Leaders in an Array

/*
 * Problem : Process an array.
 * Input   : n, then n integers.
 * Output  : The leaders.
 * Approach: Scan from the right.
 * Time    : O(n)
 * Space   : O(n)
 * Java    : 8 or newer.
 */
import java.util.Scanner;
import java.util.Arrays;
import java.util.stream.Collectors;

public class Main {
    // This method solves the task. It receives the input and returns the answer.
    static int[] solve(int[] a) {
        // Step 1: Walk from the right end. Keep the biggest value seen so far on the right side.
        int[] found = new int[a.length];
        int count = 0;
        long biggest = Long.MIN_VALUE;

        for (int i = a.length - 1; i >= 0; i--) {
            // Step 2: A value bigger than everything on its right is a leader.
            if (a[i] > biggest) {
                found[count++] = a[i];
                biggest = a[i];
            }
        }

        // Step 3: We found the leaders from right to left, so we reverse them.
        int[] result = new int[count];
        for (int i = 0; i < count; i++) {
            result[i] = found[count - 1 - i];
        }
        return result;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int size = scanner.nextInt();
        int[] a = new int[size];
        for (int i = 0; i < size; i++) {
            a[i] = scanner.nextInt();
        }
        System.out.println(Arrays.stream(solve(a)).mapToObj(String::valueOf).collect(Collectors.joining(" ")));
    }
}