1442. Count Triplets That Can Form Two Arrays of Equal XOR

#### QUESTION:

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] <= 108
``````

#### SOLUTION:

``````class Solution {
fun countTriplets(arr: IntArray): Int {
var  result:Int = 0
for (i in 0 until arr.size - 1) {
var tmpA:Int = arr[i]
for (j in i+1 until arr.size) {
tmpA = tmpA xor arr[j]
var tmpB:Int = arr[j]
for (k in j until arr.size) {
tmpB = tmpB xor arr[k]
if (tmpA == tmpB) {
result += 1
}
}
}
}
return result
}
}
``````