How to sort an IntStream in reverse order

Try negating the values before sorting and negating (back to normal) after sorting:

in = new BufferedReader(reader); int[] arr = in.lines() .mapToInt(Integer::parseInt) .map(i -> -i).sorted().map(i -> -i) .limit((long) N) .toArray(); 

Because the reverse order is not the natural order, sorted() can't be used to sort in reverse order. If you avoid the IntStream, using a Stream<Integer> instead, then you can use a Collections.reverseOrder() to sort the stream in a reverse to the natural order. Then you can call mapToInt and convert to int[] at the end.

int[] arr = in.lines() .map(Integer::valueOf) // Extract Integer, not int .sorted(Collections.reverseOrder()) // On Stream<Integer> .limit(N) .mapToInt(i -> i) // map Integer to int .toArray(); 

when using Instream, you are actually dealing with primitive, and your hands are tight (you are limited to natural ordering and you can't define custom comparator. You have two solutions:

• stick with the primitive stream and come up with hacks like the one proposed by @normanrz

• or you can convert to Integer (box) and use variety of solution like the one bellow (but be advised this boxing and unboxing might cause performance problems).

  int[] sortedArray = IntStream.of(costs).boxed() .sorted(Collections.reverseOrder()) .mapToInt(value -> value.intValue()).toArray(); 

It’s hard to say whether .sorted().limit((long) N).toArray() will get optimized in some cases (it’s implementation dependent but given Oracle’s current implementation, I wouldn’t expect it), but in this special case, the source stream is a stream of unknown size which makes optimizations even less likely.

If you want to be on the safe side, you may adapt this solution for getting the n maximum numbers of a stream efficiently. All you have to do is to reverse the order:

public static IntStream maxValuesDescending(IntStream source, int limit) { TreeMap<Integer,Integer> m=new TreeMap<>(Comparator.reverseOrder()); source.forEachOrdered(new IntConsumer() { int size, min=Integer.MIN_VALUE; public void accept(int value) { if(value<min) return; m.merge(value, 1, Integer::sum); if(size<limit) size++; else m.compute(min=m.lastKey(), (k,count)->count==1? null: count-1); } }); if(m.size()==limit)// no duplicates return m.keySet().stream().mapToInt(Integer::valueOf); return m.entrySet().stream().flatMapToInt(e->{ int value = e.getKey(), count = e.getValue(); return count==1? IntStream.of(value): IntStream.range(0, count).map(i->value); }); } 

Then you can use it like

int[] arr = maxValuesDescending(in.lines().mapToInt(Integer::parseInt), N).toArray(); 

But you are not required to create an array as you can use arbitrary IntStream operations on the result. This solution will hold at most N values, even less if there are duplicates as it only holds distinct values and their count.