How to efficiently remove duplicates from an array without using Set

Question

I was asked to write my own implementation to remove duplicated values in an array. Here is what I have created. But after tests with 1,000,000 elements it took very long ti

Answer 1

I feel Android Killer's idea is great, but I just wondered if we can leverage HashMap. So I did a little experiment. And I found HashMap seems faster than HashSet.

Here is code:

    int[] input = new int[1000000];

    for (int i = 0; i < input.length; i++) {
        Random random = new Random();
        input[i] = random.nextInt(200000);
    }

    long startTime1 = new Date().getTime();
    System.out.println("Set start time:" + startTime1);

    Set<Integer> resultSet = new HashSet<Integer>();

    for (int i = 0; i < input.length; i++) {
        resultSet.add(input[i]);
    }

    long endTime1 = new Date().getTime();
    System.out.println("Set end time:"+ endTime1);
    System.out.println("result of set:" + (endTime1 - startTime1));     
    System.out.println("number of Set:" + resultSet.size() + "\n");

    long startTime2 = new Date().getTime();
    System.out.println("Map start time:" + startTime1);

    Map<Integer, Integer> resultMap = new HashMap<Integer, Integer>();

    for (int i = 0; i < input.length; i++) {
        if (!resultMap.containsKey(input[i]))
            resultMap.put(input[i], input[i]);
    }

    long endTime2 = new Date().getTime();
    System.out.println("Map end Time:" + endTime2);
    System.out.println("result of Map:" + (endTime2 - startTime2));
    System.out.println("number of Map:" + resultMap.size());

Here is result:

Set start time:1441960583837
Set end time:1441960583917
result of set:80
number of Set:198652

Map start time:1441960583837
Map end Time:1441960583983
result of Map:66
number of Map:198652

Answer 2

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;
}

Answer 3

If you are allowed to use Java 8 streams:

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

Answer 4

I know this is kinda dead but I just wrote this for my own use. It's more or less the same as adding to a hashset and then pulling all the elements out of it. It should run in O(nlogn) worst case.

    public static int[] removeDuplicates(int[] numbers) {
    Entry[] entries = new Entry[numbers.length];
    int size = 0;
    for (int i = 0 ; i < numbers.length ; i++) {
        int nextVal = numbers[i];
        int index = nextVal % entries.length;
        Entry e = entries[index];
        if (e == null) {
            entries[index] = new Entry(nextVal);
            size++;
        } else {
            if(e.insert(nextVal)) {
                size++;
            }
        }
    }
    int[] result = new int[size];
    int index = 0;
    for (int i = 0 ; i < entries.length ; i++) {
        Entry current = entries[i];
        while (current != null) {
            result[i++] = current.value;
            current = current.next;
        }
    }
    return result;
}

public static class Entry {
    int value;
    Entry next;

    Entry(int value) {
        this.value = value;
    }

    public boolean insert(int newVal) {
        Entry current = this;
        Entry prev = null;
        while (current != null) {
            if (current.value == newVal) {
                return false;
            } else if(current.next != null) {
                prev = current;
                current = next;
            }
        }
        prev.next = new Entry(value);
        return true;
    }
}

Answer 5

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

Answer 6

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

    Set<Integer> set = new HashSet<Integer>();
    for(int i : intarray) {
        set.add(i);
    }

    Iterator<Integer> setitr = set.iterator();
    for(int pos=0; pos < intarray.length; pos ++) {
        if(pos < set.size()) {
            intarray[pos] =setitr.next();
        } else {
            intarray[pos]= 0;
        }
    }

    for(int i: intarray)
    System.out.println(i);
}