How do you solve XOR Operation in an Array on LeetCode?

XOR Operation in an Array (LeetCode #1486) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding the properties of XOR can simplify calculations.

What is the brute force solution for XOR Operation in an Array?

The brute force approach for XOR Operation in an Array is Brute Force, with O(n) time complexity and O(n) space complexity. We can create the array `nums` by calculating each element using the formula provided. Then, we can iterate through the array to compute the XOR of all elements. This is straightforward but not the most efficient way. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve XOR Operation in an Array?

XOR Operation in an Array uses the following concepts: Math, Bit Manipulation. The recommended patterns to study are: Bit Manipulation, Array.

What are the interview tips for XOR Operation in an Array?

Always consider if you can optimize space usage. Think about the mathematical properties of operations involved. Practice explaining your thought process clearly during the interview.

What are common mistakes when solving XOR Operation in an Array?

Not realizing that XOR is both commutative and associative. Overcomplicating the problem by creating unnecessary data structures.

#1486

XOR Operation in an Array

Easy

MathBit ManipulationBit ManipulationArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(n)
Space	O(n)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of creating the entire array, we can directly compute the XOR using the properties of the XOR operation. This reduces both time and space complexity.

⚙️

Algorithm

2 steps

1Step 1: Initialize a variable `result` to 0.
2Step 2: Iterate from 0 to n-1, calculating each element directly using the formula and updating `result` with the XOR of each element.

solution.py7 lines

1# Full working Python code
2
3def xorOperation(n, start):
4    result = 0
5    for i in range(n):
6        result ^= (start + 2 * i)
7    return result

ℹ

Complexity note: The time complexity remains O(n) since we still iterate through `n` elements, but the space complexity is reduced to O(1) because we no longer need to store the entire array.

1Understanding the properties of XOR can simplify calculations.
2Direct computation can save time and space.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.