# 1829. Maximum XOR for Each Query
"""
you are given a sorted array 'nums' of 'n' non negative integers and an
integer 'maximumBit'. you want to perform the following query 'n' times: find
a non-negative integer 'k < 2^maximumBit' such that 'nums[0] ^ nums[1] ^ ...
^ nums[n]' is maximised. 'k' is the answer to the i'th query. remove the last
element from the current array 'nums'. return an array 'answer' where
'answer[i]' is the answer to the i'th query.
"""
class Solution(object):
def getMaximumXor(self, nums, maximumBit):
"""
:type nums: List[int]
:type maximumBit: int
:rtype: List[int]
"""
max_xor, arr_xor, ans = (1 << maximumBit) - 1, reduce(ixor, nums), []
for i in range(len(nums) - 1, -1, -1):
ans.append(arr_xor ^ max_xor)
arr_xor ^= nums[i]
return ans
if __name__ == "__main__":
obj = Solution()
print(obj.getMaximumXor(nums=[0, 1, 1, 3], maximumBit=2))
print(obj.getMaximumXor(nums=[2, 3, 4, 7], maximumBit=3))
print(obj.getMaximumXor(nums=[0, 1, 2, 2, 5, 7], maximumBit=3))
