Question

   * Add operation counts
   * f(N) formula (show your work)
   * O(N) reduction
   */
   public static long sum3(int N)//f(N) = ; O(N) =
   {
       long opCount = 0;
       long sum = 0;
       for(int i = 0; i < N; i++)
       {
           for(int j = 0; j < N*N; j++)
           {
               sum++;
           }
       }
       System.out.println("f(N) = [your answer]");
       System.out.println("O(N) = [your answer]");
       System.out.println("OpCount : "+opCount);
       return sum;

Answer 1

        public static long sum3(int N)// f(N) = ; O(N) =
        {
                long opCount = 0;
                long sum = 0;
                for (int i = 0; i < N; i++)
                {
                        for (int j = 0; j < N * N; j++)
                        {
                                opCount++;
                                sum++;
                        }
                }
                
                /*
                 * The Outer loop runs N times.. For each outer loop, the
                 * inner loop is run N^2 times.. Hence total opCount 
                 * will be N^3.
                 * 
                 * Hence f(n) = n^3
                 * O(N) = O(N^3)
                 */

                System.out.println("f(N) = " + (N * N * N));
                System.out.println("O(N) = " + (N*N*N));
                System.out.println("OpCount : " + opCount);

                return sum;

        }

**************************************************

Thanks for your question. We try our best to help you with detailed answers, But in any case, if you need any modification or have a query/issue with respect to above answer, Please ask that in the comment section. We will surely try to address your query ASAP and resolve the issue.

Please consider providing a thumbs up to this question if it helps you. by Doing that, You will help other students, who are facing similar issue.