Second Largest Element

/*
 * Problem : Process an array.
 * Input   : n, then n integers.
 * Output  : The second largest or none.
 * Approach: Track two values.
 * Time    : O(n)
 * Space   : O(1)
 * 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: Keep the two largest distinct values. Long.MIN_VALUE means: not found yet.
        long largest = Long.MIN_VALUE;
        long second = Long.MIN_VALUE;

        // Step 2: Check every element once.
        for (int value : a) {
            if (value > largest) {
                // Step 3: A new largest value. The old largest becomes the second.
                second = largest;
                largest = value;
            } else if (value < largest && value > second) {
                // Step 4: Smaller than the largest, but bigger than the second.
                second = value;
            }
        }

        // Step 5: If the second value was never set, there is no answer.
        return second == Long.MIN_VALUE ? "none" : String.valueOf(second);
    }

    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(solve(a));
    }
}