Sort Values by Frequency

/*
 * Problem : Process an array.
 * Input   : n, then n integers.
 * Output  : One line per value.
 * Approach: Count, then sort the keys.
 * Time    : O(n log n)
 * 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(int[] a) {
        // Step 1: Count each value.
        java.util.Map<Integer, Integer> counts = new java.util.HashMap<>();
        for (int value : a) {
            counts.merge(value, 1, Integer::sum);
        }

        // Step 2: Put the entries in a list and sort. Higher count first, then smaller value first.
        java.util.List<java.util.Map.Entry<Integer, Integer>> entries = new java.util.ArrayList<>(counts.entrySet());
        entries.sort((x, y) -> {
            if (!x.getValue().equals(y.getValue())) {
                return y.getValue() - x.getValue();
            }
            return Integer.compare(x.getKey(), y.getKey());
        });

        // Step 3: Make one line for each entry.
        String[] output = new String[entries.size()];
        for (int i = 0; i < entries.size(); i++) {
            output[i] = entries.get(i).getKey() + " " + entries.get(i).getValue();
        }
        return output;
    }

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