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Fibonacci using Stream And BinaryOperator in Java

I'm student and I learning functional Java 8. I got project to do and I don't understand how this function interface work. My teacher told me "you should know that" and I'm looking for help to understand this problem. It should count Fibonacci series
I got this code
StreamUtils.generateRest(Stream.of(1, 1), (a, b) -> a + b)
.limit(7)
.forEach(System.out::println);
StreamUtils.generateRest(Stream.of("AAA", "BB", "KKKK"), (a, b) -> a + b)
.limit(7)
.forEach(System.out::println);
StreamUtils.generateRest(Stream.of(i -> 0),
(BinaryOperator<UnaryOperator<Integer>>) (f, g) -> (x -> x == 0 ? 1 : x * g.apply(x - 1)))
.limit(10)
.map(f -> f.apply(7))
.forEach(System.out::println);
I did something like this but it doesn't work
public class StreamUtils<T> {
public static <T> Stream generateRest(Stream<T> stream, BinaryOperator<T> binaryOperator) {
return Stream.of(stream.reduce((a, b) -> binaryOperator.apply(a, b)));
}
}
Someone can help me with that and explain how to solve this problem?

To make first example work you need to implement something like this:
private static class StreamUtils<T> {
public static <T> Stream generateRest(Stream<T> stream, BinaryOperator<T> binaryOperator) {
return Stream.iterate(stream.toArray(), p -> new Object[]{p[1], binaryOperator.apply((T) p[0], (T) p[1])})
.flatMap(p -> Stream.of(p[0]));
}
}
It creates array from your input stream, then applies passed function on two elements, shifts the result of previous iteration to position 0, as we need previous two values to calculate next.
And then it creates unlimited stream of calculated Fibonacci elements.
The output is:
1
1
2
3
5
8
13
And version with correct generics usage, because you initial structure produces raw types.
private static class StreamUtils {
public static <T> Stream<T> generateRest(Stream<T> stream, BinaryOperator<T> binaryOperator) {
return Stream.iterate(stream.toArray(), p -> new Object[]{p[1], binaryOperator.apply((T) p[0], (T) p[1])})
.flatMap(p -> Stream.of((T) p[0]));
}
}

I'm assuming that having more than 2 items means A, B, C, A+B, B+C, C+(A+B), (A+B)+(B+C), etc., and that having 1 item means A, A+A, A+(A+A), (A+A)+(A+(A+A)), etc., where + is the binary operator.
Basically you turn the stream into an array, then you use Stream.generate and at each step you generate the element after the ones you have, shift the array left to fit the new element, and return the old first element (which is no longer in the array). Note that since this has side effects (modifying an external array) it cannot be used with .parallel().
#SuppressWarnings("unchecked")
public static <T> Stream<T> generateRest(Stream<T> stream, BinaryOperator<T> binaryOperator) {
T[] t = (T[]) stream.toArray();
if (t.length == 1) {
t = (T[]) new Object[] { t[0], binaryOperator.apply(t[0], t[0]) };
}
final T[] items = t;
return Stream.generate(() -> {
T first = items[0];
T next = binaryOperator.apply(items[0], items[1]);
System.arraycopy(items, 1, items, 0, items.length - 1);
items[items.length - 1] = next;
return first;
});
}
Output:
1
1
2
3
5
8
13
AAA
BB
KKKK
AAABB
BBKKKK
KKKKAAABB
AAABBBBKKKK
0
0
0
0
0
0
0
0
5040
5040

Generate infinite parallel stream

Problem
Hi, I have a function where i going to return infinite stream of parallel (yes, it is much faster in that case) generated results. So obviously (or not) i used
Stream<Something> stream = Stream.generate(this::myGenerator).parallel()
It works, however ... it doesn't when i want to limit the result (everything is fine when the stream is sequential). I mean, it creates results when i make something like
stream.peek(System.out::println).limit(2).collect(Collectors.toList())
but even when peek output produces more than 10 elements, collect is still not finallized (generating is slow so those 10 can took even a minute)... and that is easy example. Actually, limiting those results is a future due the main expectation is to get only better than recent results until the user will kill the process (other case is to return first what i can make with throwing exception if nothing else will help [findFirst didn't, even when i had more elements on the console and no more results for about 30 sec]).
So, the question is...
how to copy with that? My idea was also to use RxJava, and there is another question - how to achieve similar result with that tool (or other).
Code sample
public Stream<Solution> generateSolutions() {
final Solution initialSolution = initialSolutionMaker.findSolution();
return Stream.concat(
Stream.of(initialSolution),
Stream.generate(continuousSolutionMaker::findSolution)
).parallel();
}
new Solver(instance).generateSolutions()
.map(Solution::getPurpose)
.peek(System.out::println)
.limit(5).collect(Collectors.toList());
Implementation of findSolution is not important.
It has some side effect like adding to solutions repo (singleton, sych etc..), but nothing more.

As explained in the already linked answer, the key point to an efficient parallel stream is to use a stream source already having an intrinsic size instead of using an unsized or even infinite stream and apply a limit on it. Injecting a size doesn’t work with the current implementation at all, while ensuring that a known size doesn’t get lost is much easier. Even if the exact size can’t be retained, like when applying a filter, the size still will be carried as an estimate size.
So instead of
Stream.generate(this::myGenerator).parallel()
.peek(System.out::println)
.limit(2)
.collect(Collectors.toList())
just use
IntStream.range(0, /* limit */ 2).unordered().parallel()
.mapToObj(unused -> this.myGenerator())
.peek(System.out::println)
.collect(Collectors.toList())
Or, closer to your sample code
public Stream<Solution> generateSolutions(int limit) {
final Solution initialSolution = initialSolutionMaker.findSolution();
return Stream.concat(
Stream.of(initialSolution),
IntStream.range(1, limit).unordered().parallel()
.mapToObj(unused -> continuousSolutionMaker.findSolution())
);
}
new Solver(instance).generateSolutions(5)
.map(Solution::getPurpose)
.peek(System.out::println)
.collect(Collectors.toList());

Unfortunately this is expected behavior. As I remember I've seen at least two topics on this matter, here is one of them.
The idea is that Stream.generate creates an unordered infinite stream and limit will not introduce the SIZED flag. Because of this when you spawn a parallel execution on that Stream, individual tasks have to sync their execution to see if they have reached that limit; by the time that sync happens there could be multiple elements already processed. For example this:
Stream.iterate(0, x -> x + 1)
.peek(System.out::println)
.parallel()
.limit(2)
.collect(Collectors.toList());
and this :
IntStream.of(1, 2, 3, 4)
.peek(System.out::println)
.parallel()
.limit(2)
.boxed()
.collect(Collectors.toList());
will always generate two elements in the List (Collectors.toList) and will always output two elements also (via peek).
On the other hand this:
Stream<Integer> stream = Stream.generate(new Random()::nextInt).parallel();
List<Integer> list = stream
.peek(x -> {
System.out.println("Before " + x);
})
.map(x -> {
System.out.println("Mapping x " + x);
return x;
})
.peek(x -> {
System.out.println("After " + x);
})
.limit(2)
.collect(Collectors.toList());
will generate two elements in the List, but it may process many more that later will be discarded by the limit. This is what you are actually seeing in your example.
The only sane way of going that (as far as I can tell) would be to create a custom Spliterator. I have not written many of them, but here is my attempt:
static class LimitingSpliterator<T> implements Spliterator<T> {
private int limit;
private final Supplier<T> generator;
private LimitingSpliterator(Supplier<T> generator, int limit) {
Preconditions.checkArgument(limit > 0);
this.limit = limit;
this.generator = Objects.requireNonNull(generator);
}
#Override
public boolean tryAdvance(Consumer<? super T> consumer) {
if (limit == 0) {
return false;
}
T nextElement = generator.get();
--limit;
consumer.accept(nextElement);
return true;
}
#Override
public LimitingSpliterator<T> trySplit() {
if (limit <= 1) {
return null;
}
int half = limit >> 1;
limit = limit - half;
return new LimitingSpliterator<>(generator, half);
}
#Override
public long estimateSize() {
return limit >> 1;
}
#Override
public int characteristics() {
return SIZED;
}
}
And the usage would be:
StreamSupport.stream(new LimitingSpliterator<>(new Random()::nextInt, 7), true)
.peek(System.out::println)
.collect(Collectors.toList());

Java predicate - match against first predicate [duplicate]

I've just started playing with Java 8 lambdas and I'm trying to implement some of the things that I'm used to in functional languages.
For example, most functional languages have some kind of find function that operates on sequences, or lists that returns the first element, for which the predicate is true. The only way I can see to achieve this in Java 8 is:
lst.stream()
.filter(x -> x > 5)
.findFirst()
However this seems inefficient to me, as the filter will scan the whole list, at least to my understanding (which could be wrong). Is there a better way?

No, filter does not scan the whole stream. It's an intermediate operation, which returns a lazy stream (actually all intermediate operations return a lazy stream). To convince you, you can simply do the following test:
List<Integer> list = Arrays.asList(1, 10, 3, 7, 5);
int a = list.stream()
.peek(num -> System.out.println("will filter " + num))
.filter(x -> x > 5)
.findFirst()
.get();
System.out.println(a);
Which outputs:
will filter 1
will filter 10
10
You see that only the two first elements of the stream are actually processed.
So you can go with your approach which is perfectly fine.

However this seems inefficient to me, as the filter will scan the whole list
No it won't - it will "break" as soon as the first element satisfying the predicate is found. You can read more about laziness in the stream package javadoc, in particular (emphasis mine):
Many stream operations, such as filtering, mapping, or duplicate removal, can be implemented lazily, exposing opportunities for optimization. For example, "find the first String with three consecutive vowels" need not examine all the input strings. Stream operations are divided into intermediate (Stream-producing) operations and terminal (value- or side-effect-producing) operations. Intermediate operations are always lazy.

return dataSource.getParkingLots()
.stream()
.filter(parkingLot -> Objects.equals(parkingLot.getId(), id))
.findFirst()
.orElse(null);
I had to filter out only one object from a list of objects. So i used this, hope it helps.

In addition to Alexis C's answer, If you are working with an array list, in which you are not sure whether the element you are searching for exists, use this.
Integer a = list.stream()
.peek(num -> System.out.println("will filter " + num))
.filter(x -> x > 5)
.findFirst()
.orElse(null);
Then you could simply check whether a is null.

Already answered by #AjaxLeung, but in comments and hard to find.
For check only
lst.stream()
.filter(x -> x > 5)
.findFirst()
.isPresent()
is simplified to
lst.stream()
.anyMatch(x -> x > 5)

import org.junit.Test;
import java.util.Arrays;
import java.util.List;
import java.util.Optional;
// Stream is ~30 times slower for same operation...
public class StreamPerfTest {
int iterations = 100;
List<Integer> list = Arrays.asList(1, 10, 3, 7, 5);
// 55 ms
#Test
public void stream() {
for (int i = 0; i < iterations; i++) {
Optional<Integer> result = list.stream()
.filter(x -> x > 5)
.findFirst();
System.out.println(result.orElse(null));
}
}
// 2 ms
#Test
public void loop() {
for (int i = 0; i < iterations; i++) {
Integer result = null;
for (Integer walk : list) {
if (walk > 5) {
result = walk;
break;
}
}
System.out.println(result);
}
}
}

A generic utility function with looping seems a lot cleaner to me:
static public <T> T find(List<T> elements, Predicate<T> p) {
for (T item : elements) if (p.test(item)) return item;
return null;
}
static public <T> T find(T[] elements, Predicate<T> p) {
for (T item : elements) if (p.test(item)) return item;
return null;
}
In use:
List<Integer> intList = Arrays.asList(1, 2, 3, 4, 5);
Integer[] intArr = new Integer[]{1, 2, 3, 4, 5};
System.out.println(find(intList, i -> i % 2 == 0)); // 2
System.out.println(find(intArr, i -> i % 2 != 0)); // 1
System.out.println(find(intList, i -> i > 5)); // null

Improved One-Liner answer: If you are looking for a boolean return value, we can do it better by adding isPresent:
return dataSource.getParkingLots().stream().filter(parkingLot -> Objects.equals(parkingLot.getId(), id)).findFirst().isPresent();

Find missing integer in a sequential sorted stream

Let's say I have a list
ArrayList<String> arr = new ArrayList(Arrays.asList("N1", "N2", "N3", "N5"));
How do I find "N4", I mean, how I find that the missing integer is 4?
What I've tried so far
Integer missingID = arr.stream().map(p -> Integer.parseInt(p.substring(1))).sorted()
.reduce((p1, p2) -> (p2 - p1) > 1 ? p1 + 1 : 0).get();
This doesn't work because reduce is not intended to work in the way I need in this situation, actually, I have no idea how do that.
If there's no missing number, than the next must be "N6" - or just 6 - (in this example)
It must be done with java standard stream's library, no use of third parties.

The algorithm to implement here is based from this one: to find the missing number in a sequence of integers, the trick is to:
calculate the sum of the elements in the sequence.
calculate the sum of the elements the sequence would have with the missing number: this is easy to do since we can determine the minimum, the maximum and we know that the sum from a sequence of integer going from min to max is max*(max+1)/2 - (min-1)*min/2.
find the difference between those two sums: that's our missing number
In this case, we can collect statistics on our Stream by first mapping to an IntStream formed by only the numbers themselves and then calling summaryStatistics(). This returns a IntSummaryStatistics that has all the values we want: min, max and sum:
public static void main(String[] args) {
List<String> arr = Arrays.asList("N3", "N7", "N4", "N5", "N2");
IntSummaryStatistics statistics =
arr.stream()
.mapToInt(s -> Integer.parseInt(s.substring(1)))
.summaryStatistics();
long max = statistics.getMax();
long min = statistics.getMin();
long missing = max*(max+1)/2 - (min-1)*min/2 - statistics.getSum();
System.out.println(missing); // prints "6" here
}
If there is no missing number, this will print 0.

Here's the solution involving the pairMap operation from my free StreamEx library. It prints all the missing elements of the sorted input:
ArrayList<String> arr = new ArrayList(Arrays.asList("N1", "N2", "N3", "N5"));
StreamEx.of(arr).map(n -> Integer.parseInt(n.substring(1)))
.pairMap((a, b) -> IntStream.range(a+1, b))
.flatMapToInt(Function.identity())
.forEach(System.out::println);
The pairMap operation allows you to map every adjacent pair of the stream to something else. Here we map them to the streams of the skipped numbers, then flatten these streams.
The same solution is possible without third-party library, but looks more verbose:
ArrayList<String> arr = new ArrayList(Arrays.asList("N1", "N2", "N3", "N5"));
IntStream.range(0, arr.size()-1)
.flatMap(idx -> IntStream.range(
Integer.parseInt(arr.get(idx).substring(1))+1,
Integer.parseInt(arr.get(idx+1).substring(1))))
.forEach(System.out::println);

If there's only ONE missing number in the array, and if all numbers are positive, you could use the XOR algorithm, as explained in this question and its answers:
List<String> list = Arrays.asList("N5", "N2", "N3", "N6");
int xorArray = list.stream()
.mapToInt(p -> Integer.parseInt(p.substring(1)))
.reduce(0, (p1, p2) -> p1 ^ p2);
int xorAll = IntStream.rangeClosed(2, 6)
.reduce(0, (p1, p2) -> p1 ^ p2);
System.out.println(xorArray ^ xorAll); // 4
The advantage of this approach is that you don't need to use extra data structures, all you need is a couple of ints.
EDIT as per #Holger's comments below:
This solution requires you to know the range of the numbers in advance. Although on the other hand, it doesn't require the list and stream to be sorted.
Even if the list wasn't sorted, you could still get min and max (hence, the range) with IntSummaryStatistics, but this would require an extra iteration.

You could create a state object which is used to transform a single input stream into multiple streams of missing entries. These missing entry streams can then be flat mapped to produce a single output:
public class GapCheck {
private String last;
public GapCheck(String first) {
last = first;
}
public Stream<String> streamMissing(String next) {
final int n = Integer.parseInt(next.replaceAll("N", ""));
final int l = Integer.parseInt(last.replaceAll("N", ""));
last = next;
return IntStream.range(l + 1, n).mapToObj(Integer::toString);
}
}
Usage:
final List<String> arr = new ArrayList(Arrays.asList("N1", "N3", "N5"));
arr.stream()
.flatMap(new GapCheck(arr.get(0))::streamMissing)
.forEach(System.out::println);
output:
2
4

This is more work than you might expect, but it can be done with a collect call.
public class Main {
public static void main(String[] args) {
ArrayList<String> arr = new ArrayList<String>(Arrays.asList("N1", "N2", "N3", "N5", "N7", "N14"));
Stream<Integer> st = arr.stream().map(p -> Integer.parseInt(p.substring(1))).sorted();
Holder<Integer> holder = st.collect(() -> new Holder<Integer>(),
(h, i) -> {
Integer last = h.getProcessed().isEmpty() ? null : h.getProcessed().get(h.getProcessed().size() - 1);
if (last != null) {
while (i - last > 1) {
h.getMissing().add(++last);
}
}
h.getProcessed().add(i);
},
(h, h2) -> {});
holder.getMissing().forEach(System.out::println);
}
private static class Holder<T> {
private ArrayList<T> processed;
private ArrayList<T> missing;
public Holder() {
this.processed = new ArrayList<>();
this.missing = new ArrayList<>();
}
public ArrayList<T> getProcessed() {
return this.processed;
}
public ArrayList<T> getMissing() {
return this.missing;
}
}
}
This prints
4
6
8
9
10
11
12
13
Note that this sort of thing isn't really a particularly strong fit for Streams. All of the stream processing methods will tend to pass you each item exactly one time, so you need to handle all runs of missing numbers at once, and in the end, you're writing kind of a lot of code to avoid just writing a loop.

Here is one solution using pure streams, albeit not very efficient.
public void test() {
List<String> arr = new ArrayList(
Arrays.asList("N1", "N2", "N3", "N5", "N7"));
List<Integer> list = IntStream
.range(1, arr.size())
.mapToObj(t -> new AbstractMap.SimpleEntry<Integer, Integer>(
extract(arr, t), extract(arr, t) - extract(arr, t - 1)))
.filter(t -> t.getValue() > 1)
.map(t -> t.getKey() - 1)
.collect(Collectors.toList());
System.out.println(list);
}
private int extract(List<String> arr, int t) {
return Integer.parseInt(arr.get(t).substring(1));
}
Major performance block will be because of repeated parsing of list elements. However, this solution will be able to provide all missing numbers.

How to use Java 8 streams to find all values preceding a larger value?

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);
}
}