What is the main pattern used in XOR Operation in an Array?

The main pattern is combining arithmetic sequence generation with bitwise XOR, leveraging periodicity in XOR to optimize computation.

How do I generate the array nums efficiently?

Compute each element as start + 2*i during iteration or use mathematical properties to calculate XOR without storing the array.

Can this problem be solved in constant space?

Yes, by using XOR properties and pattern analysis, the entire result can be computed without constructing the full array.

Why does the XOR of [0,2,4,6,8] equal 8?

Because XOR is associative and commutative, compute sequentially: 0^2=2, 2^4=6, 6^6=0, 0^8=8, resulting in 8.

What are common mistakes in this XOR array problem?

Forgetting the 2*i increment, attempting sum instead of XOR, or not handling repeated XOR patterns correctly.

#1486

Easy

auto_awesome数学·位运算

LeetCode 题解工作台

数组异或操作

给你两个整数， n 和 start 。数组 nums 定义为： nums[i] = start + 2*i （下标从 0 开始）且 n == nums.length 。请返回 nums 中所有元素按位异或（ XOR ）后得到的结果。示例 1：输入： n = 5, start = 0 输出： …

数学位运算

题目描述

给你两个整数，n 和 start 。

数组 nums 定义为：nums[i] = start + 2*i（下标从 0 开始）且 n == nums.length 。

请返回 nums 中所有元素按位异或（XOR）后得到的结果。

示例 1：

输入：n = 5, start = 0
输出：8
解释：数组 nums 为 [0, 2, 4, 6, 8]，其中 (0 ^ 2 ^ 4 ^ 6 ^ 8) = 8 。
     "^" 为按位异或 XOR 运算符。

示例 2：

输入：n = 4, start = 3
输出：8
解释：数组 nums 为 [3, 5, 7, 9]，其中 (3 ^ 5 ^ 7 ^ 9) = 8.

示例 3：

输入：n = 1, start = 7
输出：7

示例 4：

输入：n = 10, start = 5
输出：2

提示：

1 <= n <= 1000
0 <= start <= 1000
n == nums.length

lightbulb

解题思路

方法一：模拟

我们可以直接模拟算出数组中所有元素的异或结果。

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

1

2

3

4

class Solution:
    def xorOperation(self, n: int, start: int) -> int:
        return reduce(xor, ((start + 2 * i) for i in range(n)))

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
Looking for correct use of XOR on array elements derived from arithmetic sequence.
question_mark
Expect recognition of bit manipulation patterns that allow avoiding full array construction.
question_mark
Checking for awareness of edge cases when start or n produces repeated XOR patterns.

warning

常见陷阱

外企场景

error
Attempting to sum elements instead of XOR.
error
Forgetting to increment by 2 in the sequence leading to incorrect nums generation.
error
Overcomplicating solution instead of using XOR periodicity for optimization.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Compute XOR for an array with different step sizes instead of 2.
arrow_right_alt
Calculate XOR where nums[i] = start + i^2, blending arithmetic and bit manipulation.
arrow_right_alt
Determine XOR after filtering elements by a condition, e.g., nums[i] % 2 == 0.

help

常见问题

外企场景

继续练习

#1442 形成两个异或相等数组的三元组数目 #1342 将数字变成 0 的操作次数 #1680 连接连续二进制数字