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

'''
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
'''



class Solution:
    def countTriplets(self, arr: List[int]) -> int:
        n = len(arr)
        prefix = [0] * (n + 1)
        
        for i in range(n):
            prefix[i + 1] = prefix[i] ^ arr[i]
        
        count = 0
        for i in range(n):
            for k in range(i + 1, n):
                if prefix[i] == prefix[k + 1]:
                    count += (k - i)
        
        return count