Find the element that appears once in a sorted array

Find.java

// Divide-and-conquer
public class Find {

    public static void main(String [] args) {
//        int [] arr = {1, 1, 3, 3, 4, 5, 5, 7, 7, 8, 8};
        int [] arr = {1, 1, 3, 3, 4, 4, 5, 5, 7, 7, 8}; 

        int low = 0;
        int high = arr.length;

        if (arr.length < 3 || arr.length % 2 == 0) {
            System.err.println("Input error: array must have more than 2 elements and odd numbers of elements.");
            return;
        }

        int mid = 0;
        while (low < high) {
            mid = (high - low) / 2 + low;
            System.out.println("log: " + "low = " + low + " mid = " + mid + " high = " + high);
 
            if (mid == low) break;
    
            if ( arr[mid] == arr[mid + 1] ) { 
                if ( mid % 2 == 0 ) 
                    low = mid + 2;
                else
                    high = mid;
            }
            else if ( arr[mid] == arr[mid - 1] ) { 
                if ( mid % 2 == 0 ) 
                    high = mid - 1;
                else
                    low = mid + 1;
            }
        }

        System.out.println("arr[" + low + "] = " + arr[low]);

        return;
    }   
}

Problem.md

Given a sorted array in which all elements appear twice (one after one) and one element appears only once. Find that element in O(log n) complexity.

Example:

Input:   arr[] = {1, 1, 3, 3, 4, 5, 5, 7, 7, 8, 8}
Output:  4

Input:   arr[] = {1, 1, 3, 3, 4, 4, 5, 5, 7, 7, 8}
Output:  8