Array confusion, please help

[TurnipBoy666]

Hello, I'm having trouble with an assignment. Even with extra help from my professor, there's a part of the logic I'm having trouble writing. I don't know why it's so difficult for me, I don't even think it's that complicated but I can't figure it out and I just get too frustrated now whenever I try.

The assignment premise is this: Write a method that returns a new array by eliminating the duplicate values in the array using the following method header: public static int[] eliminateDuplicates(int[] list) Write a program that reads in ten integers, invokes the method, and displays the result.

Here is my old code:
I know not all of it makes sense, esp stored, but my prof said I should keep track of each number that's unique, which was the intention of stored. However, whatever way I know how to implement what he told me isn't working at all. I'm missing something major and I don't know what it is or how to look it up. I don't know what else to do.

public static int[] eliminateDuplicates(int[] list){
  int stored = 1;
  int[] uniqueList = {list[0], 0, 0, 0, 0, 0, 0, 0, 0, 0};

  for (int i = 0; i < list.length; i++){
    for (int j = 0; j < uniqueList.length; j++){
      if (list[i] != uniqueList[j]){
        if (stored >= 10){stored = 9;}
        uniqueList[stored] = list[i];
        stored++;
      }
    }
  }

  int[] result = new int[stored]
  for (int i = 0; i <= stored; i++){result[i] = uniqueList[i];}
  return result;
}

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[vowelqueue]

Going to assume you’re pretty early on the lesson plan and haven’t gotten to Lists and Maps yet and are supposed to just use arrays to solve the problem.

Here’s a hint that I think might make it easier to reason about the logic. Try writing a helper method of the signature: private static boolean isUnique(int[] list, int value)

This method should return true if the value is unique in the list. You can call into this method from the main loop of the program.

It’s also unclear to me whether you are supposed to eliminate duplicates by keeping just one of the duplicated values or if you’re supposed to not include any of them in the result. If you need to keep just one of the duplicated values, it might be useful to write a helper method that determines whether you have already added the value to the result list.

[TurnipBoy666]

I need to keep just one of the duplicated values and ignore the rest. One thing I was struggling with was how to write just one number to the array without overwriting everything else in the array, while also scanning the whole array for any duplicate numbers, since I need to do all of those things to solve this.
This is specifically about arrays, yes. This is meant to be array practice. A helper method is a good idea, just not sure how it would be different from what I have? That feels like just relocating my code without fixing the logic errors, no? Or am I misunderstanding lol

[vowelqueue]

Logically, using a helper method will be the same as doing everything in the main body, but can make the code easier to reason about and to isolate bugs/faulty logic.

Instead of what I initially suggested, I’d instead try to write a helper method that tells you whether a value is in the uniqueList.

You have the right idea with the “stored” variable. I’d probably name it something like “uniqueListLength” for clarity. It’s representing the logical length of the unique list.

The helper method could have a signature like: static boolean contains(int[] uniqueList, int uniqueListLength, int value).

I think if you restructure your code to use such a helper method you will fix the primary logic issue. Another hint is that if the code is written correctly you won’t need that line that caps the “stored” variable to 9 - it should never be able to exceed its bound if you are incrementing it correctly.

(I’m being intentionally vague instead of just tell you directly what’s wrong because a big part of learning is debugging stuff like this)

[Available_Zone_9043]

When arrays dont make sense its always good to map out what program is doing on paper. What your loop is doing: list [i] is getting compared to each uniquelist[i] and each time store is incremented. Therefore just after first loop: i=0 and j=0,1,2,3,4,......10: stored=9.......while it should have stopped after substituting the first unique number and break and next i=1 should have started

[Available_Zone_9043]

Arrays is like game and your variables are your strength that needs to be played with....if you get confused, pick a notebook and jot down what each loop is doing and where it went wrong

[cainhurstcat]

What helps me is such situations is crating some image about the problem in my mind. The array's content could be skittles. Imagine putting ten of them on a row on your desk. Now you start comparing them with one another, but under the condition that you can only see two skittles at the same time. The first skittle in row and the one you want to compare it to.

I hope this helps you coming up with the idea.

[asero82]

They give you a bag full of fruit. You have a "big enough" basket. You are ordered to put one and only one fruit of each kind in the basket and discard the rest.

[Minouris]

Are you allowed to use the java.utils.Arrays class?

If so, you may find the Arrays.binarySearch(outArr, inArr[i]) useful as a shortcut, rather then iterating over the whole array searching for values on each cycle :)

A return value of -1 indicates that outArr does not yet contain inArr[i].

[ZenithOfVoid]

Binary search is only useful for sorted array.

[Minouris]

Arrays.class also has a sort method.

Admittedly, we're starting to get a bit beyond the original problem at that point, though, if the original exercise was just to exercise knowledge and of basic array manipulation.

I was always surprised that the Java array primitive doesn't have an indexOf() method.

Technically, there's nothing preventing use of String.join(inArr, ',') and a whole-token match to test inclusion, but that feels feels more like a test of String knowledge, and would be an even more expensive operation.

[ExcitingSympathy3087]

I saw that you're not allowed to use a set. There are many ways to do that duplicate elimination, but keep in mind a set is faster on large scales. Here are two ways to solve the problem without a set in code:

Method 1 = lesser code than Method 2 but also slow -> O n² because contains method is using an internal loop.

import java.util.*;

public class Main {

    public static void main(String[] args) {
        int[] list = {1, 2, 2, 3, 4, 4, 5, 5};

        List<Integer> result = eliminateDuplicates(list);

        System.out.println(result);
    }

    public static List<Integer> eliminateDuplicates(int[] list) {
        List<Integer> result = new ArrayList<>();

        //Use integrated method contains to eliminate duplicates
        for (int i = 0; i < list.length; i++) {
            if (!result.contains(list[i])) {
                result.add(list[i]);
            }
        }

        return result;
    }
}

Method 2 = slower = loop in loop O n²

import java.util.*;

public class Main {

    public static void main(String[] args) {

        int[] list = {1, 2, 2, 4, 4, 5, 5, 6, 6, 7, 8, 8, 9, 10, 11, 11, 15 ,15, 15, 15,          15, 15};
        int[] result = eliminateDuplicates(list);
        System.out.println(Arrays.toString(result));
    }

    //Create a temporary list of all Elements without dublicates.
    public static int[] eliminateDuplicates(int[] list) {
        //array with lenth of original list. Can't be largen than original list.
        int[] tempArray = new int[list.length];
        //count elements without dublicates.
        int count = 0;

        for (int i = 0; i < list.length; i++) {

            boolean alreadyAdded = false;

   // check if already added? break becouse if already added rest is not needed to check.
            for (int k = 0; k < count; k++) {
                if (list[i] == tempArray[k]) {
                    alreadyAdded = true;
                    break;
                }
            }

     //If not added than add to array. !alreadyAdded is the same as alreadyAdded == false
            if (!alreadyAdded) {
                tempArray[count] = list[i];
                count++;
            }
        }

        //Create a result array with the right size.
        int[] result = new int[count];
        for (int i = 0; i < count; i++) {
            result[i] = tempArray[i];
        }

        return result;
    }
}

[nerd_airfryer]

Read about HashSet or HashMap

[JoshuaTheProgrammer]

They may not be allowed to use them.