# Solving Leetcode Interviews in Seconds with AI: XOR Operation in an Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1486" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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.   Example 1:  Input: n = 5, start = 0 Output: 8 Explanation: 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 Explanation: 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  

	# Explanation
	*   **Utilize XOR properties:** Instead of iterating and calculating XORs, derive a formula based on the properties of XOR operations to calculate the result directly.
*   **Mathematical Simplification:** Express the XOR sum in terms of `n` and `start` and simplify the calculation.
*   **XOR patterns:** Exploit repeating XOR patterns to avoid explicit array construction and iteration.

*   **Runtime Complexity:** O(1), **Storage Complexity:** O(1)

	
	# Code
	```python
	def xorOperation(n: int, start: int) -> int:
    """
    Calculates the bitwise XOR of all elements of nums where nums[i] = start + 2 * i.

    Args:
        n: The number of elements in the nums array.
        start: The starting value for the nums array.

    Returns:
        The bitwise XOR of all elements in nums.
    """

    s = start // 2
    result = xor_range(s + n - 1) ^ xor_range(s - 1)
    if n % 2 == 1:
        if start % 4 == 0:
            return result * 2
        else:
            return result * 2 + 1
    else:
        if n % 4 == 0:
            if start % 4 == 0:
                return result * 2
            else:
                return result * 2 + 1
        else:
            if start % 4 == 0:
                return result * 2 + start % 4
            else:
                return result * 2 + (start % 4) ^ 1

def xor_range(x):
    """
    Calculates the XOR sum of numbers from 0 to x.
    """
    mod = x % 4
    if mod == 0:
        return x
    elif mod == 1:
        return 1
    elif mod == 2:
        return x + 1
    else:
        return 0
	```