
## Problem statement

Given an array of numbers with no duplicates, count the number of unique triplets `(x, y, z)` such that the result of their XOR is zero. 

A **unique triplet** is a triplet where all three numbers in the triplet are unique.  

Example 1
- Input: `arr=[7, 29, 32, 54, 90, 2]`
- Output: `0`

Example 2
- Input: `arr=[1, 3, 5, 10, 14, 15]`
- Output: `2`


## Solution

The simplest method is to have three nested loops and check each one to see if the XOR of the three numbers is zero.

## Code
Let's view a code example.

public class Main{

    private static int count(int[] arr){
        int count = 0;
        int n = arr.length;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                for (int k = j + 1; k < n; k++) {
                    if ((arr[i] ^ arr[j] ^ arr[k]) == 0) {
                        count++;
                    }
                }
            }
        }
        return count;
    }

    public static void main(String[] args) {
        int[] arr = {7, 29, 32, 54, 90, 2};
        System.out.println(count(arr));
    }
}

Java


## Explanation

- Lines 3–15: We define a `count()` function that accepts an array and uses three loops to count the number of unique triplets whose XOR is zero.
- Lines 18–21: In the `main()` method, we invoke the `count()` function.

How to find the number of unique triplets whose XOR is zero

Count unique triplets in an array whose XOR is zero using three nested loops for efficiency.