Count Occurrences in a Sorted Array

/*
 * Problem : Count a number.
 * Input   : n, the array, then k.
 * Output  : The count.
 * Approach: Two lower bounds.
 * Time    : O(log n)
 * Space   : O(1)
 * Java    : 8 or newer.
 */
import java.util.Scanner;

public class Main {
        // Index of the first number that is not smaller than target (the lower bound).
        private static int lowerBound(int[] a, int target) {
            int low = 0;
            int high = a.length;
            while (low < high) {
                int middle = low + (high - low) / 2;
                if (a[middle] < target) {
                    low = middle + 1;
                } else {
                    high = middle;
                }
            }
            return low;
        }

    // This method solves the task. It receives the input and returns the answer.
    static int solve(int[] a, int k) {
        // Step 1: The first place of k is the lower bound of k. The first place after all k is the lower bound of k + 1.
        // Step 2: The count is the distance between the two places.
        return lowerBound(a, k + 1) - lowerBound(a, k);
    }

    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));
    }
}