CODEWITHSUNDEEP
javaarraysperformancejava-8
zepol
zepol

Reputation: 309

Get the element with the highest priority based on another list in Java

I have a list of priorities

List<String> priorities = Arrays.asList("NV","PH","OO","DR");

And I have a list of Items List<Item> availableItems where the Item has an attribute called type that can be one of the values from the first list.

So I am working on a method, that returns one element (the highest priority), I was doing something with a comparator and sorting, but I was reading about it and it's said it takes a lot of time since availableItems can have a lot of elements, so it might be O(NLogN).

What I am trying to achieve is if there is an element with type NV return is since is the highest priority, if not the one with PH and so on, and if there is no a single one, then return any element of the list, since they don't have, but without using comparators or the cost is not high

Upvotes: 0

Views: 1557

Answers (2)

Flown
Flown

Reputation: 11740

I would suggest using a PriorityQueue with a index-based Comparator:

public static <T> Comparator<T> comparatorByIndex(List<T> list) {
  Map<T, Integer> priority = IntStream.range(0, list.size()).boxed().collect(Collectors.toMap(list::get, Function.identity()));
  return Comparator.comparing(k -> priority.getOrDefault(k, Integer.MAX_VALUE));
}

And getting the String with highest priority like:

List<String> priorities = Arrays.asList("NV","PH","OO","DR");
Queue<String> priorityOrder = new PriorityQueue<>(comparatorByIndex(priorities));
priorityOrder.addAll(availableItems);

String highestPriority = priorityOrder.poll();

Upvotes: 1

Hadi
Hadi

Reputation: 17299

Based on @Gilbert Le Blanc comment.

TreeMap<Integer, List<Item0>> groupByType = availableItems.stream()
          .collect(Collectors.groupingBy(item -> priorities.indexOf(item.getType()),
                                                TreeMap::new, Collectors.toList()));

System.out.println(groupByType.firstEntry().getValue().get(0));

Upvotes: 1

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