How to efficiently remove duplicates from an array without using Set

Slight modification to the original code itself, by removing the innermost for loop.

public static int[] removeDuplicates(int[] arr){
    int end = arr.length;

    for (int i = 0; i < end; i++) {
        for (int j = i + 1; j < end; j++) {
            if (arr[i] == arr[j]) {                  
                /*int shiftLeft = j;
                for (int k = j+1; k < end; k++, shiftLeft++) {
                    arr[shiftLeft] = arr[k];
                }*/
                arr[j] = arr[end-1];
                end--;
                j--;
            }
        }
    }

    int[] whitelist = new int[end];
    /*for(int i = 0; i < end; i++){
        whitelist[i] = arr[i];
    }*/
    System.arraycopy(arr, 0, whitelist, 0, end);
    return whitelist;
}

If you are allowed to use Java 8 streams:

Arrays.stream(arr).distinct().toArray();

you can take the help of Set collection

int end = arr.length;
Set<Integer> set = new HashSet<Integer>();

for(int i = 0; i < end; i++){
  set.add(arr[i]);
}

now if you will iterate through this set, it will contain only unique values. Iterating code is like this :

Iterator it = set.iterator();
while(it.hasNext()) {
  System.out.println(it.next());
}

Note: I am assuming the array is sorted.

Code:

int[] input = new int[]{1, 1, 3, 7, 7, 8, 9, 9, 9, 10};
int current = input[0];
boolean found = false;

for (int i = 0; i < input.length; i++) {
    if (current == input[i] && !found) {
        found = true;
    } else if (current != input[i]) {
        System.out.print(" " + current);
        current = input[i];
        found = false;
    }
}
System.out.print(" " + current);

output:

  1 3 7 8 9 10