How to remove duplicates from a parallel array in Java?

Question

So, I started learning Java and was wondering how parallel arrays of string and int type could be stored exactly once from the source arrays. For example, I have two arrays parallel to each other, one stores the Phone number as a string and the other stores the duration of the calls as a/an int gotten from each phone number.

String[] phoneNumbers;           
    phoneNumbers = new String[100];
    int[] callDurations = new int[phoneNumbers.length];
    int size = 0;

    phoneNumbers[0] = "888-555-0000";
    callDurations[0] = 10;
    phoneNumbers[1] = "888-555-1234";
    callDurations[1] = 26;
    phoneNumbers[2] = "888-555-0000";
    callDurations[2] = 90;
    phoneNumbers[3] = "888-678-8766";
    callDurations[3] = 28;

    size = 4;

I wrote a method to find the details of a specific phone number, such as the duration of the specific call "888-555-1234" Here is the method and how I called it:

public static void findAllCalls(String[] phoneNumbers, int[] callDurations, int size, String targetNumber) {
    int match;
    System.out.println("Calls from " + targetNumber + ":");
    match = find(phoneNumbers, size, 0, targetNumber);

    while (match >= 0) {
        System.out.println(phoneNumbers[match] + " duration: " + callDurations[match] + "s");

        match = find(phoneNumbers, size, match + 1, targetNumber);

    }
}

System.out.println("\n\nAll calls from number: ");
    findAllCalls(phoneNumbers, callDurations, size, "888-555-1234");

The output of this code is:

All calls from number: 
Calls from 888-555-1234:
888-555-1234 duration: 26s
888-555-1234 duration: 28s

Process finished with exit code 0

Whereas,the output I want to get instead is:

All calls from number: 
Calls from 888-555-1234:
888-555-1234 duration: 54s


Process finished with exit code 0

(26s + 28s)

How is it possible in java to make sure there are no duplicates stored in a parallel array and get total duration for each phone number instead of having them separately in the arrays?

You're probably going to want to use a different data structures; arrays aren't enough for you. Learn about classes that implement the `Set` interface in Java https://docs.oracle.com/javase/tutorial/collections/interfaces/set.html Although, being that you want there to be a relationship between two entities, a `Map` may be of more use to you: https://docs.oracle.com/javase/tutorial/collections/interfaces/map.html — Michael, Sep 20 '15 at 20:55
Okay noted,thanks. I wanted to know if there's a way to create a method that creates a new pair of parallel arrays (phone numbers and durations, again) where each different incoming phone number is stored exactly once, and the duration is the total duration of all the calls from that phone number so It then prints these arrays. a void method perhaps? — ekeith, Sep 20 '15 at 21:12

score 0 · Answer 1 · answered Sep 20 '15 at 21:01

The question was: "How is it possible in java to make sure there are no duplicates stored in a parallel array and get total duration for each phone number instead of having them separately in the arrays?"

The answer is: There is no (inexpensive) way.

Use a hash map instead. Have a look at java.utils.HashMap. A hash map is a concept to store values (of any kind) associated to a specific key. In your case the values would be the durations, the keys would be your phone number. Therefor you should use a String-Integer hash map here.

On insert do the following:

For each phone number-duration pair do:
- Is there already an element in the HashMap of the specified key?
- No -> Add phone number and duration
- Yes ->
  - Get the duration stored
  - Add the current duration to the stored duration
  - Overwrite the existing item with the new duration calculated

Later you efficiently can perform a lookup.

score 0 · Answer 2 · answered Sep 20 '15 at 21:01

A Map is an object that maps keys to values

In your case, you want phone numbers (stored in a String) to correspond to call duration (ints). Therefore, you'd declare your HashMap as follows (Note you can't instantiate Map, it is an interface):

Map<String, Integer> callRecords = new HashMap<String, Integer>();

This is a better version because you no longer need to keep track of two different arrays. Now, instead of

phoneNumbers[0] = "888-555-0000";
callDurations[0] = 10;

You can write:

callRecords.put("888-555-0000", 10);

score 0 · Accepted Answer · edited May 23 '17 at 11:52

0

As already stated in the answers before, you can use a map - will avoid duplicates in both phoneNumber and callDuration (Java code to Prevent duplicate <Key,Value> pairs in HashMap/HashTable).

Or, if you want to stick with the String implementation, you can change the logic in the findAllCalls() method.

public static void findAllCalls(String[] phoneNumbers, int[] callDurations, int size, String targetNumber) 
{
   int match;
   System.out.println("Calls from " + targetNumber + ":");
   //match = find(phoneNumbers, size, 0, targetNumber);
   int i = 0, duration = 0;
   while (i<size)
    {
        if(phoneNumbers[i].equals(targetNumber))
            duration+=callDurations[i];
        i++;
      //System.out.println(phoneNumbers[match] + " duration: " + callDurations[match] + "s");
      //match = find(phoneNumbers, size, match + 1, targetNumber);
   }
   System.out.println(targetNumber+" duration : "+duration+"s");
}

edited May 23 '17 at 11:52

Community

1
1

answered Sep 20 '15 at 21:12

Apoorvaa Singh

26
5

Yes, i would like to stick to string implementation as I am new to Mapping. There seems to be a problem with your code. it doesn't print out the total duration, instead it prints out the first matching phone number and duration. – ekeith Sep 20 '15 at 21:17
there's still a problem. It doesn't add the two durations. Instead, it prints out the second matching phone number and duration. – ekeith Sep 20 '15 at 21:58
Tested against more cases, getting correct output. Try a dry run of your code, there might be a typo. – Apoorvaa Singh Sep 20 '15 at 22:13

How to remove duplicates from a parallel array in Java?

3 Answers3