Best Pair Sum Below k

/*
 * Problem : Best pair below k.
 * Input   : n, the array, then k.
 * Output  : The sum or -1.
 * Approach: Sort and two pointers.
 * Time    : O(n log 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 int solve(int[] a, int k) {
        // Step 1: Sort a copy. Then two pointers can walk from both ends.
        int[] sorted = a.clone();
        java.util.Arrays.sort(sorted);
        int left = 0;
        int right = sorted.length - 1;
        int best = -1;
        while (left < right) {
            long sum = (long) sorted[left] + sorted[right];
            // Step 2: A sum below k is a candidate. Move left to try a bigger sum. Otherwise move right to get a smaller sum.
            if (sum < k) {
                best = (int) Math.max(best, sum);
                left++;
            } else {
                right--;
            }
        }
        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();
        }
        int k = scanner.nextInt();
        System.out.println(solve(a, k));
    }
}