2317

2317. Maximum XOR After Operations (Medium)

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

Companies: American Express

Related Topics:
Array, Math, Bit Manipulation

Similar Questions:

• Maximum XOR of Two Numbers in an Array (Medium)

Solution 1.

A[i] x   xor and
0    0   0   0
0    1   1   0
1    0   1   1
1    1   0   0

So, we can remove bit 1s but can't make bit 0s to 1s.

As long as any A[i] as 1 at the i-th bit, the i-th bit of the answer must be one.

// OJ: https://leetcode.com/problems/maximum-xor-after-operations
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int maximumXOR(vector<int>& A) {
        int ans = 0;
        for (int n : A) ans |= n;
        return ans;
    }
};