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.
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.
nums = [1, 3]
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.
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
The time complexity is O(2^n×n) due to the generation of all subsets and the subsequent XOR calculations.
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.
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
The time complexity remains O(2^n×n) but tends to be faster in practice due to the avoidance of generating intermediate lists.
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.
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
The time complexity is O(2^n×n), similar to previous methods but generally faster due to bit operations.
The space complexity is O(1) as we only use constant extra space.
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()
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.