Java solution - passes 100% of test cases

From my HackerRank solutions.

Time Complexity: O(n) per testcase

XOR properties:
1) x ^ x = 0
2) x ^ 0 = x

We know that a number XORed with itself is 0. Instead of calculating the subarrays directly, we calculate the number of times each number appears in any subarray. If it appears an even number of times, then (x ^ x ... ^ x) where there is an even number of "x" values will give 0. If there are an odd number of "x" values, the result will be "x".

Case 1: n is even

Each element appears an even number of times throughout the subarrays, so the answer is 0.

Case 2: n is odd

We notice that the odd-indexed elements appear an even number of times throughout the subarrays, so xoring them together will give 0, so we can ignore the odd-indexed elements.

The even-indexed elements in the original subarray will appear an odd number of times throughout the subarrays. We can go ahead and XOR the values of the even-indexed elements in the original array to get our final answer.

import java.util.Scanner;

public class Solution {
    public static void main(String [] args) {
        Scanner scan = new Scanner(System.in);
        int T = scan.nextInt();
        while (T-- > 0) {
            int n = scan.nextInt();
            int [] array = new int[n];
            for (int i = 0; i < n; i++) {
                array[i] = scan.nextInt();
            }
            if (n % 2 == 0) { // Case 1
                System.out.println(0);
            } else { // Case 2
                int result = 0;
                for (int i = 0; i < n; i++) {
                    if (i % 2 == 0) {
                        result ^= array[i];
                    }
                }
                System.out.println(result);
            }
        }
        scan.close();
    }
}