Candy Distribution

/*
 * Problem : Minimum candies.
 * Input   : n, then n ratings.
 * Output  : The total.
 * Approach: Two passes.
 * 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 long solve(int[] a) {
        // Step 1: Everybody starts with one candy.
        int[] candies = new int[a.length];
        java.util.Arrays.fill(candies, 1);
        // Step 2: Left to right pass: a child with a higher rating than the left neighbour gets one more than that neighbour.
        for (int i = 1; i < a.length; i++) {
            if (a[i] > a[i - 1]) {
                candies[i] = candies[i - 1] + 1;
            }
        }
        // Step 3: Right to left pass does the same for the right neighbour. Keep the bigger number.
        for (int i = a.length - 2; i >= 0; i--) {
            if (a[i] > a[i + 1]) {
                candies[i] = Math.max(candies[i], candies[i + 1] + 1);
            }
        }
        long total = 0;
        for (int value : candies) {
            total += value;
        }
        return total;
    }

    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));
    }
}