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Formatted question description: https://leetcode.ca/all/2527.html

# 2527. Find Xor-Beauty of Array

## Description

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 of val1 and val2.
• val1 & val2 is bitwise AND of val1 and val2.

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

## Solutions

• 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;
}
};

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


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

• function xorBeauty(nums: number[]): number {
return nums.reduce((acc, cur) => acc ^ cur, 0);
}