Sum no subtraction & Product no division

 You are given an array of integers 'a' that can fit in a memory. Write a method that returns an array of the same length such that each element 'i' of this array is a sum(product) of 'a' except the element a[i]. You are not allowed to use '-' ('/') operator.

This is a very interesting problem. Basically, initialize sum = a[0],  from the second element in the rst array, let rst[i] = sum, then sum += a[i]. Note that initialize rst[0] = 1 for product. Then reset sum to a[a.length - 1]. From the second last element, let rst += sum, sum += a[i]. 

The initiative is the first loop will add(multiply) all elements from 0 to i - 1, and the second loop will add (multiply) all elements from i + 1 to a.length - 1. 

public static int[] arraySum(int[] array) {
  if (array == null || array.length == 0)
   throw new IllegalArgumentException("Null or Empty array!");
  int sum = array[0];
  int[] rst = new int[array.length];
  for (int i = 1; i < rst.length; i++) {
   rst[i] = sum;
   sum += array[i];
  }
  sum = array[array.length - 1];
  for (int i = array.length - 2; i >= 0; i--) {
   rst[i] += sum;
   sum += array[i];
  }
  return rst;
 }

public static int[] product(int[] A){
     if (A == null || A.length == 0)
         return A;
     int rst[] = new int[A.length];
     rst[0] = 1;
     //Arrays.fill(rst, 1);
     int product = A[0];
     for (int i = 1; i < A.length; i++){
         rst[i] = product;
         product *= A[i];
     }
     
     product = A[A.length - 1];
     for (int i = A.length - 2; i >= 0; i--){
         rst[i] *= product;
         product *= A[i];
     }
     return rst; 
 }