You are given a 0-indexed integer array nums
.
The effective value of three indices i
, j
, and k
is defined as ((nums[i] | nums[j]) & nums[k])
.
The xor-beauty of the array is the XORing of the effective values of all the possible triplets of indices (i, j, k)
where 0 <= i, j, k < n
.
Return the xor-beauty of nums
.
Note that:
val1 | val2
is bitwise OR ofval1
andval2
.val1 & val2
is bitwise AND ofval1
andval2
.
Example 1:
Input: nums = [1,4] Output: 5 Explanation: The triplets and their corresponding effective values are listed below: - (0,0,0) with effective value ((1 | 1) & 1) = 1 - (0,0,1) with effective value ((1 | 1) & 4) = 0 - (0,1,0) with effective value ((1 | 4) & 1) = 1 - (0,1,1) with effective value ((1 | 4) & 4) = 4 - (1,0,0) with effective value ((4 | 1) & 1) = 1 - (1,0,1) with effective value ((4 | 1) & 4) = 4 - (1,1,0) with effective value ((4 | 4) & 1) = 0 - (1,1,1) with effective value ((4 | 4) & 4) = 4 Xor-beauty of array will be bitwise XOR of all beauties = 1 ^ 0 ^ 1 ^ 4 ^ 1 ^ 4 ^ 0 ^ 4 = 5.
Example 2:
Input: nums = [15,45,20,2,34,35,5,44,32,30]
Output: 34
Explanation: The xor-beauty of the given array is 34.
Constraints:
1 <= nums.length <= 105
1 <= nums[i] <= 109
class Solution:
def xorBeauty(self, nums: List[int]) -> int:
return reduce(xor, nums)
class Solution {
public int xorBeauty(int[] nums) {
int ans = 0;
for (int x : nums) {
ans ^= x;
}
return ans;
}
}
class Solution {
public:
int xorBeauty(vector<int>& nums) {
int ans = 0;
for (auto& x : nums) {
ans ^= x;
}
return ans;
}
};
func xorBeauty(nums []int) (ans int) {
for _, x := range nums {
ans ^= x
}
return
}