CareerCup

Create a tree with these numbers and also keep the freq of each number at every node
Now do an inorder traversal to print only those with freq 1

Time complexity : nlogn

I think sorting the linked list and then returning the distinct numbers is better than maintaining a BST for that . There are 2 reason ->

1. Extra memory used to maintain the BST .
2. Worst case runtime : O(n^2)

How about using a hash.

Use Set distinctLst=new HashSet(List);

import java.util.HashSet;
import java.util.Set;

public class MainClass {
public static void main(String args[]) {
String[] name1 = { "Amy", "Jose", "Jeremy", "Alice", "Patrick" };

String[] name2 = { "Alan", "Amy", "Jeremy", "Helen", "Alexi" };

String[] name3 = { "Adel", "Aaron", "Amy", "James", "Alice" };

Set<String> letter = new HashSet<String>();

for (int i = 0; i < name1.length; i++)
letter.add(name1[i]);

for (int j = 0; j < name2.length; j++)
letter.add(name2[j]);

for (int k = 0; k < name3.length; k++)
letter.add(name3[k]);

System.out.println(letter.size() + " letters must be sent to: " + letter);

}
}

hash is best in this case, O(n)
Insert into hash with key = number and data = key.count++, //initially count is zero
if count>1 then dont print because it is already inserted and printed.

Sort - n(logn).
Iterate over the sorted list and track the value of the previous element. if previous==current, do not display the value.

def fun(arr):
       dict = default(int)
       for element in arr:
           dict[element] +=1
       for element in dict:
           if dict[element] == 1:
              print element