Reverse array by group size

Question

I was trying to solve this challenge: reverse an array of elements by groups given a group size.

Given Array: [1, 2, 3, 4, 5, 6]

Desired Result (group size of 3): [4, 5, 6, 1, 2, 3]

If last group has less elements than the group size, then just add them and finish, as follows:

Given Array: [1, 2, 3, 4, 5, 6, 7]

Desired Result: [5, 6, 7, 2, 3, 4, 1]

I tried this and it is working, but it looks kinda weird for me. Can anyone help me find a cleaner or much more intuitive solution?

extension Array {
    func reverse(groupSize: Int) -> [Element] {
        var reversed = [Element]()
        let groups = count / groupSize

        for group in 0...groups {
            let lowerBound = count - group * groupSize - groupSize
            let upperBound = count - 1 - group * groupSize

            if lowerBound >= 0 {
                reversed += Array(self[lowerBound...upperBound])
            } else {
                reversed += Array(self[0...upperBound])
            }
        }

        return reversed
    }
}

Answer 1

The following solution is based on a combo of stride+map:

let groupSize = 3
let array = [1, 2, 3, 4, 5, 6]
let reversedArray = Array(array.reversed())
let result = stride(from: 0, to: reversedArray.count, by: groupSize).map {
  reversedArray[$0 ..< min($0 + groupSize, reversedArray.count)].reversed()
}.reduce([Int](), +)
print(result) //  [4, 5, 6, 1, 2, 3]