IterativeBinarySearch.swift

// takes O(log(n) base 2 + 1) iterations where 2 ^ n is the closest to array.count 
// example 10 is closest to 2^3 which will take 3 + 1 iterations to find an element in the worst case scenario


  func iterativeBinarySearch (_ array: [Int],_ key: Int) -> Int? {
        var low = 0;
        var high = array.count - 1
        var mid : Int;
        // var iterations = 0;

        while low <= high {
            mid = (high - low) / 2 + low // high - low, gives you diff. /2 gives you mid. + low gives you exact index
            // iterations += 1;
            if(key == array[mid]) { 
                // print(iterations)
                return mid           
            } else if (key < array[mid]) {
                high = mid - 1
            } else {
                low = mid + 1
            }
        }

        return -1
    }