Apple Swift - Generate combinations with repetition

You can get rid of var sub = subcombo by writing the loop as

for subcombo in subcombos {
    ret.append([head] + subcombo)
}

This can be further simplified using the map() function:

func combos<T>(var array: Array<T>, k: Int) -> Array<Array<T>> {
    if k == 0 {
        return [[]]
    }

    if array.isEmpty {
        return []
    }

    let head = [array[0]]
    let subcombos = combos(array, k: k - 1)
    var ret = subcombos.map {head + $0}
    array.removeAtIndex(0)
    ret += combos(array, k: k)

    return ret
}

---

Update for Swift 4:

func combos<T>(elements: ArraySlice<T>, k: Int) -> [[T]] {
    if k == 0 {
        return [[]]
    }

    guard let first = elements.first else {
        return []
    }

    let head = [first]
    let subcombos = combos(elements: elements, k: k - 1)
    var ret = subcombos.map { head + $0 }
    ret += combos(elements: elements.dropFirst(), k: k)

    return ret
}

func combos<T>(elements: Array<T>, k: Int) -> [[T]] {
    return combos(elements: ArraySlice(elements), k: k)
}

Now array slices are passed to the recursive calls to avoid the creation of many temporary arrays.

Example:

print(combos(elements: [1, 2, 3], k: 2))
// [[1, 1], [1, 2], [1, 3], [2, 2], [2, 3], [3, 3]]

Your example gives combinations with repetition. For the record I have written a non-repetitive combination in Swift. I based it on the JavaScript version here: http://rosettacode.org/wiki/Combinations#JavaScript

I hope it helps others and if anyone can see improvement please do.. note this is my first attempt at Swift and was hoping for a neater way of doing the Swift equivalent of JavaScript slice.

func sliceArray(var arr: Array<Int>, x1: Int, x2: Int) -> Array<Int> {
    var tt: Array<Int> = []
    for var ii = x1; ii <= x2; ++ii {
        tt.append(arr[ii])
    }
    return tt
}

func combinations(var arr: Array<Int>, k: Int) -> Array<Array<Int>> {
    var i: Int
    var subI : Int

    var ret: Array<Array<Int>> = []
    var sub: Array<Array<Int>> = []
    var next: Array<Int> = []
    for var i = 0; i < arr.count; ++i {
        if(k == 1){
            ret.append([arr[i]])
        }else {
            sub = combinations(sliceArray(arr, i + 1, arr.count - 1), k - 1)
            for var subI = 0; subI < sub.count; ++subI {
                next = sub[subI]
                next.insert(arr[i], atIndex: 0)
                ret.append(next)
            }
        }

    }
    return ret
}


var myCombinations = combinations([1,2,3,4],2)

---

Per the OP's request, here is a version which removes the custom Array slicing routine in favor of functionality in the standard library

// Calculate the unique combinations of elements in an array
// taken some number at a time when no element is allowed to repeat
func combinations<T>(source: [T], takenBy : Int) -> [[T]] {
    if(source.count == takenBy) {
        return [source]
    }

    if(source.isEmpty) {
        return []
    }

    if(takenBy == 0) {
        return []
    }

    if(takenBy == 1) {
        return source.map { [$0] }
    }

    var result : [[T]] = []

    let rest = Array(source.suffixFrom(1))
    let sub_combos = combinations(rest, takenBy: takenBy - 1)
    result += sub_combos.map { [source[0]] + $0 }

    result += combinations(rest, takenBy: takenBy)

    return result
}

var myCombinations = combinations([1,2,3,4], takenBy: 2)
// myCombinations = [[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]]

Updated @richgordonuk answer for Swift 4 which provides a non-repetitive combination:

func combinations<T>(source: [T], takenBy : Int) -> [[T]] {
    if(source.count == takenBy) {
        return [source]
    }

    if(source.isEmpty) {
        return []
    }

    if(takenBy == 0) {
        return []
    }

    if(takenBy == 1) {
        return source.map { [$0] }
    }

    var result : [[T]] = []

    let rest = Array(source.suffix(from: 1))
    let subCombos = combinations(source: rest, takenBy: takenBy - 1)
    result += subCombos.map { [source[0]] + $0 }
    result += combinations(source: rest, takenBy: takenBy)
    return result
}