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>
Go 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
func sortedIndexOrder(values []int) []int {
}Worked examples
| Call | Result |
|---|---|
sortedIndexOrder([]int{40, 10, 30, 10}) | []int{1, 3, 2, 0} |
sortedIndexOrder([]int{20, 10, 30, 10}) | []int{1, 3, 0, 2} |
sortedIndexOrder([]int{1, 2, 3}) | []int{0, 1, 2} |
sortedIndexOrder([]int{3, 2, 1}) | []int{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 Go
func sortedIndexOrder(values []int) []int {
idx := []int{}
for i := 0; i < len(values); i++ {
idx = append(idx, i)
}
sort.SliceStable(idx, func(i, j int) bool {
if values[idx[i]] != values[idx[j]] {
return values[idx[i]] < values[idx[j]]
}
return idx[i] < idx[j]
})
return idx
}