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Given an array of integers arr.

We want to select three indices ij 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 (ij 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

## Solution 1: Brute Force (TLE)

Time complexity: O(n^4)
Space complexity: O(1)

## Solution 2: Prefix XORs

Let xors[i] = arr[0] ^ arr[1] ^ … ^ arr[i-1]
arr[i] ^ arr[i + 1] ^ … ^ arr[j – 1] = (arr[0] ^ … ^ arr[j – 1]) ^ (arr[0] ^ … ^ arr[i-1]) = xors[j] ^ xors[i]

We then can compute a and b in O(1) time.

Time complexity: O(n^3)
Space complexity: O(n)

## Solution 3: Prefix XORs II

a = arr[i] ^ arr[i + 1] ^ … ^ arr[j – 1]
b = arr[j] ^ arr[j + 1] ^ … ^ arr[k]
a == b => a ^ b == 0
XORs(i ~ k) == 0
XORS(0 ~ k) ^ XORs(0 ~ i – 1) = 0

Problem => find all pairs of (i – 1, k) such that xors[k+1] == xors[i]
For each pair (i – 1, k), there are k – i positions we can insert j.

Time complexity: O(n^2)
Space complexity: O(1)

## Solution 3: HashTable

Similar to target sum, use a hashtable to store the frequency of each prefix xors.

Time complexity: O(n)
Space complexity: O(n)

## C++

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