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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Absolute Permutation
  5. Discussions

Absolute Permutation

  • nguyenninh_uit
     + 0 comments

    Swift solution: There are only 2 cases: k == 0 n%(2*k) == 0

    Exp n=12 k=3: 4 5 6 1 2 3 10 11 12 7 8 9

    It will loop in 2 sub: [4 5 6 1 2 3] & [10 11 12 7 8 9]

    func absolutePermutation(n: Int, k: Int) -> [Int] {
    // Write your code here
    
    if k == 0 {
        var results:[Int] = []
        for i in 1...n { results.append(i) }
        return results
    } else if n%(2*k) == 0 {
        var results:[Int] = []
        let twoK = 2*k
        // Exp 12 3: 4 5 6 1 2 3 10 11 12 7 8 9
        for i in 0..<n/twoK {
            let start = i*twoK + 1
            let end = (i+1)*twoK
            let mid = (end + start)/2 + 1
            if mid <= end {
                for j in mid...end {
                    results.append(j)
                }
            }
            for j in start..<mid {
                results.append(j)
            }
        }
        return results
    }
    
    return [-1]
}