# 2317. Maximum XOR After Operations

## Description

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 <= 105
• 0 <= nums[i] <= 108

## Solutions

• class Solution {
public int maximumXOR(int[] nums) {
int ans = 0;
for (int x : nums) {
ans |= x;
}
return ans;
}
}

• class Solution {
public:
int maximumXOR(vector<int>& nums) {
int ans = 0;
for (int& x : nums) {
ans |= x;
}
return ans;
}
};

• class Solution:
def maximumXOR(self, nums: List[int]) -> int:
return reduce(or_, nums)


• func maximumXOR(nums []int) (ans int) {
for _, x := range nums {
ans |= x
}
return
}

• function maximumXOR(nums: number[]): number {
let ans = 0;
for (const x of nums) {
ans |= x;
}
return ans;
}