Maximum XOR for Each Query Python Solution

lcpy gen -n 1829   # by problem number
lcpy gen -s maximum_xor_for_each_query   # by problem name

Input: nums = [0,1,1,3], maximumBit = 2
Output: [0,3,2,3]
Explanation: The queries are answered as follows:
1st query: nums = [0,1,1,3], k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3.
2nd query: nums = [0,1,1], k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3.
3rd query: nums = [0,1], k = 2 since 0 XOR 1 XOR 2 = 3.
4th query: nums = [0], k = 3 since 0 XOR 3 = 3.

Input: nums = [2,3,4,7], maximumBit = 3
Output: [5,2,6,5]
Explanation: The queries are answered as follows:
1st query: nums = [2,3,4,7], k = 5 since 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7.
2nd query: nums = [2,3,4], k = 2 since 2 XOR 3 XOR 4 XOR 2 = 7.
3rd query: nums = [2,3], k = 6 since 2 XOR 3 XOR 6 = 7.
4th query: nums = [2], k = 5 since 2 XOR 5 = 7.

Input: nums = [0,1,2,2,5,7], maximumBit = 3
Output: [4,3,6,4,6,7]

class Solution:
    # Time: O(n)
    # Space: O(1) excluding the output array
    def get_maximum_xor(self, nums: list[int], maximum_bit: int) -> list[int]:
        mask = (1 << maximum_bit) - 1
        total = 0
        for num in nums:
            total ^= num

        answer: list[int] = []
        for num in reversed(nums):
            answer.append(total ^ mask)
            total ^= num
        return answer