Question: Question: Candidates can attempt an entrance exam as many times as they wish. The average score of all attempts is considered in the end. Given a list of scores. find the candidate with the highest score [ average score].

```
String scores[][] = {{"Ram","155"}, 
                    {"Shyam","145"},
                    {"Ram","156"},
                    {"Balram","159"},
                    {"Balram","150"},
                    {"Ram","135"},
                    {"Mira","156"},
                    {"Mira","152"},
                    {"Shyam","155"}};

```

Scores are given in two-dimensional array as above. Need suggestions to solve this in efficient way.

Answer: 
What is the reason for reducing the time and space complexity? I strongly believe the brevity and maintainability are more important than the performance.

Since you tagged [java-stream](/questions/tagged/java-stream "show questions tagged 'java-stream'"), I can suggest you the following approach.

```
Map<Object, Double> scoreMap = Arrays.stream(scores)
        .collect(
            Collectors.groupingBy(i -> i[0], 
            Collectors.averagingInt(i -> Integer.parseInt(i[1])
        )));

String winner = scoreMap.entrySet().stream()
    .max(Comparator.comparingDouble(e -> e.getValue()))
    .get().getKey().toString();

```

Thanks to @AndyTurner for the `.max(..)` parameter suggestion.
To reduce time complexity, you can use specific collections:

```
public static String findMaxScore(String[][] scores) {
    final class Candidate implements Comparable<Candidate> {

        private final String name;
        private int scoreSum;
        private int attempts;

        public Candidate(String name) {
            this.name = name;
        }

        public void score(int score) {
            scoreSum += score;
            attempts++;
        }

        public double avg() {
            return (double)scoreSum / attempts;
        }

        @Override
        public int compareTo(Candidate candidate) {
            return Double.compare(candidate.avg(), avg());
        }
    }

    Map<String, Candidate> map = new HashMap<>();

    for (String[] data : scores) {
        map.putIfAbsent(data[0], new Candidate(data[0]));
        map.get(data[0]).score(Integer.parseInt(data[1]));
    }

    return new TreeSet<>(map.values()).iterator().next().name;
}

```