Question

2317. Maximum XOR After Operations

中文文档

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

Solution 1

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

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;
  }
};

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;
}