Algorithm for removing duplicates from ArrayList

Question

I have an ArrayList<String> that contains the following:

2#3#1#0

1#0#4#1

9#2#5#0

4#2#3#2

1#1#2#1

Output: 6 different numbers.

I'm trying to write an algorithm that removes duplicates of the highlighted numbers so I can then use a counter to see how many different numbers in total in all of those locations are.

I've tried many things including some of the following: [Java remove duplicates from array using loops][1], [Java - Removing duplicates in an ArrayList][2], the first option in [How to find duplicates in Java array?][3] and many more. I've spent at least 5-10h just trying to figure what I'm doing wrong, but I can not, so I've turned to you.

Most of the time the solutions I find online seem to work on simple stuff, but not in my case. In it, when I try to print the different characters, it always returns the wrong int numbers.

I've also tried, also tried separating each line of numbers into a different int Array[] and then comparing, but it just won't catch all the different values.

In another example where I had 5 different numbers in total, I kept getting "4 different" as a result, so I even tried long n = ArrayList.stream().distinct().count(); just to see if I was doing something wrong, but even this thing returned "4 different" numbers.

I know the easiest way is using Set and Map, but I don't want that. I'd like to have an algorithm.

EDIT:

One of the many things I've tried is the following:

for (int m = 0; m < (size-1); m++){
        for (int j = m + 1; j < size; j++){
            if (ArrayList.get(j).charAt(0) != ArrayList.get(m).charAt(0)){
                continue;
            }
            current++;
            ArrayList.remove(j).charAt(0);
            j--;
            size--;
        }
    }

With this one, I'd have to use another one for the ArrayList.get().charAt(4).

EDIT2:

I've found the following code [here][1], but how would it be implemented in this case?

public static <T> ArrayList<T> uniquefy(ArrayList<T> myList) {

    ArrayList <T> uniqueArrayList = new ArrayList<T>();
    for (int i = 0; i < myList.size(); i++){
        if (!uniqueArrayList.contains(myList.get(i))){
            uniqueArrayList.add(myList.get(i));
        }
    }

    return uniqueArrayList;
}

EDIT3: I've found a possible solution, but it gives me an IndexOutOfBoundsException. I've put the numbers 2, 1, 9, 4, 1 into Array1 and 1, 4, 5, 3, 2 into Array2, but when I try to compare them, I get the mentioned error.

boolean stopSequence = false;
    for (int i = 0; i < Array1.length; i++){
        for (int a = 0; a < Array2.length && !stopSequence;){
            if (Array1[i] != Array2[a]){
                Array1[i] = 0;
                a++;
            }
            if (Array1[i] == Array2[a]){
                Array1[i] = 0;
                stopSequence = true;
            }
        }
        stopSequence = false;
    }

[1]: https://stackoverflow.com/questions/26998156/java-remove-duplicates-from-array-using-loops
[2]: https://stackoverflow.com/questions/2435156/java-removing-duplicates-in-an-arraylist
[3]: http://javarevisited.blogspot.com.es/2015/06/3-ways-to-find-duplicate-elements-in-array-java.html
[4]: https://stackoverflo

w.com/questions/203984/how-do-i-remove-repeated-elements-from-arraylist?rq=1

What would be the result required for the given input example? — Sanket Makani
– Sanket Makani, Commented May 25, 2017 at 9:08
I would use a counter and print, in this case, "6 different numbers". — Doombringer
– Doombringer, Commented May 25, 2017 at 9:09
When you have code that is not working... Show it here. See minimal reproducible example — GhostCat
– GhostCat, Commented May 25, 2017 at 9:11
What is your question exactly? We won't write the code for you, and you didn't post any. You're also rejecting the obvious solution consisting in putting all the numbers in a set, and then printing the size of the set. Why? — JB Nizet
– JB Nizet, Commented May 25, 2017 at 9:12

JB Nizet · Accepted Answer · 2017-05-25 10:59:26Z

3

The algorithm is much simpler than what you think it is:

transform every string into a pair of characters
putting all the characters into a collection or stream that removes duplicates
counting the number of characters.

Here is a complete example:

import java.util.Arrays;
import java.util.List;
import java.util.stream.IntStream;

public class Duplicates {
    public static void main(String[] args) {
        List<String> list = Arrays.asList("2#3#1#0",
                                          "1#0#4#1",
                                          "9#2#5#0",
                                          "4#2#3#2",
                                          "1#1#2#1");
        System.out.println(
            list.stream()
                .flatMapToInt(s -> IntStream.of(s.charAt(0), s.charAt(4)))
                .distinct()
                .count());
    }
}

EDIT: You seem to want to obey absurd restrictions, and thus neither use a Stream nor a Set, where these completely make sense. Here's code only using lists, but doing basically the same thing as above, but in a much less efficient way:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class Duplicates {
    public static void main(String[] args) {
        List<String> list = Arrays.asList("2#3#1#0",
                                          "1#0#4#1",
                                          "9#2#5#0",
                                          "4#2#3#2",
                                          "1#1#2#1");
        List<Character> uniqueChars = new ArrayList<>();
        for (String s : list) {
            Character c0 = s.charAt(0);
            Character c4 = s.charAt(4);

            if (!uniqueChars.contains(c0)) {
                uniqueChars.add(c0);
            }
            if (!uniqueChars.contains(c4)) {
                uniqueChars.add(c4);
            }
        }

        System.out.println(uniqueChars.size());
    }
}

edited May 25, 2017 at 10:59

answered May 25, 2017 at 9:23

JB Nizet

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10 Comments

Doombringer Over a year ago

Unfortunately, I cannot use Stream. I'm pretty sure it can all be done using just normal loops, because I've done similar things, just not when I've had 2 things in each string line that i need to compare individually to all the others.

Little Santi Over a year ago

Excellent solution! Streams just don't stop astounding me. +1

Viktor Mellgren Over a year ago

I was thinking of a similar solution, but the flatMapToInt was a really nice compact way of writing it.

JB Nizet Over a year ago

@DoombringerBG you can apply the same algorithm with loops and a Set<Integer>. But unless you are constrained by your platform (android?), you'd better start learning streams, because they're very powerful and allow for concise, expressive and efficient code.

Doombringer Over a year ago

@JBNizet If I wanted to use Sets or Maps or Streams or w/e in this case, I wouldn't have asked what I asked.

|

Oghli · Accepted Answer · 2017-05-25 10:45:24Z

2

It's not that difficult to count different numbers of the highlighted locations.you can use helper array called frequency array to get the expected result.

Try this simple algorithm using frequency array I think it worked perfectly for your case:

       ArrayList<String> numlist=new ArrayList<String>();
       int freq[] = new int [10];
       numlist.add("2#3#1#0");
       numlist.add("1#0#4#1");
       numlist.add("9#2#5#0");
       numlist.add("4#2#3#2");
       numlist.add("1#1#2#1");
       for(int i = 0; i < numlist.size(); i++){
           String row = numlist.get(i);          
           int numValue1 = Character.getNumericValue(row.charAt(0));
           int numValue2 = Character.getNumericValue(row.charAt(4));
           freq[numValue1]++;
           freq[numValue2]++;          
       }
       int count = 0;
       for(int i = 0; i < 10; i++){
           if(freq[i] > 0){
               count++;
           }
       }
       System.out.println(count + " different numbers");

Output:

6 different numbers

answered May 25, 2017 at 10:45

Oghli

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2 Comments

Doombringer Over a year ago

A little bit too late, but I think I would've used this, if JB Nizet didn't beat you to it. Either way, I'm thankful for your help. :3

Oghli Over a year ago

No problem. you're welcome. don't make your problem more complicated keep it simple as possible. try to break it down to a simpler pieces in order to get the expected result in clean way.

gil.fernandes · Accepted Answer · 2017-05-25 10:18:36Z

1

Another option with bit masks:

public static void main(String[] args) {
    List<String> arrayList = Arrays.asList("2#3#1#0", "1#0#4#1", "9#2#5#0", "4#2#3#2", "1#1#2#1");
    int mask = 0;
    for(String s : arrayList) { // Place the bits
        mask = mask | (1 << Character.getNumericValue(s.charAt(0))) | (1 << Character.getNumericValue(s.charAt(4)));
    }
    int counter = 0;
    for(int i = 0; i < 32; i++) { // count the bits
        counter += (mask & (1 << i)) == 1 << i ? 1 : 0;
    }
    System.out.println(counter);
}

Output:

This relies on the bit mask which is at the end of the execution of the code:

1000111110

Possibly this is faster than most solutions, since it does not rely on conventional data structures.

edited May 25, 2017 at 10:18

answered May 25, 2017 at 9:42

gil.fernandes

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2 Comments

Doombringer Over a year ago

Never seen bit masks in my life, but it does seem easy to read. :3

gil.fernandes Over a year ago

Just a question to whoever sees this: is there a potential way to have a "faster" solution for this problem which uses less memory? How does this compare in terms of efficiency with other solutions displayed above?

Little Santi · Accepted Answer · 2017-05-25 09:19:49Z

Well, a good practice is always to divide the problem into smaller parts:

For example, a good design would be a class with these members:

digits: This is an instance variable of array of ints to contain the number of times each digit was repeated. It must be pre-sized to the maximum allowed digit (I guess that is 9).
differentDigits: The is an instance variable to contain the number of different digits.
processList: This method shall receive the list to browse it and call processItem for each item.
processItem: This method shall receive an item String and parse the digits according to the specified format (through StringTokenizer, for example), and call storeDigit for each required digit.
storeDigit: This method shall receive an int and use it to index the instance array digits, and increment the indexed position. If the indexed position was 0, it should also increment differentDigits.

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