CareerCup

O(N), Idea: let's use data structure which avoids duplicates - hashset

public static Integer[] removeDuplicates1(Integer[] arr){
        HashSet<Integer> set = new HashSet<>();
        for (Integer elem : arr){
            set.add(elem);
        }
        Integer[] answer = new Integer[set.size()];
        set.toArray(answer);
        return answer;
}

O(NlogN), Idea: let's sort input and place all unique elements in the beggining

public static Integer[] removeDuplicates2(Integer[] arr){
        if (arr.length == 0){
            return new Integer[0];
        }
        Arrays.sort(arr);
        int lastNonRepeated = 0;
        for (int i =  1; i < arr.length; i++){
            if (arr[i] == arr[lastNonRepeated]){
                continue;
            }
            arr[++lastNonRepeated] = arr[i];
        }
        Integer[] answer = new Integer[lastNonRepeated+1];
        for (int i = 0; i <= lastNonRepeated; i++){
            answer[i] = arr[i];
        }
        return answer;
    }

def removeDup(arraylist):
	hash = {}

	for i in arraylist:
		if i not in hash:
			hash[i] = 1

	newlist = []
	for j in hash:
		newlist.append(j)
	return newlist

what does duplicate mean here

In C++ you can use std::set<int> to store the values as you scan the array. Then output the values of set.

public void RemoveDuplicates(int[] arr)
{
int[] updated = new int[arr.Length];
List<int> list = new List<int>();

for(int i=0;i<arr.Length;i++)
{

if (!list.Contains(arr[i]))
{
list.Add(arr[i]);
Console.WriteLine(arr[i]);
}

}

}

C# implementation using HashSet

public int[] RemoveDuplicates(int[] arr)
{
    HashSet<int> result = new HashSet<int>();

    for (int i = 0; i < arr.Length; i++)
    {
        if (!result.Contains(arr[i]))
        {
            result.Add(arr[i]);
        }
    }

    return result.ToArray();
}

since when did Microsoft started to ask such straight forward stuff...:)

We can have a hashmap where key and value will be integer only, and if value exist for a key means that number is already there otherwise just insert the number.
so lookup is o(1) and it will take one pas on array.
Therefor it will take o(n) time.