Could you optimize the below code?: (String anagrams)

load(
local!str1: "abca",
local!str2: "ABAC",
local!arrayList: a!forEach(
items: {
local!str1,
local!str2
},
expression: apply(
fn!index(
lower(
fv!item
),
_,
""
),
enumerate(
len(
fv!item
)
) + 1
)
),
if(
length(
intersection(
local!arrayList[1],
local!arrayList[2]
)
) = 0,
"Anagram",
"Naah!"
)
)

How about this?

load(
local!str1: "abca",
local!str2: "ABAC",

local!unicode1:code(lower(local!str1)),
local!unicode2:code(lower(local!str2)),

if(
contains(local!unicode1,local!unicode2),
"Anagram",
"Naah!!"
)
)

I believe, for an anagram, the letters should not only match but appear exactly the same number of times in both the words in different order. Check this out.:

load(
  local!str1: "abca",
  local!str2: "abA",
  
  local!unicode1:code(lower(local!str1)),
  local!uniqueUnicode1: union(local!unicode1,local!unicode1),
local!unicode2:code(lower(local!str2)),
local!uniqueUnicode2:union(local!unicode2,local!unicode2),

local!str1Details:

a!forEach(local!uniqueUnicode1, 
{character:fv!item,times:count(wherecontains(fv!item,local!unicode1))}),

local!str2Details:a!forEach(local!uniqueUnicode2, 
{character:fv!item,times:count(wherecontains(fv!item,local!unicode2))}),

local!isMatch:a!forEach(local!str1Details,
    
  fv!item.times = index(local!str2Details.times, wherecontains(fv!item.character,local!str2Details.character),{})
     
  ),
  if(contains(toboolean(local!isMatch),false()),"Not an anagram","anagram")
)

Thanks for correcting the solution Ankita, I wasn't aware about Anagram functionality and my assumption was its just the character match that is required.

Check this out:

load(
local!str1: "dabca",
local!str2: "abAcD",
if(
or(sort(code(lower(local!str1)))-sort(code(lower(local!str2)))),"Not Anagram","Anagram"
)
)

sort() is internal API and should not be used. Sorry ... With a bit of trickery you can use todatasubset() to sort your array of characters.