Description
https://leetcode.com/problems/sum-of-all-subset-xor-totals/
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]
is2 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
Explanation
Find all subsets. Then get xor total of all subsets.
Python Solution
class Solution:
def subsetXORSum(self, nums: List[int]) -> int:
subsets = []
self.find_subsets(nums, 0, [], subsets)
results = 0
for subset in subsets:
if not subset:
results += 0
else:
xor_total = reduce(lambda x, y: x ^ y, subset)
results += xor_total
return results
def find_subsets(self, nums, start, combination, results):
results.append(list(combination))
for i in range(start, len(nums)):
num = nums[i]
combination.append(num)
self.find_subsets(nums, i + 1, combination, results)
combination.pop()
- Time Complexity: O(N).
- Space Complexity: O(N).