Sum of All Subset XOR Totals

How it works

Each bit position in any number contributes independently to the XOR totals. For any bit that is set in at least one number, exactly half of all subsets will have that bit set in their XOR result (because each element either flips the bit or not, and the combinations balance out). The OR of all numbers gives us all bits that appear in at least one element. Each such bit contributes its value multiplied by 2^(n-1) subsets.

Steps

1. Compute the OR of all elements in nums. This gives a number where each bit is set if and only if that bit is set in at least one element.
2. Left shift the result by n - 1 positions (multiply by 2^(n-1)).
3. Return the result.

Complexity

• Time complexity: O(n)
• Space complexity: O(1) extra space.

Things to watch for

Forgetting the Empty Subset

The problem includes all subsets, including the empty subset which has an XOR value of 0. While this does not affect the sum, a common conceptual error is starting the subset generation from size 1 instead of size 0, or incorrectly counting the total number of subsets as 2^n - 1 instead of 2^n. This can lead to off-by-one issues in verification.

Misunderstanding XOR Properties

A frequent mistake is confusing XOR with addition or other operations. XOR has the property that a ^ a = 0 and a ^ 0 = a. When computing subset XOR totals, some assume that duplicate elements cancel out across all subsets, but each subset is computed independently. Also, the optimal solution relies on understanding that each bit appears in exactly half of all subsets, which requires grasping XOR's bit-level behavior.

from typing import List

class Solution:
    def subsetXORSum(self, nums: List[int]) -> int:
        res = 0
        for num in nums:
            res |= num
        return res << (len(nums) - 1)

class Solution {
public:
    int subsetXORSum(vector<int>& nums) {
        int res = 0;
        for (int& num : nums) {
            res |= num;
        }
        return res << (nums.size() - 1);
    }
};