2317. Maximum XOR After Operations

Problem

You are given a 0-indexed integer array nums. In one operation, select any non-negative integer x and an index i, then update nums[i] to be equal to nums[i] AND (nums[i] XOR x).

Note that AND is the bitwise AND operation and XOR is the bitwise XOR operation.

Return **the *maximum* possible bitwise XOR of all elements of nums after applying the operation any number of times**.

  Example 1:

Input: nums = [3,2,4,6]
Output: 7
Explanation: Apply the operation with x = 4 and i = 3, num[3] = 6 AND (6 XOR 4) = 6 AND 2 = 2.
Now, nums = [3, 2, 4, 2] and the bitwise XOR of all the elements = 3 XOR 2 XOR 4 XOR 2 = 7.
It can be shown that 7 is the maximum possible bitwise XOR.
Note that other operations may be used to achieve a bitwise XOR of 7.

Example 2:

Input: nums = [1,2,3,9,2]
Output: 11
Explanation: Apply the operation zero times.
The bitwise XOR of all the elements = 1 XOR 2 XOR 3 XOR 9 XOR 2 = 11.
It can be shown that 11 is the maximum possible bitwise XOR.

  Constraints:

• 1 <= nums.length <= 10^5

• 0 <= nums[i] <= 10^8

Solution (Java)

class Solution {
    public int maximumXOR(int[] nums) {
        int max = 0;
        for (int n : nums) {
            max |= n;
        }
        return max;
    }
}

Explain:

nope.

Complexity:

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