I'm trying to find all of the possible combinations of capitalization for a String in Swift. For example, given the String
"abc", I would want my method to return an Array
of String
s, like this: ["Abc", "aBc", "abC", "ABc", "abc", "ABC", "aBC", "AbC"]
. The formula for the possible number of capitalization combinations is like this:
2i
where i
is the number of Characters
in the String
.
I've tried the following function, via an extension to the String
type:
extension String {
func allPossibleCombinations() -> [String] {
let string = self
var result = [String]()
var current = string
result.append(current)
for i in 0..<string.count {
let index = string.index(string.startIndex, offsetBy: i)
if string[index].isLetter {
current = current.replacingCharacters(in: index...index, with: String(current[index]).uppercased())
result.append(current)
}
}
return result
}
}
This didn't work because it only returns capitalizations in order. For example, if I were to call this method on the String
"abc"
, it would return
["abc", "Abc", "ABc", "ABC"]
This should produce, as stated above, 8 different Strings
. I suspect that I need to factor in an exponent to my code, or potentially some form of randomly choosing a Character
.
Python
question discusses the same issue: Find all upper, lower and mixed case combinations of a string. However, this does not work for my issue because it is in Python
, not Swift
.JavaScript
question talks about finding all combinations of a String
: Find all the combinations of a string Javascript. However, this is about general combinations, not capitalization ones, and is also in the wrong language.JavaScript
question discusses the same issue: Find all lowercase and uppercase combinations of a string in Javascript. However, this is in JavaScript
, not Swift
.C
question deals with the same issue: Finding all capital letter combinations of random text. However, it is in C
, not Swift
.I am trying to make a function to get all possible capitalization forms of a String
. I have a current attempt, but it is not sufficient as it does not produce all possible options.
Compute the integers in the range 0 ..< 2^length
and use the bits of the binary representation to tell you when to capitalize a letter:
extension String {
func allPossibleCombinations() -> [String] {
guard self.count > 0 else { return [] }
var result = [String]()
let lower = self.lowercased().map(String.init)
let upper = self.uppercased().map(String.init)
let length = self.count
let limit = 1 << length
for n in 0..<limit {
var word = ""
for i in 0..<length {
if n & (1 << (length - i - 1)) != 0 {
word += upper[i]
} else {
word += lower[i]
}
}
result.append(word)
}
return result
}
}
print("abc".allPossibleCombinations())
// ["abc", "abC", "aBc", "aBC", "Abc", "AbC", "ABc", "ABC"]
print("abcd".allPossibleCombinations())
// ["abcd", "abcD", "abCd", "abCD", "aBcd", "aBcD", "aBCd", "aBCD", "Abcd", "AbcD", "AbCd", "AbCD", "ABcd", "ABcD", "ABCd", "ABCD"]
In the comments, Alex added this Swifty version using nested maps:
extension String {
func allPossibleCombinations() -> [String] {
guard self.count > 0 else { return [] }
let length = self.count
let limit = 1 << self.count
return (0..<limit) .map { n in
return self.enumerated()
.map { i, c in (n & (1 << (length - i - 1)) != 0) ? c.uppercased() : c.lowercased() }
.joined()
}
}
}