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.

Examples

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.

Solution

Method 1 - Using backtracking

This problem is very similar to subsets problem, where we generate subset of all elements in array, just that we xor the elements. We may exclude or include the element when we are doing dfs on nums:

Here is the video explanation of same:

Code

Java
public int subsetXORSum(int[] nums) {
	return dfs(nums, 0, 0);
}

private int dfs(int[] nums, int idx, int currentXor) {
	if (idx >= nums.length) {
		return currentXor;
	}
	
	int withElement = dfs(nums, idx + 1, currentXor ^ nums[idx]);
	
	int withoutElement = dfs(nums, idx + 1, currentXor);
	
	return withElement + withoutElement;
}

Complexity

  • ⏰ Time complexity: O(2^n)