# [LeetCode]#1863. Sum of All Subset XOR Totals

**Environment: Python 3.8**

**Key technique: |, pow**

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

**Analysis:**

- Observe example 2, we can see all sunsets are 1, 2, 3, 4, 5, 6, 7.
- We can covert it by formula (all bits) x 2^(n-1). n is list number.

**Solution:**

`class Solution(object):`

def subsetXORSum(self, nums):

n=len(nums)

bits=0

for i in range(n):

bits |=nums[i]

ans=bits * math.pow(2,n-1)

return int(ans)

**Submissions:**

**Reference:**

https://leetcode.com/problems/sum-of-all-subset-xor-totals/discuss/1333272/Using-OR-operator