# Get the indices of an array after sorting

Your challenge today is to write a program or function which takes a list l and gives the positions in l at which each successive element of l sorted appears.

In other words, output the index of the smallest value, followed by the index of the second smallest value, etc.

You can assume that the input array will contain only positive integers, and will contain at least one element.

Test cases:

Input                  | Output (1-indexed)
[7, 4, 5]              | [2, 3, 1]
[1, 2, 3]              | [1, 2, 3]
[2, 6, 1, 9, 1, 2, 3]  | [3, 5, 1, 6, 7, 2, 4]
[4]                    | [1]


When two or more elements with the same value appear, their indices should appear next to each other from smallest to largest.

This is , fewest bytes wins!

• -1 for a trivial challenge that can be solved with built-ins in common golfing languages, and for accepting an answer in less than 24 hours. This was neither a fair challenge, nor an interesting one. – Cody Gray Jul 30 '17 at 11:44
• Well, I get why he accepted an answer within 24 hours, it's impossible to beat. – Zacharý Jul 30 '17 at 14:28
• @CodyGray I thought downvoting when I saw the 1-2 bytes answer, but actually, I don't think it's a bad challenge for more standard programming languages. Of course, it's not a hard challenge, but still, there is definitely some golfing possibilities. Of course, it's unpleasant to see 1-byte built-ins, but I don't think that it's fair to blame the challenge for that. – Dada Jul 30 '17 at 16:07
• Using a 1 character builtin is hardly practice. Easy doesn't necessarily mean solvable using only builtins. – JAD Jul 31 '17 at 7:36
• The best solution in such cases is to forget about te accept feature, which isn't really relevant anyway here. – Mr. Xcoder Jul 31 '17 at 9:01

# Ruby, 34 bytes

->a{(0...a.size).sort_by{|i|a[i]}}


0-indexed, 1-indexing requires one more net byte:

->a{(1..a.size).sort_by{|i|a[i-1]}}


Also, a slightly golfed version of G B's answer is 37 bytes:

->a{a.each_with_index.sort.map &:pop}


# Kotlin,  44  34 bytes

not today, crossed out 44

{l->(1..l.size).sortedBy{l[it-1]}}


Try it online!

1-indexed because that's what the examples give. A 0-indexed version looks like this for 30 bytes:

{l->l.indices.sortedBy{l[it]}}


# Japt, 6 5 bytes

0-indexed

ð ñ@v


Try it

### Alternative, 6 bytes

í ñÎmÌ


Try it