Two Sum

/*
 * Problem : Process an array and a number k.
 * Input   : n, then n integers, then k.
 * Output  : Indexes i and j or none.
 * Approach: Use a hash map.
 * Time    : O(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, int k) {
        // Step 1: Remember the index of every value we have seen. Key = value, value = index.
        java.util.Map<Integer, Integer> seen = new java.util.HashMap<>();
        int bestI = -1;
        int bestJ = -1;

        for (int j = 0; j < a.length; j++) {
            // Step 2: The partner of a[j] is k - a[j]. Did we see it before?
            Integer i = seen.get(k - a[j]);
            if (i != null && (bestI == -1 || i < bestI || (i == bestI && j < bestJ))) {
                bestI = i;
                bestJ = j;
            }

            // Step 3: Keep the FIRST index of each value, so we get the smallest i.
            seen.putIfAbsent(a[j], j);
        }
        return bestI == -1 ? "none" : bestI + " " + bestJ;
    }

    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();
        }
        int k = scanner.nextInt();
        System.out.println(solve(a, k));
    }
}