Find all combination of string array in swift - arrays

Question

I have an Array of String and I want to find all the possible combinations of its element
For Example :
Array = [A,B,C,D]
should produce result as :
[A,AB,AC,AD,ABC,ABD,ACD,ABCD,B,BC,BD,BCD,C,CD,D]
Here is my Logic :
var array = ["A", "B", "C","D"]
var list = [String]()
for i in 0..<array.count{
let c = array[i]
list.append(c)
var d = c
for count in 1..<array.count{
if i+count < array.count{
for j in i+count..<array.count{
var a = d
a.appendContentsOf(array[j])
print("a : \(a)")
list.append(a)
}
}
d = c
d.appendContentsOf(array[count])
print("d : \(d)")
}
}
print(list.description)
Its Output is :
["A", "AB", "AC", "AD", "ABC", "ABD", "ACD", "B", "BC", "BD", "BBD",
"C", "CD", "D"]
This output is missing ABCD and wrongly printed BCD as BBD
Anyone Please Help me in this by Enhancing my code or suggesting your own logic for this.

Answer 1

#yannick's answer is very close.
By computing a Power Set of your set, you obtain all the possible subsets (including your original set and the empty set).
Once you have obtained the Power Set, all you have to do is join the subsets into a single string in order to obtain the result that you're looking for.
Here's the complete solution (along with updated code and plenty of comments):
extension Array {
var powerset: [[Element]] {
guard count > 0 else {
return [[]]
}
// tail contains the whole array BUT the first element
let tail = Array(self[1..<endIndex])
// head contains only the first element
let head = self[0]
// computing the tail's powerset
let withoutHead = tail.powerset
// mergin the head with the tail's powerset
let withHead = withoutHead.map { $0 + [head] }
// returning the tail's powerset and the just computed withHead array
return withHead + withoutHead
}
}
let myArray = ["A", "B", "C", "D"]
print(myArray.powerset) // -> [["D", "C", "B", "A"], ["C", "B", "A"], ["D", "B", "A"], ["B", "A"], ["D", "C", "A"], ["C", "A"], ["D", "A"], ["A"], ["D", "C", "B"], ["C", "B"], ["D", "B"], ["B"], ["D", "C"], ["C"], ["D"], []]
// joining the subsets
let myResult = myArray.powerset.map { $0.sort().joinWithSeparator("") }
print(myResult) // -> ["A", "AB", "ABC", "ABCD", "ABD", "AC", "ACD", "AD", "B", "BC", "BCD", "BD", "C", "CD", "D", ""]
PS
Note that this solution uses a recursive approach, while yours was using an iterative approach.
PPS
If you don't want the empty string "" in your solution, you can just filter it away:
let myResult = myArray.powerset.map({ $0.sort().joinWithSeparator("") }).filter({ $0 != "" })
print(myResult) // -> ["A", "AB", "ABC", "ABCD", "ABD", "AC", "ACD", "AD", "B", "BC", "BCD", "BD", "C", "CD", "D"]

Answer 2

It looks like you want to have the Power set of your array.
In mathematics, the power set (or powerset) of any set S is the set of
all subsets of S, including the empty set and S itself.
I found this Code on GitHub.
extension Array {
var powerset: [[Element]] {
if count == 0 {
return [self]
}
else {
let tail = Array(self[1..<endIndex])
let head = self[0]
let withoutHead = tail.powerset
let withHead = withoutHead.map { $0 + [head] }
return withHead + withoutHead
}
}
}
println([1,2,3,4].powerset) -> [[4, 3, 2, 1], [3, 2, 1], [4, 2, 1], [2, 1], [4, 3, 1], [3, 1], [4, 1], [1], [4, 3, 2], [3, 2], [4, 2], [2], [4, 3], [3], [4], []]

Answer 3

I find a neater answer for it.Power set of Collection.
The principle is using induction on the size of a collection, as showed on that link.
Here is the copy of code from that link. And all credits to its author.
extension Collection {
public var powerSet: [[Element]] {
guard let fisrt = self.first else {return [[]]}
return self.dropFirst().powerSet.flatMap{[$0, [fisrt] + $0]}
}
}
let s: Set<Int> = [1,2,3]
s.powerSet //[[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]
let a: Array<Int> = [1,2,3]
a.powerSet //[[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]

Answer 4

I know some good answers have been given already, but coming from a Java background, I just wanted to drop some insights using bitwise operators (which surprisingly still work in Swift).
You can try this out:
let len = stringArr.count
for i in 0 ..< (1<<len){
print("{", terminator: "")
for j in 0 ..< len {
if ((i & (1<<j)) > 0) {
print(stringArr[j], terminator: "")
}
}
print("}")
}
You can find more information on bitwise operators here

Answer 5

I will take a shot also using this logic as reference:
extension RangeReplaceableCollection {
var subSets : [SubSequence] {
guard !isEmpty else { return [] }
let count = self.count
let n = 1 << count - 1
var subSequences: [SubSequence] = .init(repeating: SubSequence(), count: n-1)
(0 ..< n).forEach { x in
autoreleasepool {
var counter = 0
for element in self {
if x & 1 << counter > 0 {
subSequences[x-1].append(element)
}
counter += 1
}
}
}
return subSequences + [self[...]]
}
}
Playground Testing:
["A", "B", "C","D"].subSets // [["A"], ["B"], ["A", "B"], ["C"], ["A", "C"], ["B", "C"], ["A", "B", "C"], ["D"], ["A", "D"], ["B", "D"], ["A", "B", "D"], ["C", "D"], ["A", "C", "D"], ["B", "C", "D"], ["A", "B", "C", "D"]]
"ABCD".subSets // ["A", "B", "AB", "C", "AC", "BC", "ABC", "D", "AD", "BD", "ABD", "CD", "ACD", "BCD", "ABCD"]