Count Triplets That Can Form Two Arrays of Equal XOR LeetCode solution in Python

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class Solution:
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    def countTriplets(self, arr: List[int]) -> int:
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        s = [0]
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        n = len(arr)
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        if n <= 1:
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            return 0
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        for i in range(n):
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            s.append(s[-1] ^ arr[i])
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        # a = s[i] ^ s[j], b = s[j] ^ s[k+1]
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        count = defaultdict(int)
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        # a = b <-> a ^ b == 0 <-> (s[i] ^ s[j]) ^ (s[j] ^ s[k+1]) == 0
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        # <-> s[i] ^ (s[j] ^ m ) = 0 (where m = s[j] ^ s[k+1])
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        # <-> s[i] ^ s[k+1] == 0 <-> s[i] == s[k+1]
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        res = 0
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        # len(s) == n+1, 0<=i<=n-2, 1<=k<=n-1, i+1<=j<=k
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        for i in range(n - 1):
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            for k in range(i + 1, n):
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                if s[i] == s[k + 1]:
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                    res += k - i
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        return res