Minimum in a Rotated Sorted Array

/*
 * Problem : Minimum of a rotated array.
 * Input   : n, then n integers.
 * Output  : The smallest.
 * Approach: Binary search.
 * 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 long solve(int[] a) {
        // Step 1: Compare the middle with the last number. If middle is bigger, the smallest number is to the right of middle.
        int low = 0;
        int high = a.length - 1;
        while (low < high) {
            int middle = low + (high - low) / 2;
            if (a[middle] > a[high]) {
                low = middle + 1;
            } else {
                // Step 2: Otherwise the right half is sorted, so the smallest number is middle or to its left.
                high = middle;
            }
        }
        return a[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();
        }
        System.out.println(solve(a));
    }
}