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
Example 3:
Input: arr = [2,3]
Output: 0
Example 4:
Input: arr = [1,3,5,7,9]
Output: 3
Example 5:
Input: arr = [7,11,12,9,5,2,7,17,22]
Output: 8
Constraints:
1 <= arr.length <= 300
1 <= arr[i] <= 10^8
class Solution:
    def countTriplets(self, arr: List[int]) -> int:
        a = []
        n = len(arr)
        for i in range(n):
            if i == 0:
                a.append(arr[i])
            else:
                a.append(arr[i] ^ a[-1])
        
        ans = 0
        for i in range(n):
            for j in range(i+1, n):
                for k in range(j, n):
                    x = a[j-1] ^ (0 if i == 0 else a[i-1])
                    y = a[k] ^ a[j-1]
                    if x == y:
                        ans += 1
        return ans
class Solution:
    def countTriplets(self, arr: List[int]) -> int:
        a = []
        n = len(arr)
        for i in range(n):
            if i == 0:
                a.append(arr[i])
            else:
                a.append(arr[i] ^ a[-1])
        
        ans = 0
        for i in range(n):
            for j in range(i+1, n):
                for k in range(j, n):
                    x = a[j-1] ^ (0 if i == 0 else a[i-1])
                    y = a[k] ^ a[j-1]
                    if x == y:
                        ans += 1
        return ans