Question

Use java programming for the following questions to have the solutions ready within today Jan 3 2020 at or before 7:00PM GMT+8 time zone:

Question 13

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.

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

// TEXT CODE

import java.util.HashMap;
import java.util.Map;

public class CompanyService {

    public static void main(String[] args){

        // Solving part A)

        // creating companyA map reference
        Map<String, Double> companyA = new HashMap<>();

        // creating companyB map reference
        Map<String, Double> companyB = new HashMap<>();

        // adding companyA service information
        companyA.put("Management Fee", 1.5);
        companyA.put("Subscription Fee", 5.0);
        companyA.put("Redemption Fee", 1.0);

        // adding companyB service information
        companyB.put("Platform Fee", 0.2);
        companyB.put("Management Fee", 1.0);
        companyB.put("Subscription Fee", 2.0);



        // solving part B)

        // creating an array to store the number of service fee types in each three fee categories
        int[] serviceTypesCount = new int[3];

        // using enhanced for loop to get the number of service fee types
        for(Map.Entry<String, Double> entry: companyA.entrySet()){

            // if category is "fee <= 0.5" store count at index 0
            if(entry.getValue() <= 0.5)
                serviceTypesCount[0]++;

            // if category is "0.5 < fee <= 1.0" store count at index 1
            if(0.5 < entry.getValue() && entry.getValue() <= 1.0 )
                serviceTypesCount[1]++;

            // if category is "fee > 1.0" store count at index 2
            if(entry.getValue() > 1.0)
                serviceTypesCount[2]++;

        }

        // printing number of service fee types of each category
        System.out.println(".......Number of service fee types in each fee category for companyA........");
        System.out.println("Category \"fee <= 0.5\" count is: " + serviceTypesCount[0]);
        System.out.println("Category \"0.5 < fee <= 1.0\" count is: " + serviceTypesCount[1]);
        System.out.println("Category \"fee > 1.0\" count is: " + serviceTypesCount[2]);



        // Solving part C)
        System.out.println("\nPrinting the name of each service fee available in both of the companies (A and B) and its lowest fee.");
        for (Map.Entry<String, Double> companyAEntry: companyA.entrySet()){

            String key1 = companyAEntry.getKey();

            // check if key1 is present in companyB or not
            if(companyB.containsKey(key1)){
                // get the lowest fee
                double minFee = 0.0;
                if(companyA.get(key1) < companyB.get(key1)){
                   minFee = companyA.get(key1);
                }else minFee = companyB.get(key1);

                //  print out the name of each service fee available in both of the companies (A and B) and its lowest fee
                System.out.println(key1 + " " + minFee);
            }
        }
    }
}

// SCREENSHOT OF CODE

import java.util.HashMap; import java.util.Map; public class Company Service { public static void main(String[] args) { // solving part A) // creating companyA map reference Map<String, Double> companyA = new HashMap<>(); // creating company map reference Map<String, Double> company B = new HashMap<>(); // adding companyA service information companyA.put(k: "Management Fee", v: 1.5); companyA.put(k: "Subscription Fee", v: 5.0); companyA.put(k: "Redemption Fee", v: 1.0); // adding company service information companyB.put( k: "Platform Fee", v: 0.2); companyB.put(k: "Management Fee", v: 1.0); companyB.put( k: "Subscription Fee", v: 2.0); // solving part B) // creating an array to store the number of service fee types in each three fee categories int[] serviceTypesCount = new int[3]; // using enhanced for Loop to get the number of service fee types for(Map. Entry<String, Double> entry: companyA.entrySet()){ // if category is "fee <= 0.5" store count at index o if(entry.getValue() <= 0.5) serviceTypesCount[@]++; // if category is "0.5 < fee <= 1.0" store count at index 1 if(0.5 < entry.getValue() && entry.getValue() <= 1.0 ) serviceTypesCount[1]++; // if category is "fee > 1.9" store count at index 2 if (entry.getValue() > 1.0)

serviceTypesCount[2]++; // printing number of service fee types of each category System.out.println(".......Number of service fee types in each fee category for companyA........"); System.out.println("Category "fee <= 0.5\" count is: " + serviceTypesCount[@]); System.out.println("Category \"0.5 < fee <= 1.0\" count is: " + serviceTypesCount[1]); System.out.println("Category "fee > 1.0\" count is: " + serviceTypesCount[2]); // Solving part C) System.out.println("\nPrinting the name of each service fee available in both of the companies (A and B) and its lowest fee."); for (Map. Entry<String, Double> companyAEntry: companyA.entrySet()){ String key1 = companyAEntry.getKey(); // check if key1 is present in companyB or not if (company.containsKey(key1)){ // get the lowest fee double minFee = 0.0; if(companyA.get(key1) < companys.get(key1)){ minFee = companyA.get(keyi); }else minFee = companyB.get(key1); // print out the name of each service fee available in both of the companies (A and B) and its Lowest fee System.out.println(key1 + " " + minFee);

// OUTPUT

.......Number of service fee types in each fee category for companyA.... Category "fee <= 0.5" count is: 0 Category "0.5 < fee <= 1.0" count is: 1 Category "fee > 1." count is: 2 Printing the name of each service fee available in both of the companies (A and B) and its lowest fee. Management Fee 1.0 Subscription Fee 2.0