How to check and compare possible combinations of array elements

Question

String [] A = {"High","Medium","Low"};
String [] B = {"High","Medium","Low"};
String [] C = {"High","Medium","Low"};
String [] D = {"High","Medium","Low"};
String [] E = {"High","Medium","Low"};
String [] F = {"High","Medium","Low"};

JComboBox Ai = new JComboBox(A); JComboBox Bi = new JComboBox(B);
JComboBox Ci = new JComboBox(C); JComboBox Di = new JComboBox(C);
JComboBox Ei = new JComboBox(E); JComboBox Fi = new JComboBox(F);

....

//add the user choice in arrayList
ArrayList<String> a = new ArrayList<String>();
a.add((String) Ai.getSelectedItem());
a.add((String) Bi.getSelectedItem());
a.add((String) Ci.getSelectedItem());
a.add((String) Di.getSelectedItem());
a.add((String) Ei.getSelectedItem());
a.add((String) Fi.getSelectedItem());

Editorial:
Scenario: There are 6 groups (Ai, Bi, Ci, Di, Ei, Fi). In each group there are 3 selections (High (H), Medium (M), Low (L)). The user needs to choose one of 6 groups

The selection may be, for example, “HHHLLL” or “MMMLLM” or “HHLLMM”, etc.

What is the best way to test and match user choices without writing many else if? eg

if(Ai=="High" && Bi=="High" && Ci=="Low" && Di=="High" && Ei=="Low" && Fi=="Medium") {
    System.out.println("Good Choice"); 
}

Thank.

Answer 1

First, you do not need to specify a new list of options for each JComboBox.

String[] choices = {"High", "Medium", "Low"};

JComboBox ai = new JComboBox(choices);
JComboBox bi = new JComboBox(choices);
JComboBox ci = new JComboBox(choices);
JComboBox di = new JComboBox(choices);
JComboBox ei = new JComboBox(choices);
JComboBox fi = new JComboBox(choices);

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JComboBoxes . , .

JComboBox[] boxes = {ai, bi, ci, di, ei, fi};

---

:

// Create an ArrayList of Strings, where each string is either "H", "M", or "L"
ArrayList<String> userChoice = new ArrayList<String>()

for (JComboBox box : boxes) {
    // Go through this code once for each JComboBox in boxes
    // The first time through, "box" means the first JComboBox
    // The second time through, "box" is the second JComboBox, etc.
    if (box.getValue().equals("High")) {
         userChoice.add("H");
    } else if (box.getValue().equals("Medium")) {
         userChoice.add("M");
    } else if (box.getValue().equals("Low")) {
         userChoice.add("L")
    }
}

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---

, ArrayList userChoice, - [ "H", "H", "M", "M", "L", "L" ].

Answer 2

.

: Ai * 3^0 + Bi * 3^1 + ... + Fi * 3^5

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Answer 3

, enum . == ( ) , . enum (toString valueOf).

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Map .

enum Setting {
   High, Medium, Low;
}

class SettingCombination {
  List<Setting> combo = ...;
  // should be immutable, with proper equals and hashCode @Override
  // should also have named getters for each of the 6 settings
}

interface SettingCombinationHandler {
   void handle(SettingCombination settings);      
}

class SettingCombinationHandlingService {
   Map<SettingCombination, SettingCombinationHandler> handlers = ...;
   // perhaps provide a default handler as well
}

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