# Solving Leetcode Interviews in Seconds with AI: Sum of All Subset XOR Totals


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1863" 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
	> The XOR total of an array is defined as the bitwise XOR of all its elements, or 0 if the array is empty.  For example, the XOR total of the array [2,5,6] is 2 XOR 5 XOR 6 = 1.  Given an array nums, return the sum of all XOR totals for every subset of nums.  Note: Subsets with the same elements should be counted multiple times. An array a is a subset of an array b if a can be obtained from b by deleting some (possibly zero) elements of b.   Example 1:  Input: nums = [1,3] Output: 6 Explanation: The 4 subsets of [1,3] are: - The empty subset has an XOR total of 0. - [1] has an XOR total of 1. - [3] has an XOR total of 3. - [1,3] has an XOR total of 1 XOR 3 = 2. 0 + 1 + 3 + 2 = 6  Example 2:  Input: nums = [5,1,6] Output: 28 Explanation: The 8 subsets of [5,1,6] are: - The empty subset has an XOR total of 0. - [5] has an XOR total of 5. - [1] has an XOR total of 1. - [6] has an XOR total of 6. - [5,1] has an XOR total of 5 XOR 1 = 4. - [5,6] has an XOR total of 5 XOR 6 = 3. - [1,6] has an XOR total of 1 XOR 6 = 7. - [5,1,6] has an XOR total of 5 XOR 1 XOR 6 = 2. 0 + 5 + 1 + 6 + 4 + 3 + 7 + 2 = 28  Example 3:  Input: nums = [3,4,5,6,7,8] Output: 480 Explanation: The sum of all XOR totals for every subset is 480.    Constraints:  1 <= nums.length <= 12 1 <= nums[i] <= 20  

	# Explanation
	Here's a breakdown of the solution:

*   **Subset Generation:** The core idea is to iterate through all possible subsets of the input array `nums`. Since each element can either be present or absent in a subset, there are 2<sup>n</sup> subsets for an array of size n.
*   **XOR Calculation:** For each subset, compute the XOR total by XORing all its elements. If the subset is empty, the XOR total is 0.
*   **Summation:** Accumulate the XOR totals of all subsets to get the final result.

*   **Time and Space Complexity:** The time complexity is O(n * 2<sup>n</sup>) because we iterate through all 2<sup>n</sup> subsets and, in the worst case, XOR up to n elements for each subset. The space complexity is O(1) as we use a constant amount of extra space.

	
	# Code
	```python
	def subsetXORSum(nums):
    """
    Calculates the sum of XOR totals for every subset of nums.

    Args:
        nums: A list of integers.

    Returns:
        The sum of all XOR totals for every subset of nums.
    """

    n = len(nums)
    total_xor_sum = 0

    for i in range(2**n):
        current_xor_sum = 0
        for j in range(n):
            if (i >> j) & 1:
                current_xor_sum ^= nums[j]
        total_xor_sum += current_xor_sum

    return total_xor_sum
	```