human
I found an interesting problem on https://leetcode.com/problems/count-triplets-that-can-form-two-arrays-of-equal-xor/:
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

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is the Python code you requested.

class Solution:
    def countTriplets(self, arr: List[int]) -> int:
        preXor = [0] * (len(arr) + 1)
        preXor[0] = arr[0]
        for i in range(1, len(arr)):
            preXor[i] = preXor[i-1] ^ arr[i]
        ret = 0
        for i in range(len(arr)-1):
            for j in range(i+1, len(arr)):
                l = preXor[i-1]
                for k in range(j, len(arr)):
                    #print(i, j, k)
                    if l == preXor[k]:
                        #print(\"match:\", i, j, k)
                        ret += 1
        return ret