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

Problem Link

Description

---

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

---

Python3

class Solution:
    def countTriplets(self, arr: List[int]) -> int:
        N = len(arr)
        prefix = [0] * (N + 1)
        res = 0
 
        # since a == b
        # a ^ a = b ^ a
        # 0 = b ^ a
        # arr[i] ^ arr[i + 1] ^ ... ^ arr[k] == 0
        # prefix[k + 1] == prefix[i]
 
        for i in range(1, N + 1):
            prefix[i] = arr[i - 1] ^ prefix[i - 1]
 
        count = Counter()
        total = Counter()
 
        for i in range(N + 1):
            res += count[prefix[i]] * (i - 1) - total[prefix[i]]
            count[prefix[i]] += 1
            total[prefix[i]] += i
        
        return res