HourRank 30 | Video Conference | Hackerrank

Source Hackerrank

Bob is making a video conference software. Whenever a new person joins the conference, Bob displays the person's name in the interface.

However, displaying full name is tedious and takes much space. So he decided to display the shortest prefix which doesn't match with any prefix of any person who has joined earlier.

Let's suppose the first person to enter the conference is alvin

=> a

Now suppose next person to join is alice. The shortest prefix of alice that doesn't match with any prefix of alvin is ali.

=> a

=> ali

If the full name of a new person matches completely with the full name of any person who has joined earlier, he will display the full name and add a suffix which indicates how many times the same name has occurred in the list so far. For example, if another person name alvin joins, the list will look like this:

=> a

=> ali

=> alvin 2

You are given the list of the persons who have joined the call in the chronological order. Your task is to figure out how the final list looks like.

Input Format

The first line contains an integer .

The subsequent  line contains a string  denoting the name of the  person to join the call.

Constraints

•  will contain only lower-case english letters.

Subtask

•  for  of the maximum score

Output Format

Return a string array with  items, the  line should contain the prefix of name of the  person which doesn't match with any other person who has joined earlier.

Sample Input 0

3
alvin
alice
alvin

Sample Output 0

a
ali
alvin 2

Sample Input 1

6
mary
stacy
sam
samuel
sam
miguel

Sample Output 1

m
s
sa
samu
sam 2
mi

Solution :

public static List<String> solve(List<String> names) {
        Map<String,Integer> data = new HashMap<>();
        
        List<String> result = new ArrayList<>();
        for(String name : names){
            
            boolean isPrefixFound = false;
            for(int i = 1; i < name.length(); i++){
                String prefix = name.substring(0,i);
                
                if(!isPrefixFound && !data.containsKey(prefix)){
                    isPrefixFound = true;
                    result.add(prefix);
                }
                
                if(!data.containsKey(prefix)){
                    data.put(prefix,0);
                }
            }
            
            int occures = data.getOrDefault(name,0);
            if(!isPrefixFound){
              String proccessed = (occures==0)?name:name + " " + (occures+1);
              result.add(proccessed);
            }
            data.put(name,occures+1);
        }
        return result;
    }