xor of first and last element of subarray should be equal to xor of remaining elements in the subarray

this means the subarray's xor=0

Ig this problem came in one of CodeChef contest

Let's consider the prefix xor array $$$p[i]$$$, then the condition can be rewritten as $$$a[l] \oplus a[r] = p[l] \oplus p[r - 1]$$$. You can rewrite the condition again as $$$a[l] \oplus p[l] = a[r] \oplus p[r - 1]$$$ and then all you have to do is iterate from the suffix, maintain a counter $$$cnt$$$ for $$$a[r] \oplus p[r - 1]$$$ and add $$$cnt[a[i] \oplus p[i]]$$$ to the answer.

Here's an $$$O(N)$$$ solution:

We want to count subarrays such that:

$$$a[l] \oplus a[r] = a[l+1] \oplus \dots \oplus a[r-1]$$$

$$$0 = a[l] \oplus \dots \oplus a[r]$$$

$$$a[1] \oplus \dots \oplus a[l-1] = a[1] \oplus \dots a[r]$$$

Now, for a fixed $$$r$$$, we just need to count the number of $$$l$$$ such that the equation is true. We can do so in $$$O(1)$$$ via prefix sums stored in a hashmap.

Code:

from collections import defaultdict

n = int(input())
a = [None] + list(map(int, input().split()))

ans = 0

l_pxor = defaultdict(int)
pxor = [0] * (n+1)
for i in range(1, n+1):
    if i >= 3:
        # We can now consider subarray with l==i-2
        l_pxor[pxor[i-3]] += 1

    pxor[i] = a[i] ^ pxor[i-1]

    ans += l_pxor[pxor[i]]

print(ans)

I guess you can just count the number of subarrays with 0 xor, by using prefix_xor and some combinatorics