Find maximum possible time HH:MM by permuting four given digits

Question

I recently took a coding test for a promotion at work. This was one of the tasks I really struggled with and was wondering what is the best way to do this. I used a load of

Answer 1

This solution is in Swift 3.0.

func returnValue (_ value :inout Int, tempArray : [Int] , compareValue : Int) -> Int {

    for i in tempArray {

        if value <= i && i <= compareValue {
            value = i
        }
    }

    return value
}

func removeValue(_ value : Int, tempArr : inout [Int]) -> Bool {
    let index = tempArr.index(of: value)
    tempArr.remove(at: index ?? 0)
    return index != nil ? true : false
}

public func solution(_ A : Int, _ B : Int, _ C : Int, _ D : Int) -> String {

    var tempArray = [A, B, C, D]

    let mainArray = [A, B, C, D]

    var H1 : Int = -1, H2: Int = -1, M1 : Int = -1, M2 : Int = -1;

    H1 = returnValue(&H1, tempArray: tempArray, compareValue: 2)

    if !removeValue(H1, tempArr: &tempArray) {
        return "NOT POSSIBLE"
    }

    for value in tempArray {

        if H1 < 2 {
            if H2 <= value && value <= 9 {
                H2 = value
            }
        } else {
            if H2 <= value && value <= 3 {
                H2 = value
            }
        }
    }

    if !removeValue(H2, tempArr: &tempArray) {
        return "NOT POSSIBLE"
    }

    M1 = returnValue(&M1, tempArray: tempArray, compareValue: 5)


    if M1 >= 0 {

        if !removeValue(M1, tempArr: &tempArray) {
            return "NOT POSSIBLE"
        }
    } else if mainArray.contains(0) || mainArray.contains(1) {

        H1 = -1

        H1 = returnValue(&H1, tempArray: mainArray, compareValue: 1)

        for value in mainArray {

            if H1 < 2 {
                if H2 <= value && value <= 9 {
                    H2 = value
                }
            } else {
                if H2 <= value && value <= 3 {
                    H2 = value
                }
            }
        }


        tempArray.removeAll()

        for value in mainArray {
            tempArray.append(value)
        }


        var index = tempArray.index(of: H1)
        tempArray.remove(at: index!)

        index = tempArray.index(of: H2)
        tempArray.remove(at: index!)

        M1 = -1
        M1 = returnValue(&M1, tempArray: tempArray, compareValue: 5)

        if !removeValue(M1, tempArr: &tempArray) {
            return "NOT POSSIBLE"
        }

    } else {
        return "NOT POSSIBLE"
    }

    // Now last we have M2 = temp.last

    if let lastValue = tempArray.last {
        M2 = lastValue
    }

    if M2 < 0 {
        return "NOT POSSIBLE"
    }

    return "\(H1)\(H2):\(M1)\(M2)"
}


print(solution(1,7,2,7))
print(solution(0,0,2,9))
print(solution(6,5,2,0))
print(solution(3,9,5,0))
print(solution(7,6,3,8))
print(solution(0,0,0,0))
print(solution(9,9,9,9))
print(solution(1,2,3,4))

 17:27
 20:09
 20:56
 09:53
 NOT POSSIBLE
 00:00
 NOT POSSIBLE
 23:41