iOS4CodeChallenge-18.11.07-mySolution.swift

/*
 Thoughts:
 - Only have to deal with positive numbers
 - Don't have to deal with zero
 - Return first solution
 
 Test cases:
 sumAndProduct(6, 9) // [3, 3]
 sumAndProduct(5, 4) // [1, 4]
 sumAndProduct(7, 12) // [3, 4]
 sumAndProduct(15, 36) // [3, 12]
 sumAndProduct(3, 11) // should fail - []

 Approach:
 - Cycle through the numbers that can possibly add up to the sum and check to see if they also equal the product.
 */


//Simple solution. Cycles through all possibilities. Kind of inelegant/inefficient. O(n^2)
func sumAndProduct(_ sum: UInt, _ product: UInt) -> [UInt] {
    var timesThroughLoop = 0
    defer { print("Times through loop: \(timesThroughLoop)") }
    
    // Check to make sure some joker didn't put zero.
    guard sum != 0, product != 0 else { return [] }
    
    // Cycle through all the numbers from 1 up to the sum
    for num in 1 ..< sum {
        // And again
        for otherNum in num ..< sum {
            timesThroughLoop += 1
            
            if num + otherNum == sum && num * otherNum == product {
                // If the two numbers add up to the sum AND multiply to the product
                // return them
                return [num, otherNum]
            }
        }
    }
    // Otherwise, return an empty array
    return []
}

sumAndProduct(6, 9) // [3, 3]
sumAndProduct(5, 4) // [1, 4]
sumAndProduct(7, 12) // [3, 4]
sumAndProduct(15, 36) // [3, 12]
sumAndProduct(3, 11) // []


/*
// Better solution.  O(n)
func sumAndProduct(_ sum: UInt, _ product: UInt) -> [UInt] {
    var timesThroughLoop = 0
    defer { print("Times through loop: \(timesThroughLoop)") }

    // Make sure some joker didn't put zero
    guard sum != 0, product != 0 else { return [] }

    // Only need to cycle through half of the sum because after that the values repeat.
    for value1 in 1 ... (sum / 2) {
        // Don't need to cycle through the second value, because there is only one value that leads to the sum we're looking for.
        let value2 = sum - value1
        timesThroughLoop += 1
        // If the two values lead to the product, return them
        if value1 * value2 == product {
            return [value1, value2]
        }
    }
    
    // Otherwise return an empty array.
    return []
}
*/