Maximum Product Subarray

/*
 * Problem : Process an array.
 * Input   : n, then n integers.
 * Output  : The largest product.
 * Approach: Track max and min.
 * 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 long solve(int[] a) {
        // Step 1: Keep the biggest and the smallest product of a subarray that ends at the current element.
        // The smallest product matters because a negative number can turn it into the biggest.
        long best = a[0];
        long maxEnding = a[0];
        long minEnding = a[0];

        for (int i = 1; i < a.length; i++) {
            long value = a[i];

            // Step 2: The new products come from the value alone, or the value times the old biggest or smallest.
            long first = maxEnding * value;
            long second = minEnding * value;
            maxEnding = Math.max(value, Math.max(first, second));
            minEnding = Math.min(value, Math.min(first, second));

            // Step 3: Keep the best result seen so far.
            best = Math.max(best, maxEnding);
        }
        return best;
    }

    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));
    }
}