查询数组异或美丽值

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

/*
 * 思路：
 * 1. 使用 Java Stream API 来实现遍历所有的三元组组合 (i, j, k)。
 * 2. 通过 mapToInt 操作将每个组合的有效值映射为整数，然后使用 reduce 操作计算所有有效值的异或结果。
 */
import java.util.stream.IntStream;

public class Solution {
    public int xorBeauty(int[] nums) {
        int n = nums.length;
        return IntStream.range(0, n)
                .flatMap(i -> IntStream.range(0, n)
                        .flatMap(j -> IntStream.range(0, n)
                                .map(k -> (nums[i] | nums[j]) & nums[k])))
                .reduce(0, (a, b) -> a ^ b);
    }
}

这个题解直接计算了数组 `nums` 中所有元素的异或总和，忽略了题目中的复杂三元组运算条件。题目要求对于数组中的任意三元组 `(i, j, k)` 计算 `((nums[i] | nums[j]) & nums[k])` 的结果，并对所有这些结果进行异或。然而，这个题解仅仅对数组中的每个元素进行了异或运算，没有涉及到按位或和与运算。这表明题解可能误解了题目要求，或者是基于某种观察到的特性进行了简化处理，但没有在题解中提供这样的说明。

O(n)

O(1)