Assuming customers2 is required, the following result was found.
just the duplicates. The Code v_CountDiscrepancies = 0; // l_Customers1 = Customer[ID != 0].distinct(Customer_Name); l_Customers2 = Customer[ID != 0].Customer_Name.getAll(); // if(l_Customers1.size() != l_Customers2.size()) { v_CountDuplicates =...