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1829. Maximum XOR for Each Query

Difficulty Topics

Description

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:

  1. 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.
  2. 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.

Solution

maximum-xor-for-each-query.py
class Solution:
    def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
        n = len(nums)
        m = (2 ** maximumBit) - 1
        res = []

        for i in range(1, n):
            nums[i] ^= nums[i - 1]

        for num in nums[::-1]:
            res.append(num ^ m)

        return res