# 1863. Sum of All Subset XOR Totals

## Description

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

## Solutions

Solution 1: Binary Enumeration

We can use binary enumeration to enumerate all subsets, and then calculate the XOR sum of each subset.

Specifically, we enumerate $i$ in the range $[0, 2^n)$, where $n$ is the length of the array $nums$. If the $j$th bit of the binary representation of $i$ is $1$, it means that the $j$th element of $nums$ is in the current subset; if the $j$th bit is $0$, it means that the $j$th element of $nums$ is not in the current subset. We can get the XOR sum of the current subset according to the binary representation of $i$, and add it to the answer.

The time complexity is $O(n \times 2^n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.

Solution 2: DFS (Depth-First Search)

We can also use depth-first search to enumerate all subsets, and then calculate the XOR sum of each subset.

We design a function $dfs(i, s)$, where $i$ represents the current search to the $i$th element of the array $nums$, and $s$ represents the XOR sum of the current subset. Initially, $i=0$, $s=0$. During the search, we have two choices each time:

• Add the $i$th element of $nums$ to the current subset, i.e., $dfs(i+1, s \oplus nums[i])$;
• Do not add the $i$th element of $nums$ to the current subset, i.e., $dfs(i+1, s)$.

When we have searched all elements of the array $nums$, i.e., $i=n$, the XOR sum of the current subset is $s$, and we can add it to the answer.

The time complexity is $O(2^n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.

• class Solution {
public int subsetXORSum(int[] nums) {
int n = nums.length;
int ans = 0;
for (int i = 0; i < 1 << n; ++i) {
int s = 0;
for (int j = 0; j < n; ++j) {
if ((i >> j & 1) == 1) {
s ^= nums[j];
}
}
ans += s;
}
return ans;
}
}

• class Solution {
public:
int subsetXORSum(vector<int>& nums) {
int n = nums.size();
int ans = 0;
for (int i = 0; i < 1 << n; ++i) {
int s = 0;
for (int j = 0; j < n; ++j) {
if (i >> j & 1) {
s ^= nums[j];
}
}
ans += s;
}
return ans;
}
};

• class Solution:
def subsetXORSum(self, nums: List[int]) -> int:
ans, n = 0, len(nums)
for i in range(1 << n):
s = 0
for j in range(n):
if i >> j & 1:
s ^= nums[j]
ans += s
return ans


• func subsetXORSum(nums []int) (ans int) {
n := len(nums)
for i := 0; i < 1<<n; i++ {
s := 0
for j, x := range nums {
if i>>j&1 == 1 {
s ^= x
}
}
ans += s
}
return
}

• function subsetXORSum(nums: number[]): number {
let ans = 0;
const n = nums.length;
for (let i = 0; i < 1 << n; ++i) {
let s = 0;
for (let j = 0; j < n; ++j) {
if ((i >> j) & 1) {
s ^= nums[j];
}
}
ans += s;
}
return ans;
}


• /**
* @param {number[]} nums
* @return {number}
*/
var subsetXORSum = function (nums) {
let ans = 0;
const n = nums.length;
for (let i = 0; i < 1 << n; ++i) {
let s = 0;
for (let j = 0; j < n; ++j) {
if ((i >> j) & 1) {
s ^= nums[j];
}
}
ans += s;
}
return ans;
};