Develop Reference

Given a string array, return all anagrams and also print the final result which has smallest string as per dictionary

I'm trying to find the smallest word in the sorted anagram of array of string. This is what I have so far:
public static List<List<String>> groupAnagrams(String[] strs) {
List<List<String>> result = new ArrayList<List<String>>();
HashMap<String, ArrayList<String>> map = new HashMap<String, ArrayList<String>>();
for(String str: strs){
char[] arr = new char[26];
for(int i=0; i<str.length(); i++){
arr[str.charAt(i)-'a']++;
}
String ns = new String(arr);
if(map.containsKey(ns)){
map.get(ns).add(str);
}else{
ArrayList<String> al = new ArrayList<String>();
al.add(str);
map.put(ns, al);
}
}
result.addAll(map.values());
return result;
}
public static void main(String[] args) {
String[] strAr1=new String[]{"cat","act","tac","good","god","dog"};
System.out.println(groupAnagrams(strAr1));
}
This yields me a result with :
O/P 1 : [[good], [god, dog], [cat, act, tac]]
But I also need to sort it like to have the desired output as with english dictionary, how can i achieve this ?
O/P 2: act, dog, good

It's a two step process: First, get the minimum value of each sub-list. Then, sort those values:
import java.util.Arrays;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;
import java.util.stream.Collectors;
public class Demo {
public static void main(String[] args) {
// Example data
List<List<String>> results = new ArrayList<>();
results.add(Arrays.asList("good"));
results.add(Arrays.asList("god", "dog"));
results.add(Arrays.asList("cat", "act", "tac"));
// Make a list of the smallest string of each list of strings
// and sort those.
List<String> smalls = results.stream()
.map(list -> list.stream()
.min(Comparator.naturalOrder())
.orElse("(empty)"))
.sorted()
.collect(Collectors.toList());
System.out.println(smalls); // [act, dog, good]
}
}
Another approach is to store your lists of anagrams already sorted, to make finding the minimum word (Or maximum) of each group simpler. Using a SortedSet in the form of a TreeSet instead of ArrayList, for example:
import java.util.Arrays;
import java.util.Collection;
import java.util.List;
import java.util.Map;
import java.util.Objects;
import java.util.SortedSet;
import java.util.TreeSet;
import java.util.stream.Collectors;
public class Demo {
public static Map<String, SortedSet<String>> groupAnagrams(Collection<String> words) {
return Objects.requireNonNull(words).stream()
.collect(Collectors.groupingBy(word -> {
int[] cps = word.codePoints().sorted().toArray();
return new String(cps, 0, cps.length);
},
Collectors.toCollection(TreeSet::new)));
}
public static void main(String[] args) {
var anagrams = groupAnagrams(Arrays.asList("cat","act","tac","good","god","dog"));
System.out.println(anagrams);
var smalls = anagrams.values().stream()
.map(SortedSet::first)
.sorted()
.collect(Collectors.toList());
System.out.println(smalls); // [act, dog, good]
}
}
Also note the streamlined anagram finding code that uses the Collectors.groupingBy() routine to group the words.

As Shawn mentions in his answer, you need to do this in a 2-step process.
A possible improvement however is to already select the smallest word (based on alphabetical order) among perfect anagrams in the first step, so only the sorting is necessary in the second step:
public static List<String> sortedUniqueAnagrams(String[] strs) {
return Arrays.stream(strs)
.collect(Collectors.toMap(
s -> s.chars().mapToObj(c -> c).collect(Collectors.groupingBy(c -> c, Collectors.counting())),
s -> s,
BinaryOperator.minBy(Comparator.<String>naturalOrder())))
.values()
.stream()
.sorted()
.collect(Collectors.toList());
}
Some notes:
the 2 steps are the 2 collect() operations
the first step is more or less identical to your groupAnagrams() implementation except:
It uses Map<Integer, Long> as map keys to avoid abusing the String type like in the question. Integer and Long are just the default types based on s.chars() and Collectors.counting()
The values are the smallest word among perfect anagrams
this is Java 8 code:
In more recent versions you could use String.codePoints() instead of String.chars() to handle 2-char special characters. Probably not an issue here as your anagrams seem to be using Latin characters.
It seems the compiler needs a little bit of hint on Comparator.<String>naturalOrder() otherwise it does not understand the expected types
Use with:
String[] strAr1 = new String[]{ "cat", "act", "tac", "good", "god", "dog" };
System.out.println(sortedUniqueAnagrams(strAr1));
Output:
[act, dog, good]

How to find the max number of an unique string element in a alphanumeric Array list in java

I have list that has alphanumeric elements. I want to find the maximum number of each elements individually.
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
public class Collect {
public static void main(String[] args) {
List<String> alphaNumericList = new ArrayList<String>();
alphaNumericList.add("Demo.23");
alphaNumericList.add("Demo.1000");
alphaNumericList.add("Demo.12");
alphaNumericList.add("Demo.12");
alphaNumericList.add("Test.01");
alphaNumericList.add("Test.02");
alphaNumericList.add("Test.100");
alphaNumericList.add("Test.99");
Collections.sort(alphaNumericList);
System.out.println("Output "+Arrays.asList(alphaNumericList));
}
I need filter only below values. For that I am sorting the list but it filters based on the string rather than int value. I want to achieve in an efficient way. Please suggest on this.
Demo.1000
Test.100
Output [[Demo.1000, Demo.12, Demo.12, Demo.23, Test.01, Test.02, Test.100, Test.99]]

You can either create a special AlphaNumericList type, wrapping the array list or whatever collection(s) you want to use internally, giving it a nice public interface to work with, or for the simplest case if you want to stick to the ArrayList<String>, just use a Comparator for sort(..):
package de.scrum_master.stackoverflow.q60482676;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import static java.lang.Integer.parseInt;
public class Collect {
public static void main(String[] args) {
List<String> alphaNumericList = Arrays.asList(
"Demo.23", "Demo.1000", "Demo.12", "Demo.12",
"Test.01", "Test.02", "Test.100", "Test.99"
);
Collections.sort(
alphaNumericList,
(o1, o2) ->
((Integer) parseInt(o1.split("[.]")[1])).compareTo(parseInt(o2.split("[.]")[1]))
);
System.out.println("Output " + alphaNumericList);
}
}
This will yield the following console log:
Output [Test.01, Test.02, Demo.12, Demo.12, Demo.23, Test.99, Test.100, Demo.1000]
Please let me know if you don't understand lambda syntax. You can also use an anonymous class instead like in pre-8 versions of Java.
Update 1: If you want to refactor the one-line lambda for better readability, maybe you prefer this:
Collections.sort(
alphaNumericList,
(text1, text2) -> {
Integer number1 = parseInt(text1.split("[.]")[1]);
int number2 = parseInt(text2.split("[.]")[1]);
return number1.compareTo(number2);
}
);
Update 2: If more than one dot "." character can occur in your strings, you need to get the numeric substring in a different way via regex match, still not complicated:
Collections.sort(
alphaNumericList,
(text1, text2) -> {
Integer number1 = parseInt(text1.replaceFirst(".*[.]", ""));
int number2 = parseInt(text2.replaceFirst(".*[.]", ""));
return number1.compareTo(number2);
}
);
Update 3: I just noticed that for some weird reason you put the sorted list into another list via Arrays.asList(alphaNumericList) when printing. I have replaced that by just alphaNumericList in the code above and also updated the console log. Before the output was like [[foo, bar, zot]], i.e. a nested list with one element.

Check below answer:
public static void main(String[] args) {
List<String> alphaNumericList = new ArrayList<String>();
alphaNumericList.add("Demo.23");
alphaNumericList.add("Demo.1000");
alphaNumericList.add("Demo.12");
alphaNumericList.add("Demo.12");
alphaNumericList.add("Test.01");
alphaNumericList.add("Test.02");
alphaNumericList.add("Test.100");
alphaNumericList.add("Test.99");
Map<String, List<Integer>> map = new HashMap<>();
for (String val : alphaNumericList) {
String key = val.split("\\.")[0];
Integer value = Integer.valueOf(val.split("\\.")[1]);
if (map.containsKey(key)) {
map.get(key).add(value);
} else {
List<Integer> intList = new ArrayList<>();
intList.add(value);
map.put(key, intList);
}
}
for (Map.Entry<String, List<Integer>> entry : map.entrySet()) {
List<Integer> valueList = entry.getValue();
Collections.sort(valueList, Collections.reverseOrder());
System.out.print(entry.getKey() + "." + valueList.get(0) + " ");
}
}

Using stream and toMap() collector.
Map<String, Long> result = alphaNumericList.stream().collect(
toMap(k -> k.split("\\.")[0], v -> Long.parseLong(v.split("\\.")[1]), maxBy(Long::compare)));
The result map will contain word part as a key and maximum number as a value of the map(in your example the map will contain {Demo=1000, Test=100})

a. Assuming there are string of type Demo. and Test. in your arraylist.
b. It should be trivial to filter out elements with String Demo. and then extract the max integer for same.
c. Same should be applicable for extracting out max number associated with Test.
Please check the following snippet of code to achieve the same.
Set<String> uniqueString = alphaNumericList.stream().map(c->c.replaceAll("\\.[0-9]*","")).collect(Collectors.toSet());
Map<String,Integer> map = new HashMap<>();
for(String s:uniqueString){
int max= alphaNumericList.stream().filter(c -> c.startsWith(s+".")).map(c -> c.replaceAll(s+"\\.","")).map(c-> Integer.parseInt(c)).max(Integer::compare).get();
map.put(s,max);
}

Java Functional Programming: How to convert a if-else ladder inside for loop to functional style?

The expectation is derive 3 lists itemIsBoth, aItems, bItems from the input list items.
How to convert code like below to functional style? (I understand this code is clear enough in an imperative style, but I want to know does declarative style really fail to deal with such a simple example). Thanks.
for (Item item: items) {
if (item.isA() && item.isB()) {
itemIsBoth.add(item);
} else if (item.isA()) {
aItems.add(item);
} else if (item.isB()){
bItems.add(item)
}
}

The question title is quite broad (convert if-else ladder), but since the actual question asks about a specific scenario, let me offer a sample that can at least illustrate what can be done.
Because the if-else structure creates three distinct lists based on a predicate applied to the item, we can express this behavior more declaratively as a grouping operation. The only extra needed to make this work out of the box would be to collapse the multiple Boolean predicates using a tagging object. For example:
class Item {
enum Category {A, B, AB}
public Category getCategory() {
return /* ... */;
}
}
Then the logic can be expressed simply as:
Map<Item.Category, List<Item>> categorized =
items.stream().collect(Collectors.groupingBy(Item::getCategory));
where each list can be retrieved from the map given its category.
If it's not possible to change class Item, the same effect can be achieved by moving the enum declaration and the categorization method outsize the Item class (the method would become a static method).

Another solution using Vavr and doing only one iteration over a list of items might be achieved using foldLeft:
list.foldLeft(
Tuple.of(List.empty(), List.empty(), List.empty()), //we declare 3 lists for results
(lists, item) -> Match(item).of(
//both predicates pass, add to first list
Case($(allOf(Item::isA, Item::isB)), lists.map1(l -> l.append(item))),
//is a, add to second list
Case($(Item::isA), lists.map2(l -> l.append(item))),
//is b, add to third list
Case($(Item::isB), lists.map3(l -> l.append(item)))
))
);
It will return a tuple containing three lists with results.

Of course, you can. The functional way is to use declarative ways.
Mathematically you are setting an Equivalence relation, then, you can write
Map<String, List<Item>> ys = xs
.stream()
.collect(groupingBy(x -> here your equivalence relation))
A simple example show this
public class Main {
static class Item {
private final boolean a;
private final boolean b;
Item(boolean a, boolean b) {
this.a = a;
this.b = b;
}
public boolean isB() {
return b;
}
public boolean isA() {
return a;
}
}
public static void main(String[] args) {
List<Item> xs = asList(new Item(true, true), new Item(true, true), new Item(false, true));
Map<String, List<Item>> ys = xs.stream().collect(groupingBy(x -> x.isA() + "," + x.isB()));
ys.entrySet().forEach(System.out::println);
}
}
With output
true,true=[com.foo.Main$Item#64616ca2, com.foo.Main$Item#13fee20c]
false,true=[com.foo.Main$Item#4e04a765]

Another way you can get rid of the if-else is to to replace them with Predicate and Consumer:
Map<Predicate<Item>, Consumer<Item>> actions =
Map.of(item.predicateA(), aItems::add, item.predicateB(), bItems::add);
actions.forEach((key, value) -> items.stream().filter(key).forEach(value));
Therefore you need to enhace your Item with the both mehods predicateA() and predicateB() using the logic you have implemented in your isA() and isB()
Btw I would still suggest to use your if-else logic.

Since you've mentioned vavr as a tag, I'm gonna provide a solution using vavr collections.
import static io.vavr.Predicates.allOf;
import static io.vavr.Predicates.not;
...
final Array<Item> itemIsBoth = items.filter(allOf(Item::isA, Item::isB));
final Array<Item> aItems = items.filter(allOf(Item::isA, not(Item::isB)));
final Array<Item> bItems = items.filter(allOf(Item::isB, not(Item::isA)));
The advantage of this solution that it's simple to understand at a glance and it's as functional as you can get with Java. The drawback is that it will iterate over the original collections three times instead of once. That's still an O(n), but with a constant multiplier factor of 3. On non-critical code paths and with small collections it might be worth to trade a few CPU cycles for code clarity.
Of course, this works with all the other vavr collections too, so you can replace Array with List, Vector, Stream, etc.

Not (functional in the sense of) using lambda's or so, but quite functional in the sense of using only functions (as per mathematics) and no local state/variabels anywhere :
/* returns 0, 1, 2 or 3 according to isA/isB */
int getCategory(Item item) {
return item.isA() ? 1 : 0 + 2 * (item.isB() ? 1 : 0)
}
LinkedList<Item>[] lists = new LinkedList<Item> { initializer for 4-element array here };
{
for (Item item: items) {
lists[getCategory(item)].addLast(item);
}
}

The question is somewhat controversial, as it seems (+5/-3 at the time of writing this).
As you mentioned, the imperative solution here is most likely the most simple, appropriate and readable one.
The functional or declarative style does not really "fail". It's rather raising questions about the exact goals, conditions and context, and maybe even philosophical questions about language details (like why there is no standard Pair class in core Java).
You can apply a functional solution here. One simple, technical question is then whether you really want to fill the existing lists, or whether it's OK to create new lists. In both cases, you can use the Collectors#groupingBy method.
The grouping criterion is the same in both cases: Namely, any "representation" of the specific combination of isA and isB of one item. There are different possible solutions for that. In the examples below, I used an Entry<Boolean, Boolean> as the key.
(If you had further conditions, like isC and isD, then you could in fact also use a List<Boolean>).
The example shows how you can either add the item to existing lists (as in your question), or create new lists (which is a tad simpler and cleaner).
import java.util.AbstractMap.SimpleEntry;
import java.util.ArrayList;
import java.util.LinkedHashMap;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.stream.Collectors;
public class FunctionalIfElse
{
public static void main(String[] args)
{
List<Item> items = new ArrayList<Item>();
items.add(new Item(false, false));
items.add(new Item(false, true));
items.add(new Item(true, false));
items.add(new Item(true, true));
fillExistingLists(items);
createNewLists(items);
}
private static void fillExistingLists(List<Item> items)
{
System.out.println("Filling existing lists:");
List<Item> itemIsBoth = new ArrayList<Item>();
List<Item> aItems = new ArrayList<Item>();
List<Item> bItems = new ArrayList<Item>();
Map<Entry<Boolean, Boolean>, List<Item>> map =
new LinkedHashMap<Entry<Boolean, Boolean>, List<Item>>();
map.put(entryWith(true, true), itemIsBoth);
map.put(entryWith(true, false), aItems);
map.put(entryWith(false, true), bItems);
items.stream().collect(Collectors.groupingBy(
item -> entryWith(item.isA(), item.isB()),
() -> map, Collectors.toList()));
System.out.println("Both");
itemIsBoth.forEach(System.out::println);
System.out.println("A");
aItems.forEach(System.out::println);
System.out.println("B");
bItems.forEach(System.out::println);
}
private static void createNewLists(List<Item> items)
{
System.out.println("Creating new lists:");
Map<Entry<Boolean, Boolean>, List<Item>> map =
items.stream().collect(Collectors.groupingBy(
item -> entryWith(item.isA(), item.isB()),
LinkedHashMap::new, Collectors.toList()));
List<Item> itemIsBoth = map.get(entryWith(true, true));
List<Item> aItems = map.get(entryWith(true, false));
List<Item> bItems = map.get(entryWith(false, true));
System.out.println("Both");
itemIsBoth.forEach(System.out::println);
System.out.println("A");
aItems.forEach(System.out::println);
System.out.println("B");
bItems.forEach(System.out::println);
}
private static <K, V> Entry<K, V> entryWith(K k, V v)
{
return new SimpleEntry<K, V>(k, v);
}
static class Item
{
private boolean a;
private boolean b;
public Item(boolean a, boolean b)
{
this.a = a;
this.b = b;
}
public boolean isA()
{
return a;
}
public boolean isB()
{
return b;
}
#Override
public String toString()
{
return "(" + a + ", " + b + ")";
}
}
}

How to compare two Streams in Java 8

What would be a good way to compare two Stream instances in Java 8 and find out whether they have the same elements, specifically for purposes of unit testing?
What I've got now is:
#Test
void testSomething() {
Stream<Integer> expected;
Stream<Integer> thingUnderTest;
// (...)
Assert.assertArrayEquals(expected.toArray(), thingUnderTest.toArray());
}
or alternatively:
Assert.assertEquals(
expected.collect(Collectors.toList()),
thingUnderTest.collect(Collectors.toList()));
But that means I'm constructing two collections and discarding them. It's not a performance issue, given the size of my test streams, but I'm wondering whether there's a canonical way to compare two streams.

static void assertStreamEquals(Stream<?> s1, Stream<?> s2) {
Iterator<?> iter1 = s1.iterator(), iter2 = s2.iterator();
while(iter1.hasNext() && iter2.hasNext())
assertEquals(iter1.next(), iter2.next());
assert !iter1.hasNext() && !iter2.hasNext();
}

Collecting the stream under test (as you show) is a straightforward and effective way of performing the test. You may create the list of expected results in the easiest way available, which might not be collecting a stream.
Alternatively, with most libraries for creating mock collaborators, one could mock a Consumer that "expects" a series of accept() calls with particular elements. Consume the Stream with it, and then "verify" that its configured expectations were met.

Using the elementsEqual method in the Guava library:
Iterators.elementsEqual(s1.iterator(), s2.iterator())

You can assert the stream's content without creating a Stream<> expected.
AssertJ has fluent and readable solutions for this.
import static org.assertj.core.api.Assertions.assertThat;
import java.util.stream.Stream;
import org.junit.jupiter.api.Test;
class MyTests {
#Test
void test() {
Stream<Integer> actual = Stream.of(0, 8, 15); // your thingUnderTest
assertThat(actual).containsExactly(0, 8, 15);
}
}

public static boolean equalStreams(Stream<?> ... streams) {
List<Iterator<?>> is = Arrays.stream(streams)
.map(Stream::iterator)
.collect(Collectors.toList());
while (is.stream().allMatch(Iterator::hasNext))
if (is.stream().map(Iterator::next).distinct().limit(2).count() > 1)
return false;
return is.stream().noneMatch(Iterator::hasNext);
}

If order of elements doesn't matter, then comparison can be done for any number of streams using stream groupingBy() and then checking count for each element.
Example is for IntStream but you can get the idea behind this method:
IntStream stream1 = IntStream.range(1, 9);
IntStream stream2 = IntStream.range(1, 10);
boolean equal = IntStream.concat(stream1, stream2)
.boxed()
.collect(Collectors.groupingBy(Function.identity()))
.entrySet().stream().noneMatch(e -> e.getValue().size() != 2);

Another way of doing the same thing:
static void assertStreamEquals(Stream<?> s1, Stream<?> s2) {
Iterator<?> it2 = s2.iterator();
assert s1.allMatch(o -> it2.hasNext() && Objects.equals(o, it2.next())) && !it2.hasNext();
}

How to Compare Two Streams in java 8 and above: with the example of Comparing IntStream
package com.techsqually.java.language.generics.basics;
import java.util.Iterator;
import java.util.stream.IntStream;
import java.util.stream.Stream;
public class TwoStreamComparision {
public static void main(String[] args) {
String a = "Arpan";
String b = "Arpen";
IntStream s1 = a.chars();
IntStream s2 = b.chars();
Iterator<Integer> s1It = s1.iterator();
Iterator<Integer> s2It = s2.iterator();
//Code to check how many characters are not equal in both the string at their Respective Position
int count = 0;
while (s2It.hasNext()){
if (!s1It.next().equals(s2It.next())){
count++;
}
}
System.out.println(count);
}
}

How to pop the first element in a java stream? [duplicate]

Java 8 introduced a Stream class that resembles Scala's Stream, a powerful lazy construct using which it is possible to do something like this very concisely:
def from(n: Int): Stream[Int] = n #:: from(n+1)
def sieve(s: Stream[Int]): Stream[Int] = {
s.head #:: sieve(s.tail filter (_ % s.head != 0))
}
val primes = sieve(from(2))
primes takeWhile(_ < 1000) print // prints all primes less than 1000
I wondered if it is possible to do this in Java 8, so I wrote something like this:
IntStream from(int n) {
return IntStream.iterate(n, m -> m + 1);
}
IntStream sieve(IntStream s) {
int head = s.findFirst().getAsInt();
return IntStream.concat(IntStream.of(head), sieve(s.skip(1).filter(n -> n % head != 0)));
}
IntStream primes = sieve(from(2));
Fairly simple, but it produces java.lang.IllegalStateException: stream has already been operated upon or closed because both findFirst() and skip() are terminal operations on Stream which can be done only once.
I don't really have to use up the stream twice since all I need is the first number in the stream and the rest as another stream, i.e. equivalent of Scala's Stream.head and Stream.tail. Is there a method in Java 8 Stream that I can use to achieve this?
Thanks.

Even if you hadn’t the problem that you can’t split an IntStream, you code didn’t work because you are invoking your sieve method recursively instead of lazily. So you had an infinity recursion before you could query your resulting stream for the first value.
Splitting an IntStream s into a head and a tail IntStream (which has not yet consumed) is possible:
PrimitiveIterator.OfInt it = s.iterator();
int head = it.nextInt();
IntStream tail = IntStream.generate(it::next).filter(i -> i % head != 0);
At this place you need a construct of invoking sieve on the tail lazily. Stream does not provide that; concat expects existing stream instances as arguments and you can’t construct a stream invoking sieve lazily with a lambda expression as lazy creation works with mutable state only which lambda expressions do not support. If you don’t have a library implementation hiding the mutable state you have to use a mutable object. But once you accept the requirement of mutable state, the solution can be even easier than your first approach:
IntStream primes = from(2).filter(i -> p.test(i)).peek(i -> p = p.and(v -> v % i != 0));
IntPredicate p = x -> true;
IntStream from(int n)
{
return IntStream.iterate(n, m -> m + 1);
}
This will recursively create a filter but in the end it doesn’t matter whether you create a tree of IntPredicates or a tree of IntStreams (like with your IntStream.concat approach if it did work). If you don’t like the mutable instance field for the filter you can hide it in an inner class (but not in a lambda expression…).

My StreamEx library has now headTail() operation which solves the problem:
public static StreamEx<Integer> sieve(StreamEx<Integer> input) {
return input.headTail((head, tail) ->
sieve(tail.filter(n -> n % head != 0)).prepend(head));
}
The headTail method takes a BiFunction which will be executed at most once during the stream terminal operation execution. So this implementation is lazy: it does not compute anything until traversal starts and computes only as much prime numbers as requested. The BiFunction receives a first stream element head and the stream of the rest elements tail and can modify the tail in any way it wants. You may use it with predefined input:
sieve(IntStreamEx.range(2, 1000).boxed()).forEach(System.out::println);
But infinite stream work as well
sieve(StreamEx.iterate(2, x -> x+1)).takeWhile(x -> x < 1000)
.forEach(System.out::println);
// Not the primes till 1000, but 1000 first primes
sieve(StreamEx.iterate(2, x -> x+1)).limit(1000).forEach(System.out::println);
There's also alternative solution using headTail and predicate concatenation:
public static StreamEx<Integer> sieve(StreamEx<Integer> input, IntPredicate isPrime) {
return input.headTail((head, tail) -> isPrime.test(head)
? sieve(tail, isPrime.and(n -> n % head != 0)).prepend(head)
: sieve(tail, isPrime));
}
sieve(StreamEx.iterate(2, x -> x+1), i -> true).limit(1000).forEach(System.out::println);
It interesting to compare recursive solutions: how many primes they capable to generate.
#John McClean solution (StreamUtils)
John McClean solutions are not lazy: you cannot feed them with infinite stream. So I just found by trial-and-error the maximal allowed upper bound (17793) (after that StackOverflowError occurs):
public void sieveTest(){
sieve(IntStream.range(2, 17793).boxed()).forEach(System.out::println);
}
#John McClean solution (Streamable)
public void sieveTest2(){
sieve(Streamable.range(2, 39990)).forEach(System.out::println);
}
Increasing upper limit above 39990 results in StackOverflowError.
#frhack solution (LazySeq)
LazySeq<Integer> ints = integers(2);
LazySeq primes = sieve(ints); // sieve method from #frhack answer
primes.forEach(p -> System.out.println(p));
Result: stuck after prime number = 53327 with enormous heap allocation and garbage collection taking more than 90%. It took several minutes to advance from 53323 to 53327, so waiting more seems impractical.
#vidi solution
Prime.stream().forEach(System.out::println);
Result: StackOverflowError after prime number = 134417.
My solution (StreamEx)
sieve(StreamEx.iterate(2, x -> x+1)).forEach(System.out::println);
Result: StackOverflowError after prime number = 236167.
#frhack solution (rxjava)
Observable<Integer> primes = Observable.from(()->primesStream.iterator());
primes.forEach((x) -> System.out.println(x.toString()));
Result: StackOverflowError after prime number = 367663.
#Holger solution
IntStream primes=from(2).filter(i->p.test(i)).peek(i->p=p.and(v->v%i!=0));
primes.forEach(System.out::println);
Result: StackOverflowError after prime number = 368089.
My solution (StreamEx with predicate concatenation)
sieve(StreamEx.iterate(2, x -> x+1), i -> true).forEach(System.out::println);
Result: StackOverflowError after prime number = 368287.
So three solutions involving predicate concatenation win, because each new condition adds only 2 more stack frames. I think, the difference between them is marginal and should not be considered to define a winner. However I like my first StreamEx solution more as it more similar to Scala code.

The solution below does not do state mutations, except for the head/tail deconstruction of the stream.
The lazyness is obtained using IntStream.iterate. The class Prime is used to keep the generator state
import java.util.PrimitiveIterator;
import java.util.stream.IntStream;
import java.util.stream.Stream;
public class Prime {
private final IntStream candidates;
private final int current;
private Prime(int current, IntStream candidates)
{
this.current = current;
this.candidates = candidates;
}
private Prime next()
{
PrimitiveIterator.OfInt it = candidates.filter(n -> n % current != 0).iterator();
int head = it.next();
IntStream tail = IntStream.generate(it::next);
return new Prime(head, tail);
}
public static Stream<Integer> stream() {
IntStream possiblePrimes = IntStream.iterate(3, i -> i + 1);
return Stream.iterate(new Prime(2, possiblePrimes), Prime::next)
.map(p -> p.current);
}
}
The usage would be this:
Stream<Integer> first10Primes = Prime.stream().limit(10)

You can essentially implement it like this:
static <T> Tuple2<Optional<T>, Seq<T>> splitAtHead(Stream<T> stream) {
Iterator<T> it = stream.iterator();
return tuple(it.hasNext() ? Optional.of(it.next()) : Optional.empty(), seq(it));
}
In the above example, Tuple2 and Seq are types borrowed from jOOλ, a library that we developed for jOOQ integration tests. If you don't want any additional dependencies, you might as well implement them yourself:
class Tuple2<T1, T2> {
final T1 v1;
final T2 v2;
Tuple2(T1 v1, T2 v2) {
this.v1 = v1;
this.v2 = v2;
}
static <T1, T2> Tuple2<T1, T2> tuple(T1 v1, T2 v2) {
return new Tuple<>(v1, v2);
}
}
static <T> Tuple2<Optional<T>, Stream<T>> splitAtHead(Stream<T> stream) {
Iterator<T> it = stream.iterator();
return tuple(
it.hasNext() ? Optional.of(it.next()) : Optional.empty,
StreamSupport.stream(Spliterators.spliteratorUnknownSize(
it, Spliterator.ORDERED
), false)
);
}

If you don't mind using 3rd party libraries cyclops-streams, I library I wrote has a number of potential solutions.
The StreamUtils class has large number of static methods for working directly with java.util.stream.Streams including headAndTail.
HeadAndTail<Integer> headAndTail = StreamUtils.headAndTail(Stream.of(1,2,3,4));
int head = headAndTail.head(); //1
Stream<Integer> tail = headAndTail.tail(); //Stream[2,3,4]
The Streamable class represents a replayable Stream and works by building a lazy, caching intermediate data-structure. Because it is caching and repayable - head and tail can be implemented directly and separately.
Streamable<Integer> replayable= Streamable.fromStream(Stream.of(1,2,3,4));
int head = repayable.head(); //1
Stream<Integer> tail = replayable.tail(); //Stream[2,3,4]
cyclops-streams also provides a sequential Stream extension that in turn extends jOOλ and has both Tuple based (from jOOλ) and domain object (HeadAndTail) solutions for head and tail extraction.
SequenceM.of(1,2,3,4)
.splitAtHead(); //Tuple[1,SequenceM[2,3,4]
SequenceM.of(1,2,3,4)
.headAndTail();
Update per Tagir's request -> A Java version of the Scala sieve using SequenceM
public void sieveTest(){
sieve(SequenceM.range(2, 1_000)).forEach(System.out::println);
}
SequenceM<Integer> sieve(SequenceM<Integer> s){
return s.headAndTailOptional().map(ht ->SequenceM.of(ht.head())
.appendStream(sieve(ht.tail().filter(n -> n % ht.head() != 0))))
.orElse(SequenceM.of());
}
And another version via Streamable
public void sieveTest2(){
sieve(Streamable.range(2, 1_000)).forEach(System.out::println);
}
Streamable<Integer> sieve(Streamable<Integer> s){
return s.size()==0? Streamable.of() : Streamable.of(s.head())
.appendStreamable(sieve(s.tail()
.filter(n -> n % s.head() != 0)));
}
Note - neither Streamable of SequenceM have an Empty implementation - hence the size check for Streamable and the use of headAndTailOptional.
Finally a version using plain java.util.stream.Stream
import static com.aol.cyclops.streams.StreamUtils.headAndTailOptional;
public void sieveTest(){
sieve(IntStream.range(2, 1_000).boxed()).forEach(System.out::println);
}
Stream<Integer> sieve(Stream<Integer> s){
return headAndTailOptional(s).map(ht ->Stream.concat(Stream.of(ht.head())
,sieve(ht.tail().filter(n -> n % ht.head() != 0))))
.orElse(Stream.of());
}
Another update - a lazy iterative based on #Holger's version using objects rather than primitives (note a primitive version is also possible)
final Mutable<Predicate<Integer>> predicate = Mutable.of(x->true);
SequenceM.iterate(2, n->n+1)
.filter(i->predicate.get().test(i))
.peek(i->predicate.mutate(p-> p.and(v -> v%i!=0)))
.limit(100000)
.forEach(System.out::println);

There are many interesting suggestions provided here, but if someone needs a solution without dependencies to third party libraries I came up with this:
import java.util.AbstractMap;
import java.util.Optional;
import java.util.Spliterators;
import java.util.stream.StreamSupport;
/**
* Splits a stream in the head element and a tail stream.
* Parallel streams are not supported.
*
* #param stream Stream to split.
* #param <T> Type of the input stream.
* #return A map entry where {#link Map.Entry#getKey()} contains an
* optional with the first element (head) of the original stream
* and {#link Map.Entry#getValue()} the tail of the original stream.
* #throws IllegalArgumentException for parallel streams.
*/
public static <T> Map.Entry<Optional<T>, Stream<T>> headAndTail(final Stream<T> stream) {
if (stream.isParallel()) {
throw new IllegalArgumentException("parallel streams are not supported");
}
final Iterator<T> iterator = stream.iterator();
return new AbstractMap.SimpleImmutableEntry<>(
iterator.hasNext() ? Optional.of(iterator.next()) : Optional.empty(),
StreamSupport.stream(Spliterators.spliteratorUnknownSize(iterator, 0), false)
);
}

To get head and tail you need a Lazy Stream implementation. Java 8 stream or RxJava are not suitable.
You can use for example LazySeq as follows.
Lazy sequence is always traversed from the beginning using very cheap
first/rest decomposition (head() and tail())
LazySeq implements java.util.List interface, thus can be used in
variety of places. Moreover it also implements Java 8 enhancements to
collections, namely streams and collectors
package com.company;
import com.nurkiewicz.lazyseq.LazySeq;
public class Main {
public static void main(String[] args) {
LazySeq<Integer> ints = integers(2);
LazySeq primes = sieve(ints);
primes.take(10).forEach(p -> System.out.println(p));
}
private static LazySeq<Integer> sieve(LazySeq<Integer> s) {
return LazySeq.cons(s.head(), () -> sieve(s.filter(x -> x % s.head() != 0)));
}
private static LazySeq<Integer> integers(int from) {
return LazySeq.cons(from, () -> integers(from + 1));
}
}

Here is another recipe using the way suggested by Holger.
It use RxJava just to add the possibility to use the take(int) method and many others.
package com.company;
import rx.Observable;
import java.util.function.IntPredicate;
import java.util.stream.IntStream;
public class Main {
public static void main(String[] args) {
final IntPredicate[] p={(x)->true};
IntStream primesStream=IntStream.iterate(2,n->n+1).filter(i -> p[0].test(i)).peek(i->p[0]=p[0].and(v->v%i!=0) );
Observable primes = Observable.from(()->primesStream.iterator());
primes.take(10).forEach((x) -> System.out.println(x.toString()));
}
}

This should work with parallel streams as well:
public static <T> Map.Entry<Optional<T>, Stream<T>> headAndTail(final Stream<T> stream) {
final AtomicReference<Optional<T>> head = new AtomicReference<>(Optional.empty());
final var spliterator = stream.spliterator();
spliterator.tryAdvance(x -> head.set(Optional.of(x)));
return Map.entry(head.get(), StreamSupport.stream(spliterator, stream.isParallel()));
}

If you want to get head of a stream, just:
IntStream.range(1, 5).first();
If you want to get tail of a stream, just:
IntStream.range(1, 5).skip(1);
If you want to get both head and tail of a stream, just:
IntStream s = IntStream.range(1, 5);
int head = s.head();
IntStream tail = s.tail();
If you want to find the prime, just:
LongStream.range(2, n)
.filter(i -> LongStream.range(2, (long) Math.sqrt(i) + 1).noneMatch(j -> i % j == 0))
.forEach(N::println);
If you want to know more, go to get abacus-common
Declaration： I'm the developer of abacus-common.