## Description

https://leetcode.com/problems/sum-of-all-subset-xor-totals/

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:6Explanation: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:28Explanation: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:480Explanation:The sum of all XOR totals for every subset is 480.

**Constraints:**

`1 <= nums.length <= 12`

`1 <= nums[i] <= 20`

## Explanation

Find all subsets. Then get xor total of all subsets.

## Python Solution

```
class Solution:
def subsetXORSum(self, nums: List[int]) -> int:
subsets = []
self.find_subsets(nums, 0, [], subsets)
results = 0
for subset in subsets:
if not subset:
results += 0
else:
xor_total = reduce(lambda x, y: x ^ y, subset)
results += xor_total
return results
def find_subsets(self, nums, start, combination, results):
results.append(list(combination))
for i in range(start, len(nums)):
num = nums[i]
combination.append(num)
self.find_subsets(nums, i + 1, combination, results)
combination.pop()
```

- Time Complexity: O(N).
- Space Complexity: O(N).