ios - Calculate all permutations of a string in Swift

Question

For the string "ABC" the code snippet below calculates 5 of the 6 total permutations. My strategy was to insert each character at each index possible index. But the function never gets "CBA" as a possible permutation. What am I missing?

var permutationArray:[String] = [];
let string: String = "ABC"

func permute(input: String) -> Array<String>
{
    var permutations: Array<String> = []

    /*   Convert the input string into characters      */
    var inputArray: Array<String>
    inputArray = input.characters.map { String($0) }
    print(inputArray)

    /*   For each character in the input string...     */
    for var i = 0; i < inputArray.count; i++
    {

        /*       Insert it at every index              */
        let characterInArray: String = inputArray[i]
        var inputArrayCopy: Array<String> = []
        for var y = 0; y < inputArray.count; y++
        {

            inputArrayCopy = inputArray
            inputArrayCopy.removeAtIndex(i)
            inputArrayCopy.insert(characterInArray, atIndex:y)

            let joiner = ""
            let permutation = inputArrayCopy.joinWithSeparator(joiner)
            if !permutations.contains(permutation) {
                permutations.insert(permutation, atIndex: 0)
            }
        }
    }

    return permutations
}

var permutations = permute(string)
print(permutations)

Answer 1

While Stefan and Matt make a good point about using Heap's algorithm, I think you have an important question about why your code doesn't work and how you would debug that.

In this case, the algorithm is simply incorrect, and the best way to discover that is with pencil and paper IMO. What you are doing is picking each element, removing it from the array, and then injecting it into each possible location. Your code does what you have asked it to do. But it's not possible to get to "CBA" that way. You're only moving one element at a time, but "CBA" has two elements out of order. If you expanded to ABCD, you'd find many more missing permutations (it only generates 10 of the 24).

While Heap's algorithm is nicely efficient, the deeper point is that it walks through the entire array and swaps every possible pair, rather than just moving a single element through the array. Any algorithm you choose must have that property.

And just to throw my hat into the ring, I'd expand on Matt's implementation this way:

// Takes any collection of T and returns an array of permutations
func permute<C: Collection>(items: C) -> [[C.Iterator.Element]] {
    var scratch = Array(items) // This is a scratch space for Heap's algorithm
    var result: [[C.Iterator.Element]] = [] // This will accumulate our result

    // Heap's algorithm
    func heap(_ n: Int) {
        if n == 1 {
            result.append(scratch)
            return
        }

        for i in 0..<n-1 {
            heap(n-1)
            let j = (n%2 == 1) ? 0 : i
            scratch.swapAt(j, n-1)
        }
        heap(n-1)
    }

    // Let's get started
    heap(scratch.count)

    // And return the result we built up
    return result
}

// We could make an overload for permute() that handles strings if we wanted
// But it's often good to be very explicit with strings, and make it clear
// that we're permuting Characters rather than something else.

let string = "ABCD"
let perms = permute(string.characters) // Get the character permutations
let permStrings = perms.map() { String($0) } // Turn them back into strings
print(permStrings) // output if you like