Question

Some service fees (% of asset amount) of two investment companies are shown in the following tables:

Company A:

Service Fee Type	Fee
Management Fee	1.5
Subscription Fee	5.0
Redemption Fee	1.0

Company B:

Service Fee Type	Fee
Platform Fee	0.2
Management Fee	1.0
Subscription Fee	2.0

The above information is to be stored in two maps (Map<String, Double>) referenced by companyA and companyB for Company A and Company B respectively. The service fee type is assumed to be unique in each map and it is the key of it.

1. Write a program segment to store the above information of Company A in a map companyA. Remember to create it first.

2. Assume that there are many entries in the map referenced by companyA. Using an enhanced for loop and an array, write a program segment to store the number of service fee types in each fee category using an element of the array. The fee categories are "fee <= 0.5", "0.5 < fee <= 1.0", and "fee > 1.0".

3. Assume there are many entries in the maps and one company can have some service fee types not available in the other company. Write a program segment with an enhanced for loop to print out the name of each service fee available in both of the companies (A and

B) and its lowest fee. For example, the string "Subscription Fee, 2.0" is printed when we are considering "Subscription Fee".

Answer 1

a) program segment to create the 2 maps
       Map<String, Double> mapA = new HashMap<String, Double>();
       Map<String, Double> mapB = new HashMap<String, Double>();

       mapA.put("Management Fee", 1.5);
       mapA.put("Subscription Fee", 5.0);
       mapA.put("Redemption Fee", 1.0);
      
       mapB.put("Platform Fee", 0.2);
       mapB.put("Management Fee", 1.0);
       mapB.put("Subscription Fee", 2.0);

==========================================================================
b)program segment to store the fees in 3 categories in an array
       int[] counts = new int[3];
       Double fee;
       for(String s : mapA.keySet()) {
           fee = mapA.get(s);
           if(fee <= 0.5)
               counts[0]++;
           else if(fee <= 1.0)
               counts[1]++;
           else
               counts[2]++;
       }
      
===============================================
c) program segment to print out fees for service types available in both companies
      
       for(String s : mapA.keySet()) {
           if(mapB.containsKey(s)) {
               double f1 = mapA.get(s);
               double f2 = mapB.get(s);
               if(f1 < f2)
                   System.out.println(s + ", " + f1);
               else
                   System.out.println(s + ", " + f2);
               }
           }
       }