Misc

Array

To initialise array of custom class

Entry<K, V>[] data = (Entry<K, V>[]) Array.newInstance(Entry.class, capacity);

When accessing an element from array like this without intialising, depends on the thing it will be different, for example.

int[] intergers = new int[10];
integers[0] // == 0

String[] strings = new String[10];  
System.out.println(strings[2]); // == null

Object[] objects = new Object[10];  
System.out.println(objects[2]); // == null

ArrayList

If we do the same for arraylist however we get index out of bound

List<Integer> test = new ArrayList<>(100);  
System.out.println(test.get(3)); // IndexOutOfBoundException

Don't use toList() as it returns an unmodifiable list. Use .collect(Collectors.toList())

Array vs ArrayList

Try use ArrayList as possible, if need a fixed size list, use

List<Entry<K, V>> entries = Collections.nCopies(capacity, null);  
List<Entry<K, V>> fixedSizeList = entries.stream().collect(Collectors.toList());

this is because nCopies will return immutableList

It's either that or

new ArrayList<>(Collections.nCopies(100, null));

or

new ArrayList<>(new Integer[100]) // 100 element of Integer[] array

natural

(a, b) -> a - b

5,4,3,2,1

reverse

(b, a) -> b - a
1,2,3,4,5

Nested comparator

Comparator<Candidate> compareVotes = (a, b) -> b.votes() - a.votes();  
Comparator<Candidate> compareNames = (a, b) -> a.name().compareTo(b.name());  
  
sortedCandidates = new PriorityBlockingQueue<>(10,  compareVotes.thenComparing(compareNames));

Modifiable 2D ArrayList

List<List<Integer>> arrays = Stream.of(  
	List.of(1, 1, 1, 1, 1),  
	List.of(0, 1, 0, 0, 0),  
	List.of(0, 1, 0, 1, 1),  
	List.of(0, 1, 1, 1, 1)  
).map(ArrayList::new).collect(Collectors.toList());

BinarySearch

List<Integer> integerList = List.of(1,2,3,4,5,6,7,8,9);  
System.out.println(Collections.binarySearch(integerList, 9));

Sort but return new

List<Integer> integerList = List.of(5,3,1,9,7,10,13);  
System.out.println(integerList.stream().sorted().collect(Collectors.toList()));

BinarySearch

public static int binarySearch(List<Integer> array, Integer target) {
  int startIndex = 0;
  int endIndex = array.size() - 1;

  while (startIndex < endIndex) {
    int middleIndex = startIndex + (endIndex - startIndex) / 2;

    if (array.get(middleIndex) >= target) {
      endIndex = middleIndex;
    } else {
      startIndex = middleIndex + 1;
    }
  }

  return startIndex;
}

SortedSet

public class SetDemo {
  public static void main(String[] args) {
    SortedSet<Integer> sortedSet = new TreeSet<>();
    Set<Integer> set = new HashSet<>();

    Stream.of(5,5,3,2,9,15,-30).forEach(sortedSet::add);
    Stream.of(5,5,3,2,9,15,-30).forEach(set::add);
    System.out.println("sortedSet: " + sortedSet);
    System.out.println("set: " + set);
  }
}

sortedSet: [-30, 2, 3, 5, 9, 15]
set: [2, 3, 5, 9, -30, 15]

For SortedSet, time complexity is log(n) for everything since it's a Tree

For Set it's O(1)