Floor and Ceiling in a Sorted Set

/*
 * Problem : Process lines.
 * Input   : Several lines.
 * Output  : One answer per command.
 * Approach: A sorted set and binary search.
 * Time    : O(log n) per query (O(n) insert in arrays)
 * Space   : O(n)
 * 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: A TreeSet keeps its values sorted. floor and ceiling find the nearest values in O(log n).
        java.util.TreeSet<Long> set = new java.util.TreeSet<>();
        String[] output = new String[lines.length];

        for (int i = 0; i < lines.length; i++) {
            String[] part = lines[i].split(" ");
            long value = Long.parseLong(part[1]);

            if (part[0].equals("insert")) {
                set.add(value);
                output[i] = "ok";
            } else if (part[0].equals("floor")) {
                // Step 2: floor = the biggest value that is at most the query. null means there is none.
                Long found = set.floor(value);
                output[i] = found == null ? "none" : String.valueOf(found);
            } else {
                // Step 3: ceiling = the smallest value that is at least the query.
                Long found = set.ceiling(value);
                output[i] = found == null ? "none" : String.valueOf(found);
            }
        }
        return output;
    }

    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(String.join("\n", solve(lines)));
    }
}