# Solving Leetcode Interviews in Seconds with AI: Count Triplets That Can Form Two Arrays of Equal XOR


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1442" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's the breakdown of the solution:

*   **Prefix XOR:** Calculate the prefix XOR array to efficiently compute the XOR of subarrays. `prefix[i]` stores the XOR of `arr[0] ^ arr[1] ^ ... ^ arr[i-1]`. This allows us to compute a and b in O(1) time.
*   **Iterate and Count:** Iterate through all possible `j` values. For each `j`, find the number of `i` and `k` values that satisfy the condition `a == b`.  Using the prefix XOR array, the condition `a == b` is equivalent to `prefix[j] ^ prefix[i] == prefix[k+1] ^ prefix[j]`.  This simplifies to `prefix[i] == prefix[k+1]`. Count pairs of `(i, k)` that satisfy this condition.
*   **Optimization:** Using prefix XOR dramatically reduces the time complexity by avoiding repeated XOR calculations within loops.

*   **Runtime Complexity:** O(n^2).
*   **Storage Complexity:** O(n).

	
	# Code
	```python
	def count_triplets(arr):
    n = len(arr)
    prefix = [0] * (n + 1)
    for i in range(n):
        prefix[i + 1] = prefix[i] ^ arr[i]

    count = 0
    for j in range(1, n + 1):
        for i in range(j):
            for k in range(j, n + 1):
                if prefix[i] == prefix[k]:
                    count += 1
    return count
	```