Count Triplets That Can Form Two Arrays of Equal XOR

Problem

Given an array of integers arr.

We want to select three indices i, j and k where (0 <= i < j <= k < arr.length).

Let's define a and b as follows:

a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]
b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]

Note that ^ denotes the bitwise-xor operation.

Return the number of triplets (i, j and k) Where a == b.

Example 1:

Input: arr = [2,3,1,6,7]
Output: 4
Explanation: The triplets are (0,1,2), (0,2,2), (2,3,4) and (2,4,4)

Example 2:

Input: arr = [1,1,1,1,1]
Output: 10

Constraints:

1 <= arr.length <= 300
1 <= arr[i] <= 10^8

Solution (Java)

class Solution {
    public int countTriplets(int[] arr) {
        int count = 0;
        for (int i = 0; i < arr.length; i++) {
            int xor = arr[i];
            for (int k = i + 1; k < arr.length; k++) {
                xor ^= arr[k];
                if (xor == 0) {
                    count += k - i;
                }
            }
        }
        return count;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).