package ch.claude_martin;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Set;
import java.util.TreeSet;

class SomeClass {

	static boolean test(int a, int b) {
		return a + 2 == b;
	}

	public static void main(final String... args) {
		// Take the data from the arguments or use the example list:
		final var data = args.length > 0 ? Arrays.stream(args).mapToInt(Integer::parseInt).sorted().toArray()
				: new int[] { 1, 2, 5, 7, 8, 9, 11, 12, 14 };
		System.out.println(Arrays.toString(data));// [5, 7, 9, 11, 12, 14]
		final var size = data.length;
		// three pointers are needed for this. Initially we look at the first two
		// elements of the data.
		// When third is used below to find longer progressions and add each sublist to
		// the result set.
		var p0 = 0;
		var p1 = 1;
		// The result set is a collection of sets. Each of those sets is a progression.
		var resultSet = new ArrayList<Set<Integer>>();
		// This loops through all possible pairs, but p1 is always < p1:
		while (p0 < size && p1 < size) {
			// The set contains at least 2 elements if it is part of the result set:
			var set = new TreeSet<Integer>();
			var a = data[p0];
			var b = data[p1];
			if (test(a, b)) {
				set.add(a);
				set.add(b);
				resultSet.add(new TreeSet<>(set));
				// Now we look for longer progressions that start with the one we just found:
				for (var p2 = p1 + 1; p2 < size; p2++) {
					var c = data[p2];
					if (test(b, c)) {
						set.add(b = c);
						resultSet.add(new TreeSet<>(set));
					}
				}
			}
			p1++; // increment pointer for "b":
			// But it we go out of bounds we instead increment p0:
			if (p1 == size) { 
				// And p1 points to the one next to the new p0:
				p1 = ++p0 + 1;
			}
		}
		// Show the result set:
		System.out.println(resultSet);
	}
}