The order that sorts
A report lists items by rank without disturbing the source rows: instead of the values, it wants their positions.
- Return the original positions (counting from zero) that would put the values in ascending order.
- When values are equal, their original-index order is kept.
sortedIndexOrder(values: list<int>) → list<int>
Java needs a compiler and Drill does not host one yet, so this page is the reference rather than an exercise: the problem, worked examples, and the solution in full. To type it out, the same problem runs in Python.
Where you start
List<Integer> sortedIndexOrder(List<Integer> values) {
}Worked examples
| Call | Result |
|---|---|
sortedIndexOrder(Main.<Integer>ls(40, 10, 30, 10)) | Main.<Integer>ls(1, 3, 2, 0) |
sortedIndexOrder(Main.<Integer>ls(20, 10, 30, 10)) | Main.<Integer>ls(1, 3, 0, 2) |
sortedIndexOrder(Main.<Integer>ls(1, 2, 3)) | Main.<Integer>ls(0, 1, 2) |
sortedIndexOrder(Main.<Integer>ls(3, 2, 1)) | Main.<Integer>ls(2, 1, 0) |
Hint
Sort a list of the indices with a comparator that looks the values up, falling back to the index itself on a tie.
Reference solution in Java
List<Integer> sortedIndexOrder(List<Integer> values) {
List<Integer> idx = new ArrayList<>();
for (int i = 0; i < values.size(); i++) idx.add(i);
Collections.sort(idx, (a, b) -> {
int d = values.get(a) - values.get(b);
return d != 0 ? d : a - b;
});
return idx;
}