CODEWITHSUNDEEP
javacollectionsjava-8java-streamcontains
Nolequen
Nolequen

Reputation: 4337

Is there a way to check if a Stream contains all collection elements?

For example, I need something like:

Collection<String> collection = /* ... */;
Stream<Object> stream = /* ... */;
boolean containsAll = stream.map(Object::toString).containsAll(collection);

Of course, I could accumulate all elements of the stream into another Collection using collect() method and the call Collection.containsAll(), but what if the stream is too big and it is inefficient to process all its elements?

Upvotes: 14

Views: 22103

Answers (3)

Eklavya
Eklavya

Reputation: 18480

Create a Set from Collection<String> to make search operation faster O(1)

Set<String> set = new HashSet<>(collection);

Then use allMatch to check every item in stream contains in set or not

boolean containsAll = stream.map(Object::toString)
                            .allMatch(s -> set.contains(s));

Another way :

Filter by not contained in set and use limit(1) for optimization

boolean isContains = stream.map(Object::toString)
                           .filter(s -> !set.contains(s))
                           .limit(1)
                           .count() > 0;

Upvotes: 3

ETO
ETO

Reputation: 7299

This should do the trick:

Set<String> set = new HashSet<>(collection);
boolean containsAll = set.isEmpty() || stream.map(Object::toString)
                                             .anyMatch(s -> set.remove(s) && set.isEmpty());

The solution might look confusing, but the idea is straightforward:

  1. In order to prevent multiple iterations over collection we wrap it into a HashSet. (In case your stream is a parallel one, then you will have to use a concurrent hash set. See this post for more details)
  2. If the collection (or set) is empty then we return true without processing the stream
  3. For each entry of stream we try to remove it from set. In case the result of Set::remove is true (hence it was contained by set) and the set is empty after removal, we can conclude that stream contained all the elements of initial collection.
  4. The terminal operation Stream::anyMatch is a short-circuiting one. So it will stop iterating over stream once the set is empty. In worst case we will process the entire stream.

Perhaps this is a bit more readable form:

Set<String> set = new HashSet<>(collection);
boolean containsAll = set.isEmpty() || stream.map(Object::toString)
                                             .filter(set::remove)
                                             .anyMatch(__ -> set.isEmpty());

If the collection can contain duplicates and there is a requirement to check if stream contains all of them, then we will need to maintain a concurrent map of counters.

Map<String, AtomicLong> map = new ConcurrentHashMap<>();
collection.forEach(s -> map.computeIfAbsent(s, __ -> new AtomicLong()).incrementAndGet());
boolean containsAll = map.isEmpty() || stream.map(Object::toString)
                                             .filter(map::containsKey)
                                             .filter(s -> map.get(s).decrementAndGet() == 0)
                                             .filter(s -> map.remove(s) != null)
                                             .anyMatch(__ -> map.isEmpty());

The code slightly changed but the idea is the same.

Upvotes: 11

Eran
Eran

Reputation: 394156

Regardless of how big the Stream is, you will have to process all of its elements if it does not contain all the elements of the Collection.

You could save processing time if a small prefix of the Stream contains all the elements of Collection, and the Collection is much smaller than the Stream.

boolean containsAll = 
    stream.map(Object::toString)
          .filter(s -> collection.contains(s)) // it would be wise to convert collection to a Set
          .limit(collection.size())
          .count() == collection.size();

Note that if the Stream may contain multiple copies of the same element of the Collection, you may have to add a .distinct() operation after the filter().

Upvotes: 4

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