My solutions to a "Google" Coding Question: https://www.youtube.com/watch?v=uQdy914JRKQ

GoogleQuestion.cs

namespace GoogleQuestion
{
    class Program
    {
        static void Main(string[] args)
        {
            // Solution one - cheap way

            int[] arr = new[] { 9, 9, 9 };

            int tmp = 0;
            string convertMe = string.Empty;

            for (int i = 0; i < arr.Length; i++)
                convertMe += arr[i];

            tmp = int.Parse(convertMe) + 1;

            string back = tmp.ToString();
            int[] newArr = new int[back.Length];

            for (int i = 0; i < newArr.Length; i++)
                newArr[i] = int.Parse(back[i].ToString());

            // Solution two - more complicated, but more logical

            // Original array
            arr = new[] {9,9,9};

            //  10^3, 10^2, 10^1, 10^0

            int originalNumber = 0;

            for (int i = 0; i < arr.Length; i++)
            {
                // Here I think of each position of the number in powers of 10
                // multiplier will be based on the position of the number in the array
                // Then multiply the mulitplier by the number in the array, and add up the values to get the actual number representation
                int multiplier = (int)Math.Pow(10, arr.Length - 1 - i);
                int value = arr[i] * multiplier;
                originalNumber += value;
            }

            // Increment array value by one
            originalNumber += 1;

            // If last number in original array is 9, adding one must increase new array size by one
            bool sizeChange = arr[arr.Length - 1] == 9;

            // If size change is needed, set new array size to be 1 element larger, otherwise keep same size
            int arrSize = sizeChange ? arr.Length + 1 : arr.Length;

            // Declare our new array
            int[] newArray = new int[arrSize];


            for (int i = 0; i < newArray.Length; i++)
            {
                // Here we do the opposite of the first loop.
                // Get the remainder of dividing the new number by 10 to get the individual digits.

                int digit = originalNumber % 10;
                originalNumber = originalNumber / 10;

                // Numbers will be pinched off in reverse order, so insertion to the newarray must be done reversed as well.
                newArray[newArray.Length - 1 - i] = digit;
            }

        }
    }
}