The expression checks whether the given word reads the same forward and backward.

Example: MADAM

len(ri!word) returns 5, so enumerate(...) + 1 generates positions {1, 2, 3, 4, 5}.
mid() extracts each character: {M, A, D, A, M}.
reverse() reverses the character array: {M, A, D, A, M}.
joinarray() combines the characters into "MADAM".
upper() converts both strings to uppercase before comparison.
Since "MADAM" equals "MADAM", the expression returns "Palindrome".

Solution 2 — Palindrome Without reverse()

a!localVariables(

local!word: ri!word,

local!length: len(local!word),

if(

and(

a!forEach(

items: enumerate(floor(local!length / 2)) + 1,

expression: mid(

local!word,

fv!item,

1

) = mid(local!word, local!length - fv!item + 1, 1)

)

),

concat(ri!word, " is Palindrome"),

concat(ri!word, " is not Palindrome")

)

)

This solution checks whether the word is a palindrome without creating a reversed string.

For example, with MADAM:

Advantage: This approach avoids reverse() and avoids building another string, making the logic more efficient.