to have class which holds both values only one sort is required.

In case this is not possible please check the solution from @NiravHR, seems legit.

public class Tester {

    static class Holder implements  Comparable<Holder> {

        private Holder(String letter, Double value){
            this.letter = letter;
            this.value  = value;
        }

        String letter;

        Double value;

        static Holder of(String letter, Double value){
            return new Holder(letter,value);
        }

        @Override
        public int compareTo(Holder o) {
            return value.compareTo(o.value);
        }

        @Override
        public String toString() {
            return String.valueOf(this.letter);
        }
    }


    public static void main(String[] args) {

       List<Holder> data = Arrays.asList(
         Holder.of("b",2.0),
         Holder.of("d",4.0),
         Holder.of("a",1.0),
         Holder.of("c",3.0)
       );
       Collections.sort(data, Collections.reverseOrder());


       System.out.println(data);
    }
}