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

Question

load(
local!str1: "abca",
local!str2: "ABAC",
local!splitstr1: apply(
fn!index(
lower(local!str1),
_,
""
),
enumerate(
len(
local!str1
)
) + 1
),
local!splitstr2: apply(
fn!index(
lower(local!str2),
_,
""
),
enumerate(
len(
local!str2
)
) + 1
),
local!checkData: and(
a!forEach(
items: local!splitstr2,
expression: contains(
local!splitstr1,
fv!item
)
)
),
if(
local!checkData,
"Anagram",
"Naah!!"
)
)

ankitab0001 · Accepted Answer

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")
)

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

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