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

Solution 1: Bit Manipulation

We first consider the case where \(i\) and \(j\) are not equal. In this case, ((nums[i] | nums[j]) & nums[k]) and ((nums[j] | nums[i]) & nums[k]) produce the same result, and their XOR result is \(0\).

Therefore, we only need to consider the case where \(i\) and \(j\) are equal. In this case, ((nums[i] | nums[j]) & nums[k]) = (nums[i] & nums[k]). If \(i \neq k\), then this is the same as the result of nums[k] & nums[i], and the XOR result of these values is \(0\).

Therefore, we ultimately only need to consider the case where \(i = j = k\), and the answer is the XOR result of all \(nums[i]\).

The time complexity is \(O(n)\) and the space complexity is \(O(1)\), where \(n\) is the length of the array.

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

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class Solution {
    public int xorBeauty(int[] nums) {
        int ans = 0;
        for (int x : nums) {
            ans ^= x;
        }
        return ans;
    }
}

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class Solution {
public:
    int xorBeauty(vector<int>& nums) {
        int ans = 0;
        for (auto& x : nums) {
            ans ^= x;
        }
        return ans;
    }
};

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

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

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