1829. Maximum XOR for Each Query

Problem

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 <= 10^5

• 1 <= maximumBit <= 20

• 0 <= nums[i] < 2maximumBit

• nums​​​ is sorted in ascending order.

Solution (Java)

class Solution {
    public int[] getMaximumXor(int[] nums, int maximumBit) {
        int[] result = new int[nums.length];
        int val = nums[0];
        int target = (1 << maximumBit) - 1;
        for (int i = 1; i < nums.length; i++) {
            val ^= nums[i];
        }
        for (int i = 0; i < result.length; i++) {
            result[i] = target ^ val;
            val ^= nums[nums.length - i - 1];
        }
        return result;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).