Pipeline of Operations

/*
 * Problem : Process lines.
 * Input   : First line numbers, then one operation per line.
 * Output  : The numbers left, or empty.
 * Approach: Lazy pipeline.
 * Time    : O(n * stages)
 * Space   : O(1) extra
 * Java    : 8 or newer.
 */
import java.util.Scanner;

public class Main {
    // This method solves the task. It receives the input and returns the answer.
    static String solve(String[] lines) {
        // Step 1: Read the numbers into a stream. Streams are lazy: each number goes through all steps before the next number starts.
        java.util.stream.LongStream stream = java.util.Arrays.stream(lines[0].split(" ")).mapToLong(Long::parseLong);
        // Step 2: Add one stage for each operation line.
        for (int i = 1; i < lines.length; i++) {
            String[] part = lines[i].split(" ");
            final long x = part.length > 1 ? Long.parseLong(part[1]) : 0;
            if (part[0].equals("add")) {
                stream = stream.map(v -> v + x);
            } else if (part[0].equals("mul")) {
                stream = stream.map(v -> v * x);
            } else if (part[0].equals("even")) {
                stream = stream.filter(v -> v % 2 == 0);
            } else if (part[0].equals("gt")) {
                stream = stream.filter(v -> v > x);
            } else {
                stream = stream.filter(v -> v < x);
            }
        }
        String result = stream.mapToObj(String::valueOf).collect(java.util.stream.Collectors.joining(" "));
        return result.isEmpty() ? "empty" : result;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        java.util.List<String> all = new java.util.ArrayList<>();
        while (scanner.hasNextLine()) {
            String line = scanner.nextLine().trim();
            if (!line.isEmpty()) {
                all.add(line);
            }
        }
        String[] lines = all.toArray(new String[0]);
        System.out.println(solve(lines));
    }
}