CODEWITHSUNDEEP
javaarrayssortingmaps
user8569908
user8569908

Reputation:

Java sort second array depending on indexes of first array

I have two int arrays

int[] sum=new int[n];
int[] newTime=new int[n];
  1. First: 1 5 3
  2. Second 10 15 13

Arrays.sort(sum);

  1. Prints 1 3 5

What I want is even for the second array to be sorted with the same indexes of the

  1. first=>second : 10 13 15

I have tried with maps:

SortedMap<Integer, Integer> m = new TreeMap<Integer, Integer>();
        for(int i = 0; i < priorities.length; i++){
            m.put(sum[i],newTime[i]);
        }

It just sorts the first array only,the indexes of the second array don't change.Help is appreciated! Thank You!

Upvotes: 4

Views: 1255

Answers (4)

Eugene
Eugene

Reputation: 120848

You could do it with java-8 like this for example:

    int[] left = new int[] { 1, 5, 3 };
    int[] right = new int[] { 10, 15, 13 };

    IntStream.range(0, left.length)
            .boxed()
            .map(x -> new AbstractMap.SimpleEntry<>(left[x], right[x]))
            .sorted(Comparator.comparing(SimpleEntry::getKey))
            .forEach(System.out::println);

EDIT

to actually get the second array:

Integer[] second = IntStream.range(0, left.length)
            .boxed()
            .map(x -> new AbstractMap.SimpleEntry<>(left[x], right[x]))
            .sorted(Comparator.comparing(SimpleEntry::getKey))
            .map(SimpleEntry::getValue)
            .toArray(Integer[]::new);

Upvotes: 2

Arnaud
Arnaud

Reputation: 17524

You may sort both arrays at the same time, following the reordering of the first array like this :

int[] sum = { 1, 5, 3 };
int[] newTime = { 10, 15, 13 };

for (int i = 0; i < sum.length; i++) {
    for (int j = 0; j < sum.length; j++) {
        if (sum[i] < sum[j]) {

            int temp = sum[i];
            int temp2 = newTime[i];

            sum[i] = sum[j];
            sum[j] = temp;

            newTime[i] = newTime[j];
            newTime[j] = temp2;
        }
    }
}

System.out.println(Arrays.toString(sum));
System.out.println(Arrays.toString(newTime));

Upvotes: 0

mlecz
mlecz

Reputation: 1026

Here is the solution, which works when there are duplicates in first array. You keep values from both arrays together, sort them using indices from one array, and make a list, from corresponding indices from second array.

  static Integer[] sort(int[] arr1, int[] arr2) {
    assert arr1.length == arr2.length;
    class Tuple{
      int a1;
      int a2;
    }

    List<Tuple> tuples = new ArrayList<>();
    for(int i=0;i<arr1.length;i++){
      Tuple t = new Tuple();
      t.a1=arr1[i];
      t.a2=arr2[i];
      tuples.add(t);
    }
    tuples.sort((t1,t2)->Integer.compare(t1.a1,t2.a1));

    return (Integer[]) tuples.stream().map(t->t.a2).collect(Collectors.toList()).toArray(new Integer[arr1.length]);
  }

Upvotes: 0

Eran
Eran

Reputation: 393781

Your TreeMap approach leads to what you need:

SortedMap<Integer, Integer> m = new TreeMap<Integer, Integer>();
for(int i = 0; i < priorities.length; i++){
    m.put(sum[i],newTime[i]);
}
// this will print the elements of the second array in the required order
for (Integer i : m.values()) {
    System.out.println (i);
}

Of course you can assign the elements back to the original array if you want:

int count = 0;
for (Integer i : m.values()) {
    newTime[count] = i;
}

As correctly commented by mlecz, this solution will only work if the first array (sum) has no duplicates.

Upvotes: 1

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