The XOR total of an array is defined as the bitwise XOR of all its elements, or 0 if the array is empty.
[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 <= 121 <= nums[i] <= 20class Solution {
public int subsetXORSum(int[] nums) {
return solve(nums, 0, 0);
}
private int solve(int[] nums, int i, int xor) {
if (i >= nums.length)
return xor;
return solve(nums, i + 1, xor ^ nums[i]) + solve(nums, i + 1, xor);
}
}