Leetcode #1863: Sum of All Subset XOR Totals
In this guide, we solve Leetcode #1863 Sum of All Subset XOR Totals in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Bit Manipulation, Array, Math, Backtracking, Combinatorics, Enumeration
Intuition
We must explore combinations of choices, but many branches can be pruned early.
Backtracking enumerates valid candidates while keeping the search space under control.
Approach
Use DFS to build candidates step by step, and backtrack when constraints are violated.
Pruning keeps the exploration practical for typical constraints.
Steps:
- Define the decision tree.
- DFS through choices and backtrack.
- Prune invalid paths early.
Example
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
Python Solution
class Solution:
def subsetXORSum(self, nums: List[int]) -> int:
ans, n = 0, len(nums)
for i in range(1 << n):
s = 0
for j in range(n):
if i >> j & 1:
s ^= nums[j]
ans += s
return ans
Complexity
The time complexity is , where is the length of the array . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.