# Leetcode – Sum of All Subset XOR Totals Solution

The “Sum of All Subset XOR Totals” problem on Leetcode offers an opportunity to delve deep into bitwise operations, combinatorial algorithms, and optimization techniques. While the problem may appear to be simple at first glance, solving it efficiently provides a gateway to understanding essential computer science concepts like bitwise XOR and subset generation. In this article, we’ll explore multiple approaches to solving this problem, with a focus on their time and space complexities.

## Problem Statement

You are given an array nums. We are allowed to select any subset of nums (including an empty set). Then we take the bitwise XOR of all the elements in the subset. Our objective is to find the sum of all possible results for such subsets.

### Example

Input: nums = [1, 3]

Output: 6

Explanation: The subsets are , , [1,3]. The XOR totals are 1 XOR 1 = 1, 3 XOR 3 = 3, 1 XOR 3 = 2. Thus, the sum is 1 + 3 + 2 = 6.

## Approach 1: Brute-Force Subset Generation and XOR

The most straightforward way to solve this problem is to generate all possible subsets of nums and then calculate the XOR of each subset. Finally, we can sum up all these XOR totals.

Python Code:

from itertools import combinations

def subsetXORSum(nums):
total_sum = 0
n = len(nums)

for r in range(1, n+1):
for subset in combinations(nums, r):
xor_total = 0
for num in subset:
xor_total ^= num
total_sum += xor_total

return total_sum

### Time Complexity

The time complexity is O(2^n×n) due to the generation of all subsets and the subsequent XOR calculations.

### Space Complexity

The space complexity is O(n) due to the use of combinations.

## Approach 2: Backtracking

We can generate all subsets using a backtracking algorithm and calculate the XOR total for each subset along the way.

Python Code:

def subsetXORSum(nums):
total_sum = 0

def backtrack(start=0, xor_total=0):
nonlocal total_sum
total_sum += xor_total
for i in range(start, len(nums)):
backtrack(i+1, xor_total ^ nums[i])

backtrack()
return total_sum

### Time Complexity

The time complexity remains O(2^n×n) but tends to be faster in practice due to the avoidance of generating intermediate lists.

### Space Complexity

The space complexity is O(n) due to the recursion stack.

## Approach 3: Bit Manipulation

A more efficient approach involves using bit manipulation to generate subsets and calculate their XOR totals.

Python Code:

def subsetXORSum(nums):
total_sum = 0
n = len(nums)

# Total number of subsets
num_subsets = 1 << n

for i in range(num_subsets):
xor_total = 0
for j in range(n):
if i & (1 << j):
xor_total ^= nums[j]
total_sum += xor_total

return total_sum

### Time Complexity

The time complexity is O(2^n×n), similar to previous methods but generally faster due to bit operations.

### Space Complexity

The space complexity is O(1) as we only use constant extra space.

## Unit Testing

Unit tests are crucial for validating the correctness of the solution.

def test_subsetXORSum():
assert subsetXORSum([1, 3]) == 6
assert subsetXORSum([5, 1, 6]) == 28
assert subsetXORSum([3, 4, 5, 6, 7, 8]) == 480
print("All test cases pass")

test_subsetXORSum()

## Conclusion

The “Sum of All Subset XOR Totals” problem offers a unique blend of combinatorial algorithms and bitwise operations. While the brute-force approach is easy to implement, the backtracking and bit manipulation methods offer more optimized solutions.