Suppose:

ri!text = "HELLO"

Step 1: Find the length

len(ri!text)

Result:

5

Step 2: Generate positions

enumerate(len(ri!text))

Result:

{0, 1, 2, 3, 4}

Step 3: Calculate the indexes from the end

len(ri!text) - enumerate(len(ri!text))

Result:

5 - {0, 1, 2, 3, 4}

which gives:

{5, 4, 3, 2, 1}

Step 4: Use those indexes to access the string

ri!text[{5, 4, 3, 2, 1}]

The characters are accessed as:

H E L L O

↓

O L L E H

Final output:

"OLLEH"

Why this is a good solution

It is very concise and uses only:

len()
enumerate()
Appian indexing []

No reverse(), a!forEach(), split(), or temporary variables are required.

Solution 2 — Using todatasubset() to Sort Characters

a!localVariables(

local!sortMap: a!forEach(

enumerate(len(ri!text)) + 1,

a!map(

index: fv!item,

currentLetter: ri!text[fv!item]

)

),

local!sortedSet: todatasubset(

arrayToPage: local!sortMap,

pagingConfiguration: a!pagingInfo(

1,

-1,

a!sortInfo(

field: "index",

ascending: false()

)

)

).data,

concat(local!sortedSet.currentLetter)

)

Explanation

Suppose:

ri!text = "HELLO"

Step 1 — Create an index and character map

a!forEach(

enumerate(len(ri!text)) + 1,

a!map(

index: fv!item,

currentLetter: ri!text[fv!item]

)

)

This creates:

{

{index: 1, currentLetter: "H"},

{index: 2, currentLetter: "E"},

{index: 3, currentLetter: "L"},

{index: 4, currentLetter: "L"},

{index: 5, currentLetter: "O"}

}

Step 2 — Sort the map in descending order

a!sortInfo(

field: "index",

ascending: false()

)

The indexes become:

5 → 4 → 3 → 2 → 1

Therefore, the characters become:

O → L → L → E → H

Step 3 — Get the sorted data

todatasubset(

arrayToPage: local!sortMap,

pagingConfiguration: a!pagingInfo(...)

).data

This returns the records in the sorted order.

Step 4 — Concatenate the characters

concat(local!sortedSet.currentLetter)

Final result:

"OLLEH"