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]
is2 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
class Solution:
def subsetXORSum(self, nums: List[int]) -> int:
def dfs(nums, depth, prev):
self.res += prev
for num in nums[depth:]:
prev ^= num
depth += 1
dfs(nums, depth, prev)
prev ^= num
self.res = 0
dfs(nums, 0, 0)
return self.res
class Solution {
private int res;
public int subsetXORSum(int[] nums) {
dfs(nums, 0, 0);
return res;
}
private void dfs(int[] nums, int depth, int prev) {
res += prev;
for (int i = depth; i < nums.length; ++i) {
prev ^= nums[i];
dfs(nums, ++depth, prev);
prev ^= nums[i];
}
}
}
/**
* @param {number[]} nums
* @return {number}
*/
var subsetXORSum = function (nums) {
let res = [];
let prev = 0;
dfs(nums, 0, prev, res);
return res.reduce((a, c) => a + c, 0);
};
function dfs(nums, depth, prev, res) {
res.push(prev);
for (let i = depth; i < nums.length; i++) {
prev ^= nums[i];
depth++;
dfs(nums, depth, prev, res);
// bracktrack
prev ^= nums[i];
}
}