I am working with java 8 stream and trying to modify the object content in the forEach terminal operation.
The issues which i am facing here is that i am able to modify the List<Employee> object contents but not able to modify the contents of List<Integer>
The code snippet is as follows:
public static void streamExample() {
List<Employee> listEmp = Arrays.asList(new Employee());
listEmp.stream().forEach(a -> a.setEmptName("John Doe"));
listEmp.stream().forEach(System.out::println);
List<Integer> listInteger = Arrays.asList(2, 4, 6, 8, 12,17, 1234);
listInteger.stream().filter(v -> v % 2 == 0).forEach(a -> a=a+1);
listInteger.stream().forEach(System.out::println);
}
I am wondering the change is not reflecting back in the list because of unboxing the Integer object while performing the a=a+1 operation but not sure.
You use not optimal approach of Stream. Do think of each step in Stream as modify existed (or create new) element and return it back to the Stream. Finally you receive final result and you can use one of final method to finalize (and actually run the whole stream working) the Stream:
List<Integer> listInteger = Arrays.asList(2, 4, 6, 8, 12, 17, 1234);
listInteger.stream().filter(v -> v % 2 == 0).forEach(a -> a = a + 1);
listInteger.stream().forEach(System.out::println);
Here you have initial array. You want to do following:
Filter out some elements (this is not final step);
Print filtered elements out (this is final step).
To do so, you do not have to create Streams two times. Do use one:
Stream.of(2, 4, 6, 8, 12, 17, 1234) // create stream (there're many way to do it)
.filter(v -> v % 2 == 0) // filter out required elements
.map(v -> v + 1) // transform elements using given rule
.forEach(System.out::println); // finalize stream with printing out each element separately
Note: Stream.of(...) creates a Stream, then we add two steps to the stream filter and map and then finalize or START created stream with forEach.
You are assigning a new value to a local variable (a), so that has no affect on the source of the second Stream (your List<Integer>). Note that this is not what you are doing with your List<Employee>, where you are calling a setter method to mutate the elements of the List.
Since Integers are immutable, you can't mutate the elements of your input List<Integer>.
Instead, you can create a new List:
List<Integer> newList =
listInteger.stream()
.map(v -> v % 2 == 0 ? v + 1 : v)
.collect(Collectors.toList());
Or you can stream over the indices of your List, and replace some of the elements of that List:
IntStream.range(0,listInteger.size())
.filter(i -> listInteger.get(i) % 2 == 0)
.forEach(i -> listInteger.set(i, listInteger.get(i + 1));
Related
As I couldn't find anything related to this, I am wondering if streams even allow this.
In my answer to another question, I have following code to add elements to a result list, only if the result list doesn't already contain it:
List<Entry<List<Integer>, Integer>> list = new ArrayList<>(diffMap.entrySet());
list.sort(Entry.comparingByValue());
List<List<Integer>> resultList = new ArrayList<>();
for (Entry<List<Integer>, Integer> entry2 : list) {
if (!checkResultContainsElement(resultList, entry2.getKey()))
resultList.add(entry2.getKey());
}
checkResultContainsElement method:
private static boolean checkResultContainsElement(List<List<Integer>> resultList, List<Integer> key) {
List<Integer> vals = resultList.stream().flatMap(e -> e.stream().map(e2 -> e2))
.collect(Collectors.toList());
return key.stream().map(e -> e).anyMatch(e -> vals.contains(e));
}
Now I am wondering, if this for-loop:
for (Entry<List<Integer>, Integer> entry2 : list) {
if (!checkResultContainsElement(resultList, entry2.getKey()))
resultList.add(entry2.getKey());
}
can be realized using streams. I don't think that .filter() method would work, as it would remove data from List<Entry<List<Integer>, Integer>> list while I don't even know if an element should be considered. I guess that a custom collector could work, but I also wouldn't know how to implement one, as the result is constantly changing with each newly added element.
I am looking for something like this (can be different if something else is better):
list.stream().sorted(Entry.comparingByValue()).collect(???);
where ??? would filter the data and return it as a list.
The values of one result list may not be contained in another one. So these lists are valid:
[1, 2, 3, 4]
[5, 6, 7, 8]
[12, 12, 12, 12]
but of these, only the first is valid:
[1, 2, 3, 4] <-- valid
[5, 3, 7, 8] <-- invalid: 3 already exists
[12, 12, 2, 12] <-- invalid: 2 already exists
If we put aside for a moment the details on whether implementation will be stream-based or not, the existing implementation of how uniqueness of the values of incoming lists is being checked can be improved.
We can gain a significant performance improvement by maintaining a Set of previously encountered values.
I.e. values from each list that was added to the resulting list would be stored in a set. And in order to ensure uniqueness of every incoming list, its values would be checked against the set.
Since operations of a stream pipeline should be stateless, as well as collector shouldn't hold a state (i.e. changes should happen only inside its mutable container). We can approach this problem by defining a container that will encompass a resulting list of lists of Foo and a set of foo-values.
I've implemented this container as a Java 16 record:
public record FooContainer(Set<Integer> fooValues, List<List<Foo>> foosList) {
public void tryAdd(List<Foo> foos) {
if (!hasValue(foos)) {
foos.forEach(foo -> fooValues.add(foo.getValue()));
foosList.add(foos);
}
}
public boolean hasValue(List<Foo> foos) {
return foos.stream().map(Foo::getValue).anyMatch(fooValues::contains);
}
}
The record shown above would is used as a mutable container of a custom collector created with Colloctors.of(). Collector's accumulator make's use of tryAdd() method defined by the container. And the finisher extracts the resulting list from the container.
Note that this operation is not parallelizable, hence combiner of the collector throws an AssertionError.
public static void main(String[] args) {
Map<List<Foo>, Integer> diffMap =
Map.of(List.of(new Foo(1), new Foo(2), new Foo(3)), 1,
List.of(new Foo(1), new Foo(4), new Foo(5)), 2,
List.of(new Foo(7), new Foo(8), new Foo(9)), 3);
List<List<Foo>> result = diffMap.entrySet().stream()
.sorted(Map.Entry.comparingByValue())
.map(Map.Entry::getKey)
.collect(Collector.of(
() -> new FooContainer(new HashSet<>(), new ArrayList<>()),
FooContainer::tryAdd,
(left, right) -> {throw new AssertionError("The operation isn't parallelizable");},
FooContainer::foosList
));
System.out.println(result);
}
Output:
[[Foo{1}, Foo{2}, Foo{3}], [Foo{7}, Foo{8}, Foo{9}]]
May be something like this:-
list.stream().
sorted(Entry.comparingByValue()).
collect(ArrayList<List<Foo>>::new,(x,y)->!checkResultContainsElement(x, y.getKey()),(x,y)->x.add(y.getKey()));
Is there an elegant way to interoperate two streams?
/**
* e.g. add([1, -1, 5], [2, 3, 4]) -> [3, 2, 9]
*/
Stream<Integer> add(Stream<Integer> a, Stream<Integer> b) {
//?
}
I would hate to collect both streams to do the thing in a for-loop, just to stream the result again
Stream<Integer> add(Stream<Integer> a, Stream<Integer> b) {
Iterator<Integer> i=a.iterator();
return b.map(bz-> i.hasNext() ? bz + i.next() : bz);
}
I think this question really highlights why you probably shouldn't be doing what you're doing.
1st. You're adding three pairs of int's, and your using a Stream so you have to use Integer. That's huge overhead relative to the operation.
2nd. The concept of Stream is not limited to ordered sets of data.
For that reason collecting to a list makes sense because you're explicitly operating on a finite ordered dataset.
List<Integer> la = a.collect( Collectors.toList());
List<Integer> lb = b.collect( Collectors.toList());
Stream<Integer> result = IntStream.range( 0, la.size() ).mapToObj(
i -> la.get(i) && lb.get(j)
);
It might be more stream-like to use a.iterator() instead of lists because you wouldn't be limited to finite datasets.
Iterator<Integer> ia = a.iterator();
Iterator<Integer> ib = b.iterator();
if( ! ia.hasNext() || ! ib.hasNext() ) return Stream.empty();
return Stream.iterate(
ia.next() + ib.next(),
last -> ia.hasNext() && ib.hasNext(),
last -> ia.next() + ib.next() );
I’m trying to obtain a only duplicated numbers list from a list of integers:
final Set<Integer> setOfNmums = new HashSet<>();
Arrays.asList(5,6,7,7,7,6,2,4,2,4).stream()
.peek(integer -> System.out.println("XX -> " + integer))
.filter(n -> !setOfNmums.add(n))
.peek(System.out::println)
.map(String::valueOf)
.sorted()
.collect(Collectors.toList());
The output is 2,4,6,7,7
Expected : 2,4,6,7
I don’t understand how that’s happening.. is this running in parallel? how am I getting two '7'?
The hashset should return false if it exists and that used by the filter?
Yes I can use distinct, but I’m curious to know why would the filter fail.. is it being done in parallel?
Your filter rejects the first occurrence of each element and accepts all subsequent occurrences. Therefore, when an element occurs n times, you’ll add it n-1 times.
Since you want to accept all elements which occur more than once, but only accept them a single time, you could use .filter(n -> !setOfNmums.add(n)) .distinct() or you enhance the set to a map, to be able to accept an element only on its second occurrence.
Map<Integer, Integer> occurrences = new HashMap<>();
List<String> result = Stream.of(5,6,7,7,7,6,2,4,2,4)
.filter(n -> occurrences.merge(n, 1, Integer::sum) == 2)
.map(String::valueOf)
.sorted()
.collect(Collectors.toList());
But generally, using stateful filters with streams is discouraged.
A cleaner solution would be
List<String> result = Stream.of(5,6,7,7,7,6,2,4,2,4)
.collect(Collectors.collectingAndThen(
Collectors.toMap(String::valueOf, x -> true, (a,b) -> false, TreeMap::new),
map -> { map.values().removeIf(b -> b); return new ArrayList<>(map.keySet()); }));
Note that this approach doesn’t count the occurrences but only remembers whether an element is unique or has seen at least a second time. This works by mapping each element to true with the second argument to the toMap collector, x -> true, and resolving multiple occurrences with a merge function of (a,b) -> false. The subsequent map.values().removeIf(b -> b) will remove all unique elements, i.e. those mapped to true.
You can use .distinct() function in your stream check this out.
Since Holger already explained why your solution didn't work, I'll just provide an alternative.
Why not use Collections.frequency(collection, element) together with distinct()?
The solution would be quite simple(i apologize for the formatting, i just copied it from my ide and there doesn't seem to be an autoformat feature in SOF):
List<Integer> numbers = List.of(5, 6, 7, 7, 7, 6, 2, 4, 2, 4);
List<String> onlyDuplicates = numbers.stream()
.filter(n -> Collections.frequency(numbers, n) > 1)
.distinct()
.sorted()
.map(String::valueOf)
.toList();
This simply keeps all elements that occur more than once and then filters out the duplicates before sorting, converting each element to a string and collecting to a list since that seems to be what you want.
if you need a mutable list you can use collect(toCollection(ArrayList::new)) instead of toList()
Given a list of elements, I want to get the element with a given property and remove it from the list. The best solution I found is:
ProducerDTO p = producersProcedureActive
.stream()
.filter(producer -> producer.getPod().equals(pod))
.findFirst()
.get();
producersProcedureActive.remove(p);
Is it possible to combine get and remove in a lambda expression?
To Remove element from the list
objectA.removeIf(x -> conditions);
eg:
objectA.removeIf(x -> blockedWorkerIds.contains(x));
List<String> str1 = new ArrayList<String>();
str1.add("A");
str1.add("B");
str1.add("C");
str1.add("D");
List<String> str2 = new ArrayList<String>();
str2.add("D");
str2.add("E");
str1.removeIf(x -> str2.contains(x));
str1.forEach(System.out::println);
OUTPUT:
A
B
C
Although the thread is quite old, still thought to provide solution - using Java8.
Make the use of removeIf function. Time complexity is O(n)
producersProcedureActive.removeIf(producer -> producer.getPod().equals(pod));
API reference: removeIf docs
Assumption: producersProcedureActive is a List
NOTE: With this approach you won't be able to get the hold of the deleted item.
Consider using vanilla java iterators to perform the task:
public static <T> T findAndRemoveFirst(Iterable<? extends T> collection, Predicate<? super T> test) {
T value = null;
for (Iterator<? extends T> it = collection.iterator(); it.hasNext();)
if (test.test(value = it.next())) {
it.remove();
return value;
}
return null;
}
Advantages:
It is plain and obvious.
It traverses only once and only up to the matching element.
You can do it on any Iterable even without stream() support (at least those implementing remove() on their iterator).
Disadvantages:
You cannot do it in place as a single expression (auxiliary method or variable required)
As for the
Is it possible to combine get and remove in a lambda expression?
other answers clearly show that it is possible, but you should be aware of
Search and removal may traverse the list twice
ConcurrentModificationException may be thrown when removing element from the list being iterated
The direct solution would be to invoke ifPresent(consumer) on the Optional returned by findFirst(). This consumer will be invoked when the optional is not empty. The benefit also is that it won't throw an exception if the find operation returned an empty optional, like your current code would do; instead, nothing will happen.
If you want to return the removed value, you can map the Optional to the result of calling remove:
producersProcedureActive.stream()
.filter(producer -> producer.getPod().equals(pod))
.findFirst()
.map(p -> {
producersProcedureActive.remove(p);
return p;
});
But note that the remove(Object) operation will again traverse the list to find the element to remove. If you have a list with random access, like an ArrayList, it would be better to make a Stream over the indexes of the list and find the first index matching the predicate:
IntStream.range(0, producersProcedureActive.size())
.filter(i -> producersProcedureActive.get(i).getPod().equals(pod))
.boxed()
.findFirst()
.map(i -> producersProcedureActive.remove((int) i));
With this solution, the remove(int) operation operates directly on the index.
Use can use filter of Java 8, and create another list if you don't want to change the old list:
List<ProducerDTO> result = producersProcedureActive
.stream()
.filter(producer -> producer.getPod().equals(pod))
.collect(Collectors.toList());
I'm sure this will be an unpopular answer, but it works...
ProducerDTO[] p = new ProducerDTO[1];
producersProcedureActive
.stream()
.filter(producer -> producer.getPod().equals(pod))
.findFirst()
.ifPresent(producer -> {producersProcedureActive.remove(producer); p[0] = producer;}
p[0] will either hold the found element or be null.
The "trick" here is circumventing the "effectively final" problem by using an array reference that is effectively final, but setting its first element.
With Eclipse Collections you can use detectIndex along with remove(int) on any java.util.List.
List<Integer> integers = Lists.mutable.with(1, 2, 3, 4, 5);
int index = Iterate.detectIndex(integers, i -> i > 2);
if (index > -1) {
integers.remove(index);
}
Assert.assertEquals(Lists.mutable.with(1, 2, 4, 5), integers);
If you use the MutableList type from Eclipse Collections, you can call the detectIndex method directly on the list.
MutableList<Integer> integers = Lists.mutable.with(1, 2, 3, 4, 5);
int index = integers.detectIndex(i -> i > 2);
if (index > -1) {
integers.remove(index);
}
Assert.assertEquals(Lists.mutable.with(1, 2, 4, 5), integers);
Note: I am a committer for Eclipse Collections
The below logic is the solution without modifying the original list
List<String> str1 = new ArrayList<String>();
str1.add("A");
str1.add("B");
str1.add("C");
str1.add("D");
List<String> str2 = new ArrayList<String>();
str2.add("D");
str2.add("E");
List<String> str3 = str1.stream()
.filter(item -> !str2.contains(item))
.collect(Collectors.toList());
str1 // ["A", "B", "C", "D"]
str2 // ["D", "E"]
str3 // ["A", "B", "C"]
When we want to get multiple elements from a List into a new list (filter using a predicate) and remove them from the existing list, I could not find a proper answer anywhere.
Here is how we can do it using Java Streaming API partitioning.
Map<Boolean, List<ProducerDTO>> classifiedElements = producersProcedureActive
.stream()
.collect(Collectors.partitioningBy(producer -> producer.getPod().equals(pod)));
// get two new lists
List<ProducerDTO> matching = classifiedElements.get(true);
List<ProducerDTO> nonMatching = classifiedElements.get(false);
// OR get non-matching elements to the existing list
producersProcedureActive = classifiedElements.get(false);
This way you effectively remove the filtered elements from the original list and add them to a new list.
Refer the 5.2. Collectors.partitioningBy section of this article.
As others have suggested, this might be a use case for loops and iterables. In my opinion, this is the simplest approach. If you want to modify the list in-place, it cannot be considered "real" functional programming anyway. But you could use Collectors.partitioningBy() in order to get a new list with elements which satisfy your condition, and a new list of those which don't. Of course with this approach, if you have multiple elements satisfying the condition, all of those will be in that list and not only the first.
the task is: get ✶and✶ remove element from list
p.stream().collect( Collectors.collectingAndThen( Collector.of(
ArrayDeque::new,
(a, producer) -> {
if( producer.getPod().equals( pod ) )
a.addLast( producer );
},
(a1, a2) -> {
return( a1 );
},
rslt -> rslt.pollFirst()
),
(e) -> {
if( e != null )
p.remove( e ); // remove
return( e ); // get
} ) );
resumoRemessaPorInstrucoes.removeIf(item ->
item.getTipoOcorrenciaRegistro() == TipoOcorrenciaRegistroRemessa.PEDIDO_PROTESTO.getNome() ||
item.getTipoOcorrenciaRegistro() == TipoOcorrenciaRegistroRemessa.SUSTAR_PROTESTO_BAIXAR_TITULO.getNome());
Combining my initial idea and your answers I reached what seems to be the solution
to my own question:
public ProducerDTO findAndRemove(String pod) {
ProducerDTO p = null;
try {
p = IntStream.range(0, producersProcedureActive.size())
.filter(i -> producersProcedureActive.get(i).getPod().equals(pod))
.boxed()
.findFirst()
.map(i -> producersProcedureActive.remove((int)i))
.get();
logger.debug(p);
} catch (NoSuchElementException e) {
logger.error("No producer found with POD [" + pod + "]");
}
return p;
}
It lets remove the object using remove(int) that do not traverse again the
list (as suggested by #Tunaki) and it lets return the removed object to
the function caller.
I read your answers that suggest me to choose safe methods like ifPresent instead of get but I do not find a way to use them in this scenario.
Are there any important drawback in this kind of solution?
Edit following #Holger advice
This should be the function I needed
public ProducerDTO findAndRemove(String pod) {
return IntStream.range(0, producersProcedureActive.size())
.filter(i -> producersProcedureActive.get(i).getPod().equals(pod))
.boxed()
.findFirst()
.map(i -> producersProcedureActive.remove((int)i))
.orElseGet(() -> {
logger.error("No producer found with POD [" + pod + "]");
return null;
});
}
A variation of the above:
import static java.util.function.Predicate.not;
final Optional<MyItem> myItem = originalCollection.stream().filter(myPredicate(someInfo)).findFirst();
final List<MyItem> myOtherItems = originalCollection.stream().filter(not(myPredicate(someInfo))).toList();
private Predicate<MyItem> myPredicate(Object someInfo) {
return myItem -> myItem.someField() == someInfo;
}
Use Case
Through some coding Katas posted at work, I stumbled on this problem that I'm not sure how to solve.
Using Java 8 Streams, given a list of positive integers, produce a
list of integers where the integer preceded a larger value.
[10, 1, 15, 30, 2, 6]
The above input would yield:
[1, 15, 2]
since 1 precedes 15, 15 precedes 30, and 2 precedes 6.
Non-Stream Solution
public List<Integer> findSmallPrecedingValues(final List<Integer> values) {
List<Integer> result = new ArrayList<Integer>();
for (int i = 0; i < values.size(); i++) {
Integer next = (i + 1 < values.size() ? values.get(i + 1) : -1);
Integer current = values.get(i);
if (current < next) {
result.push(current);
}
}
return result;
}
What I've Tried
The problem I have is I can't figure out how to access next in the lambda.
return values.stream().filter(v -> v < next).collect(Collectors.toList());
Question
Is it possible to retrieve the next value in a stream?
Should I be using map and mapping to a Pair in order to access next?
Using IntStream.range:
static List<Integer> findSmallPrecedingValues(List<Integer> values) {
return IntStream.range(0, values.size() - 1)
.filter(i -> values.get(i) < values.get(i + 1))
.mapToObj(values::get)
.collect(Collectors.toList());
}
It's certainly nicer than an imperative solution with a large loop, but still a bit meh as far as the goal of "using a stream" in an idiomatic way.
Is it possible to retrieve the next value in a stream?
Nope, not really. The best cite I know of for that is in the java.util.stream package description:
The elements of a stream are only visited once during the life of a stream. Like an Iterator, a new stream must be generated to revisit the same elements of the source.
(Retrieving elements besides the current element being operated on would imply they could be visited more than once.)
We could also technically do it in a couple other ways:
Statefully (very meh).
Using a stream's iterator is technically still using the stream.
That's not a pure Java8, but recently I've published a small library called StreamEx which has a method exactly for this task:
// Find all numbers where the integer preceded a larger value.
Collection<Integer> numbers = Arrays.asList(10, 1, 15, 30, 2, 6);
List<Integer> res = StreamEx.of(numbers).pairMap((a, b) -> a < b ? a : null)
.nonNull().toList();
assertEquals(Arrays.asList(1, 15, 2), res);
The pairMap operation internally implemented using custom spliterator. As a result you have quite clean code which does not depend on whether the source is List or anything else. Of course it works fine with parallel stream as well.
Committed a testcase for this task.
It's not a one-liner (it's a two-liner), but this works:
List<Integer> result = new ArrayList<>();
values.stream().reduce((a,b) -> {if (a < b) result.add(a); return b;});
Rather than solving it by "looking at the next element", this solves it by "looking at the previous element, which reduce() give you for free. I have bent its intended usage by injecting a code fragment that populates the list based on the comparison of previous and current elements, then returns the current so the next iteration will see it as its previous element.
Some test code:
List<Integer> result = new ArrayList<>();
IntStream.of(10, 1, 15, 30, 2, 6).reduce((a,b) -> {if (a < b) result.add(a); return b;});
System.out.println(result);
Output:
[1, 15, 2]
The accepted answer works fine if either the stream is sequential or parallel but can suffer if the underlying List is not random access, due to multiple calls to get.
If your stream is sequential, you might roll this collector:
public static Collector<Integer, ?, List<Integer>> collectPrecedingValues() {
int[] holder = {Integer.MAX_VALUE};
return Collector.of(ArrayList::new,
(l, elem) -> {
if (holder[0] < elem) l.add(holder[0]);
holder[0] = elem;
},
(l1, l2) -> {
throw new UnsupportedOperationException("Don't run in parallel");
});
}
and a usage:
List<Integer> precedingValues = list.stream().collect(collectPrecedingValues());
Nevertheless you could also implement a collector so thats works for sequential and parallel streams. The only thing is that you need to apply a final transformation, but here you have control over the List implementation so you won't suffer from the get performance.
The idea is to generate first a list of pairs (represented by a int[] array of size 2) which contains the values in the stream sliced by a window of size two with a gap of one. When we need to merge two lists, we check the emptiness and merge the gap of the last element of the first list with the first element of the second list. Then we apply a final transformation to filter only desired values and map them to have the desired output.
It might not be as simple as the accepted answer, but well it can be an alternative solution.
public static Collector<Integer, ?, List<Integer>> collectPrecedingValues() {
return Collectors.collectingAndThen(
Collector.of(() -> new ArrayList<int[]>(),
(l, elem) -> {
if (l.isEmpty()) l.add(new int[]{Integer.MAX_VALUE, elem});
else l.add(new int[]{l.get(l.size() - 1)[1], elem});
},
(l1, l2) -> {
if (l1.isEmpty()) return l2;
if (l2.isEmpty()) return l1;
l2.get(0)[0] = l1.get(l1.size() - 1)[1];
l1.addAll(l2);
return l1;
}), l -> l.stream().filter(arr -> arr[0] < arr[1]).map(arr -> arr[0]).collect(Collectors.toList()));
}
You can then wrap these two collectors in a utility collector method, check if the stream is parallel with isParallel an then decide which collector to return.
If you're willing to use a third party library and don't need parallelism, then jOOλ offers SQL-style window functions as follows
System.out.println(
Seq.of(10, 1, 15, 30, 2, 6)
.window()
.filter(w -> w.lead().isPresent() && w.value() < w.lead().get())
.map(w -> w.value())
.toList()
);
Yielding
[1, 15, 2]
The lead() function accesses the next value in traversal order from the window.
Disclaimer: I work for the company behind jOOλ
You can achieve that by using a bounded queue to store elements which flows through the stream (which is basing on the idea which I described in detail here: Is it possible to get next element in the Stream?
Belows example first defines instance of BoundedQueue class which will store elements going through the stream (if you don't like idea of extending the LinkedList, refer to link mentioned above for alternative and more generic approach). Later you just examine the two subsequent elements - thanks to the helper class:
public class Kata {
public static void main(String[] args) {
List<Integer> input = new ArrayList<Integer>(asList(10, 1, 15, 30, 2, 6));
class BoundedQueue<T> extends LinkedList<T> {
public BoundedQueue<T> save(T curElem) {
if (size() == 2) { // we need to know only two subsequent elements
pollLast(); // remove last to keep only requested number of elements
}
offerFirst(curElem);
return this;
}
public T getPrevious() {
return (size() < 2) ? null : getLast();
}
public T getCurrent() {
return (size() == 0) ? null : getFirst();
}
}
BoundedQueue<Integer> streamHistory = new BoundedQueue<Integer>();
final List<Integer> answer = input.stream()
.map(i -> streamHistory.save(i))
.filter(e -> e.getPrevious() != null)
.filter(e -> e.getCurrent() > e.getPrevious())
.map(e -> e.getPrevious())
.collect(Collectors.toList());
answer.forEach(System.out::println);
}
}