What is the xor-beauty of an array?

The xor-beauty of an array is the result of XORing the effective values of all possible triplets of indices in the array, where each triplet's effective value is calculated as ((nums[i] | nums[j]) & nums[k]).

How do I simplify the XOR expression in this problem?

You can simplify the XOR expression by leveraging properties of bitwise operations, focusing on reducing redundant calculations and optimizing the evaluation of each triplet.

What is the time complexity of the brute force solution?

The brute force solution has a time complexity of O(n^3) since it evaluates all possible triplets of indices in the array.

How does GhostInterview help with the bitwise optimization for this problem?

GhostInterview guides you in applying bitwise operations efficiently, helping you identify redundant steps and optimize the solution to avoid unnecessary recalculations.

What is the optimal solution for the Find Xor-Beauty of Array problem?

The optimal solution involves reducing the time complexity by recognizing patterns in bitwise operations, enabling you to calculate the effective values more efficiently than brute force.

#2527

Medium

auto_awesome数组·数学

LeetCode 题解工作台

查询数组异或美丽值

给你一个下标从 0 开始的整数数组 nums 。三个下标 i ， j 和 k 的有效值定义为 ((nums[i] | nums[j]) & nums[k]) 。一个数组的异或美丽值是数组中所有满足 0 的三元组 (i, j, k) 的有效值的异或结果。请你返回 nums 的异或美丽…

数组数学位运算

题目描述

给你一个下标从 0 开始的整数数组 nums 。

三个下标 i ，j 和 k 的 有效值 定义为 ((nums[i] | nums[j]) & nums[k]) 。

一个数组的 异或美丽值 是数组中所有满足 0 <= i, j, k < n 的三元组 (i, j, k) 的 有效值 的异或结果。

请你返回 nums 的异或美丽值。

注意：

val1 | val2 是 val1 和 val2 的按位或。
val1 & val2 是 val1 和 val2 的按位与。

示例 1：

输入：nums = [1,4]
输出：5
解释：
三元组和它们对应的有效值如下：
- (0,0,0) 有效值为 ((1 | 1) & 1) = 1
- (0,0,1) 有效值为 ((1 | 1) & 4) = 0
- (0,1,0) 有效值为 ((1 | 4) & 1) = 1
- (0,1,1) 有效值为 ((1 | 4) & 4) = 4
- (1,0,0) 有效值为 ((4 | 1) & 1) = 1
- (1,0,1) 有效值为 ((4 | 1) & 4) = 4
- (1,1,0) 有效值为 ((4 | 4) & 1) = 0
- (1,1,1) 有效值为 ((4 | 4) & 4) = 4 
数组的异或美丽值为所有有效值的按位异或 1 ^ 0 ^ 1 ^ 4 ^ 1 ^ 4 ^ 0 ^ 4 = 5 。

示例 2：

输入：nums = [15,45,20,2,34,35,5,44,32,30]
输出：34
解释：数组的异或美丽值为 34 。

提示：

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹

lightbulb

解题思路

方法一：位运算

我们首先考虑 $i$ 与 $j$ 不相等的情况，此时 ((nums[i] | nums[j]) & nums[k]) 与 ((nums[j] | nums[i]) & nums[k]) 的结果是相同的，两者的异或结果为 $0$ 。

因此，我们只需要考虑 $i$ 与 $j$ 相等的情况。此时 ((nums[i] | nums[j]) & nums[k]) = (nums[i] & nums[k])，如果 $i \neq k$ ，那么与 nums[k] & nums[i] 的结果是相同的，这些值的异或结果为 $0$ 。

因此，我们最终只需要考虑 $i = j = k$ 的情况，那么答案就是所有 $nums[i]$ 的异或结果。

时间复杂度 $O(n)$ ，空间复杂度 $O(1)$ 。其中 $n$ 为数组的长度。

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

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Check for the candidate's understanding of bitwise operations and their ability to identify optimization opportunities.
question_mark
Evaluate whether the candidate can recognize the trade-off between brute force and optimized solutions.
question_mark
Assess the candidate's ability to explain their approach clearly and in detail.

warning

常见陷阱

外企场景

error
Not simplifying the XOR expression leads to inefficient solutions.
error
Overcomplicating the solution by missing opportunities for bitwise optimization.
error
Neglecting edge cases such as when the array contains only one element.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Consider variants where the XOR expression is changed to use other bitwise operators.
arrow_right_alt
Implement a solution where the array is sorted first to see how it affects performance.
arrow_right_alt
Change the problem to compute a different bitwise combination or condition between indices.

help

常见问题

外企场景

继续练习

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