Smallest Ship Capacity for D Days

/*
 * Problem : Smallest capacity.
 * Input   : n, the weights, then D.
 * Output  : The capacity.
 * Approach: Binary search on the answer.
 * Time    : O(n log(sum))
 * Space   : O(1)
 * Java    : 8 or newer.
 */
import java.util.Scanner;

public class Main {
        // Count the days needed when the ship can carry "capacity" per day.
        private static int daysNeeded(int[] a, long capacity) {
            int days = 1;
            long current = 0;
            for (int weight : a) {
                if (current + weight > capacity) {
                    days++;
                    current = 0;
                }
                current += weight;
            }
            return days;
        }

    // This method solves the task. It receives the input and returns the answer.
    static long solve(int[] a, int k) {
        // Step 1: The capacity is at least the heaviest package and at most the total weight.
        long low = 0;
        long high = 0;
        for (int weight : a) {
            low = Math.max(low, weight);
            high += weight;
        }
        // Step 2: Binary search on the answer. A bigger capacity never needs more days, so the test is monotone.
        while (low < high) {
            long middle = low + (high - low) / 2;
            if (daysNeeded(a, middle) <= k) {
                high = middle;
            } else {
                low = middle + 1;
            }
        }
        return low;
    }

    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));
    }
}