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XOR Operation in an Array

Difficulty: Easy

Problem Description

You are given an integer n and an integer start. Define an array nums where nums[i] = start + 2 * i (0-indexed) and n == nums.length. Return the bitwise XOR of all elements of nums.

Key Insights

  • The array elements are generated using a simple formula based on the start value and the index.
  • The XOR operation has properties that can simplify calculations, such as x ^ x = 0 and x ^ 0 = x.
  • The XOR operation is commutative and associative, meaning the order of operations does not affect the result.

Space and Time Complexity

Time Complexity: O(n) - We need to iterate through the array to compute the XOR. Space Complexity: O(1) - We only use a constant amount of extra space for the result.

Solution

To solve the problem, we can directly compute the XOR of the elements in the array without explicitly creating the array. We initialize a variable to hold the XOR result and iterate through the indices, applying the formula to calculate each element on-the-fly. This approach utilizes the properties of XOR to compute the final result efficiently.

Code Solutions

def xorOperation(n: int, start: int) -> int:
    result = 0
    for i in range(n):
        result ^= (start + 2 * i)  # Calculate the value at index i and perform XOR
    return result
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