Java 8 Stream mixing two elements

Question

I have a many objects of type Slot in an array list.

Slot class is as shown below-

Slot{
   int start;
   int end;
}

let the list of type List<Slot> be called slots. The slots are sorted based on start time. End time of one slot may be equal to start time of next slot, but they would never overlap.

Is there any possible way in which I can iterate over this list using Java 8 streams, and combine two slots if end time of one matches start time of next and output them into an ArrayList?

Answer 1

Such scenario is perfectly supported by my free StreamEx library which enhances standard Stream API. There's an intervalMap intermediate operation which is capable to collapse several adjacent stream elements to the single element. Here's complete example:

// Slot class and sample data are taken from @Andreas answer
List<Slot> slots = Arrays.asList(new Slot(3, 5), new Slot(5, 7), 
                new Slot(8, 10), new Slot(10, 11), new Slot(11, 13));

List<Slot> result = StreamEx.of(slots)
        .intervalMap((s1, s2) -> s1.end == s2.start,
                     (s1, s2) -> new Slot(s1.start, s2.end))
        .toList();
System.out.println(result);
// Output: [3-7, 8-13]

The intervalMap method takes two parameters. The first is a BiPredicate accepting two adjacent elements from the input stream and returns true if they must be merged (here the condition is s1.end == s2.start). The second parameter is a BiFunction which takes the first and the last elements from the merged series and produces the resulting element.

Note that if you have, for example 100 adjacent slots which should be combined into one, this solution does not create 100 intermediate objects (like in @Misha's answer, which is nevertheless very interesting), it tracks first and last slot in the series immediately forgetting about intermediate onces. Of course this solution is parallel friendly. If you have many thousands of input slots, using .parallel() may improve the performance.

Note that current implementation will recreate the Slot even if it's not merged with anything. In this case the BinaryOperator receives the same Slot parameter twice. If you want to optimize this case, you can make additional check like s1 == s2 ? s1 : ...:

List<Slot> result = StreamEx.of(slots)
        .intervalMap((s1, s2) -> s1.end == s2.start,
                     (s1, s2) -> s1 == s2 ? s1 : new Slot(s1.start, s2.end))
        .toList();