CareerCup

use TRIE
===============
wwwDOTgeeksforgeeks.org/trie-insert-and-search/+&cd=1&hl=en&ct=clnk&gl=in
==================
Using trie, search complexities can be brought to optimal limit (key length). If we store keys in binary search tree, a well balanced BST will need time proportional to M * log N, where M is maximum string length and N is number of keys in tree. Using trie, we can search the key in O(M) time. However the penalty is on trie storage requirements.

use hashset as it does't allow duplicates..time complexity to add elements would be O(n).

Python=>
list =[“Ted”, “John”, “Mark”, “Ted”, “David”]
dict={}
while i in list:
dict[i]=1
del list
while i in dict.keys:
list+=i

'''Python3.3:
Removing duplicates on the fly
E.g.:
a
v
a 

then answer: 
a, v
'''

if __name__ == "__main__":
    name_space = {}
    ordered_list = []
    while True:
        _name = input("")
        if(_name == ""): break
        
        if _name not in name_space:
            name_space[_name] = 1
            ordered_list.append(_name)
    
    print(list(name_space.keys()))
    print("ordered list: ", ordered_list)

Here is my solution in C++:

void unique_stream(vector<string> &stream)
{
    unordered_map<string, int> strMap;
    vector<string> vecResult;
    for(auto& str: stream)
    {    
        strMap.find(str) != strMap.end() ? ++strMap[str] : strMap[str] = 1 ;
    }
    for(auto& str: stream)
    {   
        if(strMap[str] == 1)
            vecResult.push_back(str);
    }
    stream = vecResult;
}

List<String> lstStr = new ArrayList<String>();
lstStr.add("Nuno");
lstStr.add("Paulo");
lstStr.add("Carlos");
lstStr.add("Ana");
lstStr.add("Nuno");
lstStr.add("Paulo");
lstStr.add("Nuno");
lstStr.add("Paulo");
lstStr.add("Sofia");
lstStr.add("Sofia");
lstStr.add("Sofia");
lstStr.add("Sofia");
lstStr.add("Rita");

for(int i=0; i< lstStr.size(); i++)
{
for(int j=1;j<lstStr.size();j++)
{
if(i!= j && lstStr.get(i) == lstStr.get(j))
{
lstStr.remove(j);
j--;
}
if(j>=lstStr.size())
break;
}
if(i>=lstStr.size())
break;
}
for(int k = 0; k<lstStr.size();k++)
{
System.out.println(lstStr.get(k))
}

Here is the pseudo code-

String[]={"ted","mark","mark",,"ted"}
String pre=s[0];
list.add(pre);
for(int i=1;i<s.len;i++)
{
if(pre!=s(i))
list.add(s(1));
pre=s(i);
}
return list;

O(n)