#1486

Easy

LeetCode Problem Workspace

XOR Operation in an Array

Compute the bitwise XOR of a dynamically generated array using a combination of math and bit manipulation techniques efficiently.

math bit manipulation

Problem Statement

Given two integers n and start, define an array nums of length n where nums[i] = start + 2 * i for 0 <= i < n. Return the bitwise XOR of all elements in nums. Use a combination of arithmetic computation and bitwise operations to solve efficiently.

For example, if n = 5 and start = 0, nums would be [0,2,4,6,8], and the XOR of these elements is 8. Constraints include 1 <= n <= 1000 and 0 <= start <= 1000, emphasizing math plus bit manipulation for an optimal solution.

Examples

Example 1

Input: n = 5, start = 0

Output: 8

Array nums is equal to [0, 2, 4, 6, 8] where (0 ^ 2 ^ 4 ^ 6 ^ 8) = 8. Where "^" corresponds to bitwise XOR operator.

Example 2

Input: n = 4, start = 3

Output: 8

Array nums is equal to [3, 5, 7, 9] where (3 ^ 5 ^ 7 ^ 9) = 8.

Constraints

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

Solution Approach

Simulate and XOR Directly

Generate the nums array by iterating from 0 to n-1, compute each element as start + 2*i, and XOR each element into a result variable. This approach ensures correctness but runs in O(n) time and O(1) extra space.

Use XOR Properties with Patterns

Notice that XOR has patterns for consecutive even numbers. Use the property that XOR of a sequence can be reduced using XOR from 0 to a number. Transform nums[i] = start + 2*i into a pattern that allows computing XOR without storing the full array, reducing memory usage.

Optimized Math-Based XOR Computation

Derive a formula by separating start into even/odd and leveraging XOR periodicity. Compute XOR over a sequence of integers adjusted by start/step to get the final XOR in constant time. This fully eliminates iteration, ideal for larger n within constraints.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The naive approach is O(n) time and O(1) space. Using XOR properties and patterns, computation can be reduced to O(1) time and O(1) space by exploiting arithmetic sequence structures and bitwise periodicity.

What Interviewers Usually Probe

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

Common Pitfalls or Variants

Common pitfalls

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

Follow-up variants

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

FAQ

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.

terminal

Solution

Solution 1: Simulation

We can directly simulate to calculate the XOR result of all elements in the array.

1

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3

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

Continue Practicing

#1442 count triplets that can form two arrays of equal xor #1342 number of steps to reduce a number to zero #1680 concatenation of consecutive binary numbers