How would you solve the letter changer in Julia? - string

Question

I found this challenge:
Using your language, have the function LetterChanges(str) take the str parameter being passed and modify it using the following algorithm. Replace every letter in the string with the letter following it in the alphabet (ie. c becomes d, z becomes a). Then capitalize every vowel in this new string (a, e, i, o, u) and finally return this modified string.
I am new in Julia, and I was challenging myself in this challenge. I found this challenge very hard in Julia lang and I could not find a solution.
Here I tried to solve in the way below, but I got error: the x value is not defined
How would you solve this?
function LetterChanges(stringis::AbstractString)
alphabet = "abcdefghijklmnopqrstuvwxyz"
vohels = "aeiou"
for Char(x) in split(stringis, "")
if x == 'z'
x = 'a'
elseif x in vohels
uppercase(x)
else
Int(x)+1
Char(x)
println(x)
end
end
end
Thank you

Answer 1

As a side note:
The proposed solution works properly. However, if you would need high performance (which you probably do not given the source of your problem) it is more efficient to use string builder:
function LetterChanges2(str::AbstractString)
v = Set("aeiou")
#sprint(sizehint=sizeof(str)) do io # use on Julia 0.7 - new keyword argument
sprint() do io # use on Julia 0.6.2
for c in str
c = c == 'z' ? 'a' : c+1 # we assume that we got only letters from 'a':'z'
print(io, c in v ? uppercase(c) : c)
end
end
end
it is over 10x faster than the above.
EDIT: for Julia 0.7 this is a bit faster:
function LetterChanges2(str::AbstractString)
v = BitSet(collect(Int,"aeiouy"))
sprint(sizehint=sizeof(str)) do io # use on Julia 0.7 - new keyword argument
for c in str
c = c == 'z' ? 'a' : c+1 # we assume that we got only letters from 'a':'z'
write(io, Int(c) in v ? uppercase(c) : c)
end
end
end

Answer 2

There is a logic error. It says "Replace every letter in the string with the letter following it in the alphabet. Then capitalize every vowel in this new string". Your code checks, if it is a vowel. Then it capitalizes it or replaces it. That's different behavior. You have to first replace and then to check if it is a vowel.
You are replacing 'a' by 'Z'. You should be replacing 'z' by 'a'
The function split(stringis, "") returns an array of strings. You can't store these strings in Char(x). You have to store them in x and then you can transform theses string to char with c = x[1].
After transforming a char you have to store it in the variable: c = uppercase(c)
You don't need to transform a char into int. You can add a number to a char: c = c + 1
You have to store the new characters in a string and return them.
function LetterChanges(stringis::AbstractString)
# some code
str = ""
for x in split(stringis, "")
c = x[1]
# logic
str = "$str$c"
end
return str
end

Answer 3

Here's another version that is a bit faster than #BogumilKaminski's answer on version 0.6, but that might be different on 0.7. On the other hand, it might be a little less intimidating than the do-block magic ;)
function changeletters(str::String)
vowels = "aeiouy"
carr = Vector{Char}(length(str))
i = 0
for c in str
newchar = c == 'z' ? 'a' : c + 1
carr[i+=1] = newchar in vowels ? uppercase(newchar) : newchar
end
return String(carr)
end

Answer 4

At the risk of being accused of cheating, this is a dictionary-based approach:
function change_letters(s::String)::String
k = collect('a':'z')
v = vcat(collect('b':'z'), 'A')
d = Dict{Char, Char}(zip(k, v))
for c in Set("eiou")
d[c - 1] = uppercase(d[c - 1])
end
b = IOBuffer()
for c in s
print(b, d[c])
end
return String(take!(b))
end
It seems to compare well in speed terms with the other Julia 0.6 methods for long strings (e.g. 100,000 characters). There's a bit of unnecessary overhead in constructing the dictionary which is noticeable on small strings, but I'm far too lazy to type out the 'a'=>'b' construction long-hand!