method that removes all twice duplicate characters in String - ios

Question

I wrote a method that removes all 2 duplicate characters in String, for example
I need delete only char that contains twice, for example
"bndkss" -> "bndk"
"nnmmhj" - > "hj"
"aaabbaac" -> "ac
"abba" -> ""
I wrote on objc and everything works, but Swift is not working, help please, where did I go wrong?
override func viewDidLoad() {
super.viewDidLoad()
let string = "baab"
print("before: \(string)")
let stringAfter = checkString(string: string)
print("after: \(stringAfter)")
}
func checkString(string : String) -> String {
var tempString = string
for (index, element) in string.characters.enumerated() {
for (index2, element2) in string.characters.enumerated() {
if element == element2 && index != index2 {
if index > index2 {
tempString.remove(at: tempString.index(tempString.startIndex, offsetBy: index))
tempString.remove(at: tempString.index(tempString.startIndex, offsetBy: index2))
} else {
tempString.remove(at: tempString.index(tempString.startIndex, offsetBy: index2))
tempString.remove(at: tempString.index(tempString.startIndex, offsetBy: index))
}
if tempString.characters.count < 1 {
return ""
} else {
checkString(string: tempString)
}
} else {
if index == tempString.characters.count - 1 && index2 == tempString.characters.count - 1 {
return tempString
}
}
}
}
return ""
}
Updates:
just need
return checkString(string: tempString)
instead
checkString(string: tempString)

Answer 1

There are two problems in your code, such as
After removing characters in tempString, the indices index and index2 to long refer to the original characters in tempString.
Wrong characters are removed as a consequence.
You call checkString() recursively but discard the result.
Update: As you already noticed in the meantime, return checkString(string: tempString) solves these problems.
Here is an alternative implementation. The idea is to use a dictionary
to remember where a character has been seen last, and an index set
which keeps track of the positions of the characters which are to
be preserved. Instead of two nested loops and recursion, two "simple"
loops are used here, plus the cost of the dictionary and set operations.
func removeDuplicateCharacters(string: String) -> String {
var seen = [Character: Int]()
var keep = IndexSet(integersIn: 0..<string.characters.count)
for (idx, c) in string.characters.enumerated() {
if let prevIndex = seen[c] {
keep.remove(prevIndex)
keep.remove(idx)
seen.removeValue(forKey: c)
} else {
seen[c] = idx
}
}
return String(keep.map { string[string.index(string.startIndex, offsetBy: $0)] })
}
Examples:
print(removeDuplicateCharacters(string: "bndkss")) // ""bndk"
print(removeDuplicateCharacters(string: "nnmmhj")) // "jh"
print(removeDuplicateCharacters(string: "abba")) // ""
print(removeDuplicateCharacters(string: "aaabbaac")) // "ac"

Answer 2

Martin wrote a much cleaner version than myself, but I worked on this for a little while so I figured I'd post it to show you another way it could have been accomplished.
func removeDuplicates(from original: String) -> String {
var originalString = original
var newString = ""
for character in originalString.characters {
if !newString.contains("\(character)") {
newString.append(character)
originalString = originalString.characters.filter { $0.description != "\(character)" }.map { "\($0)" }.joined(separator: "")
} else {
newString = newString.characters.filter { $0.description != "\(character)" }.map { "\($0)" }.joined(separator: "")
}
}
return newString
}