Group Anagrams

/*
 * Problem : Group anagrams.
 * Input   : One word per line.
 * Output  : One group per line.
 * Approach: Sorted letters as key.
 * Time    : O(n * k log k)
 * Space   : O(n * k)
 * 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: Two words are anagrams if their sorted letters are equal. We use the sorted letters as the key of a map.
        java.util.Map<String, java.util.List<String>> groups = new java.util.HashMap<>();
        for (String word : lines) {
            char[] letters = word.toCharArray();
            java.util.Arrays.sort(letters);
            groups.computeIfAbsent(new String(letters), key -> new java.util.ArrayList<>()).add(word);
        }

        // Step 2: Sort the words inside each group and join them.
        java.util.List<String> output = new java.util.ArrayList<>();
        for (java.util.List<String> group : groups.values()) {
            java.util.Collections.sort(group);
            output.add(String.join(" ", group));
        }

        // Step 3: Sort the groups. The first word decides the order, so we compare up to the first space.
        output.sort((x, y) -> x.split(" ")[0].compareTo(y.split(" ")[0]));
        return output.toArray(new String[0]);
    }

    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)));
    }
}