Sum of All Subset XOR Totals - LeetCode solutions

Problem

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

Solution (Java)

class Solution {
    public int subsetXORSum(int[] nums) {
        if (nums.length == 0) {
            return 0;
        }
        return subsetXORSum(nums, 0, 0);
    }

    private int subsetXORSum(int[] nums, int currIndex, int res) {
        if (currIndex == nums.length) {
            return res;
        }
        int sum1 = subsetXORSum(nums, currIndex + 1, nums[currIndex] ^ res);
        int sum2 = subsetXORSum(nums, currIndex + 1, res);
        return sum1 + sum2;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).