1940 Maximum Xor For Each Query

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 < 2maximumBit such that nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k is **maximized**. k is the answer to the ith query.
Remove the **last **element from the current array nums.

Return an array answer, where *answer[i] is the answer to the ith query*.

 

Example 1:

**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.

Example 2:

**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.

Example 3:

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

 

Constraints:

nums.length == n
1 <= n <= 105
1 <= maximumBit <= 20
0 <= nums[i] < 2maximumBit
nums​​​ is sorted in **ascending** order.

class Solution:
    def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
        largestnum = 2 ** maximumBit - 1
        res = []


        a =[nums[0]]

        for idx, k in enumerate(nums):
            if idx == 0:
                continue
            a.append(a[-1] ^ k)


        for l in a:
            res.append(largestnum - l)
        res.reverse()
        return res