First and Last Position of a Value

/*
 * Problem : Search in an array.
 * Input   : n, then n integers, then k.
 * Output  : Two indexes.
 * Approach: Two binary searches.
 * Time    : O(log 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, int k) {
        // Step 1: Find the first index with a[i] >= target.
        int low = 0;
        int high = a.length;
        while (low < high) {
            int mid = low + (high - low) / 2;
            if (a[mid] < k) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        int first = low;

        // Step 2: Find the first index with a[i] > target. One step before it is the last index of k.
        low = 0;
        high = a.length;
        while (low < high) {
            int mid = low + (high - low) / 2;
            if (a[mid] <= k) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        int last = low - 1;

        // Step 3: If k does not exist, first is bigger than last.
        return first > last ? "-1 -1" : first + " " + last;
    }

    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));
    }
}