JSON : Array & Compare

Question

I have a set of JSON array :

listSession: [h0y78u93, h0y78u93, h0y78u93, h0y78u93, h0y78u93, 9i88u93, 9i88u93, 9i88u93, 9i88u93, 9i88u93]

I've created the array using the below code:

ArrayList<String> listSession = new ArrayList<String>(); 
            for(int u=1; u < k+1; u++) {
                String str = Integer.toString(u);

                JSONArray arrTime=(JSONArray)mergedJSON2.get(str);
                JSONObject objSession;
                StringsessionName;
                for (Object ro : arrTime) {

                    objSession = (JSONObject) ro;
                    sessionName = String.valueOf(objSession.get("sessionID"));

                    listSession.add(sessionName);
                }
            }

May I get your advice or opinion on how am I going to compare the value from each of the attributes in the list. If it is the same, I should it as ONE. Meaning from the above sample, the count should be only TWO instead of TEN.

Thank You.

Answer 1

You can utilize Arraylist.contains() method like below:

ArrayList<String> listSession = new ArrayList<String>(); 
for(int u=1; u < k+1; u++) {
    String str = Integer.toString(u);

    JSONArray arrTime=(JSONArray)mergedJSON2.get(str);
    JSONObject objSession;
    StringsessionName;
    for (Object ro : arrTime) {

        objSession = (JSONObject) ro;
        sessionName = String.valueOf(objSession.get("sessionID"));

        if (!listSession.contains(sessionName)) {
            listSession.add(sessionName);
        }
    }
}

OR

You can use a Set implementation which doesn't allow duplicate values instead of ArrayList. There's no need to compare explicitly.

// initialize
Set sessionsSet = new HashSet(); 
//add like below
sessionsSet.add(sessionName);
sessionsSet.size() // getting the length which should be what you expect to be 2

Answer 2

I would recommend to use a Set over ArrayList here. You can use ArrayList and check the list whether it contains the element and add it. ArrayList.contains() takes O(n) time because it maintains a dynamic array inside. Where as a HashSet or TreeSet can do that check in O(1) and you also don't have to do that compare yourself.

Set<String> setSession = new HashSet<String>(); 
        for(int u=1; u < k+1; u++) {
            String str = Integer.toString(u);
            JSONArray arrTime=(JSONArray)mergedJSON2.get(str);
            JSONObject objSession;
            StringsessionName;
            for (Object ro : arrTime) {

                objSession = (JSONObject) ro;
                sessionName = String.valueOf(objSession.get("sessionID"));
                setSession.add(sessionName);
            }
        }

Answer 3

If you're okay using Java 8, then you can use shorthand implementation like this:

Example:

ArrayList<String> data = new ArrayList<String>(Arrays.asList("A", "A", "A", "B", "B", "B"));

// This will be required if your target SDK < Android N
if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.N) {
    List<String> uniqueData = data.stream().distinct().collect(Collectors.toList()); // Results ["A", "B"]
}