How to remove duplicates from a list using an auxiliary array in Java?

Question

I am trying to remove duplicates from a list by creating a temporary array that stores the indices of where the duplicates are, and then copies off the original array into a

Answer 1

You can use a set for removing multiples.

Answer 2

Here's another alternative without the use of sets, only primitive types:

public static double [] removeDuplicates(double arr[]) {
    double [] tempa = new double[arr.length];
    int uniqueCount = 0;
    for (int i=0;i<arr.length;i++) {
        boolean unique = true;
        for (int j=0;j<uniqueCount && unique;j++) {
            if (arr[i] == tempa[j]) {
                unique = false;
            }
        }
        if (unique) {
            tempa[uniqueCount++] = arr[i];
        }
    }

    return Arrays.copyOf(tempa,  uniqueCount);
}

It does require a temporary array of double objects on the way towards getting your actual result.

Answer 3

Instead of doing it in array, you can simply use java.util.Set.

Here an example:

public static void main(String[] args)
{
    Double[] values = new Double[]{ 1.0, 2.0, 2.0, 2.0, 3.0, 10.0, 10.0 };
    Set<Double> singleValues = new HashSet<Double>();

    for (Double value : values)
    {
        singleValues.add(value);
    }
    System.out.println("singleValues: "+singleValues);
    // now convert it into double array
    Double[] dValues = singleValues.toArray(new Double[]{});
}

Answer 4

You have already used num_items to bound your loop. Use that variable to set your array size for tempa also.

double tempa [] = new double [num_items];

Answer 5

Imagine this was your input data:

Index: 0, 1, 2, 3, 4, 5, 6, 7, 8
Value: 1, 2, 3, 3, 3, 3, 3, 3, 3

Then according to your algorithm, tempa would need to be:

Index: 0, 1, 2, 3, 4, 5, 6, 7, 8, ....Exception!!!
Value: 3, 4, 5, 6, 7, 8, 4, 5, 6, 7, 8, 5, 6, 7, 8, 6, 7, 8, 7, 8, 8

Why do you have this problem? Because the first set of nested for loops does nothing to prevent you from trying to insert duplicates of the duplicate array indices!

What is the best solution?

Use a Set! Sets guarantee that there are no duplicate entries in them. If you create a new Set and then add all of your array items to it, the Set will prune the duplicates. Then it is just a matter of going back from the Set to an array.

Alternatively, here is a very C-way of doing the same thing:

//duplicates will be a truth table indicating which indices are duplicates.
//initially all values are set to false
boolean duplicates[] = new boolean[items.length];
for ( int i = 0; i< numItems ; i++) {
    if (!duplicates[i]) { //if i is not a known duplicate
        for(int j = i + 1; j < numItems; j++) {
            if(items[i] ==items[j]) {
                duplicates[j] = true; //mark j as a known duplicate
            }
        }
    }
}

I leave it to you to figure out how to finish.

Answer 6

Done for int arrays, but easily coud be converted to double.

1) If you do not care about initial array elements order:

private static int[] withoutDuplicates(int[] a) {
    Arrays.sort(a);
    int hi = a.length - 1;
    int[] result = new int[a.length];
    int j = 0;
    for (int i = 0; i < hi; i++) {
        if (a[i] == a[i+1]) {
            continue;
        }
        result[j] = a[i];
        j++;            
    }
    result[j++] = a[hi];
    return Arrays.copyOf(result, j);
}

2) if you care about initial array elements order:

private static int[] withoutDuplicates2(int[] a) {
    HashSet<Integer> keys = new HashSet<Integer>();
    int[] result = new int[a.length];
    int j = 0;
    for (int i = 0 ; i < a.length; i++) {
        if (keys.add(a[i])) {
            result[j] = a[i];
            j++;
        }
    }
    return Arrays.copyOf(result, j);
}

3) If you do not care about initial array elements order:

private static Object[] withoutDuplicates3(int[] a) {
    HashSet<Integer> keys = new HashSet<Integer>();
    for (int value : a) {
        keys.add(value);
    }
    return keys.toArray();
}