LeetCode 1863. Sum of All Subset XOR Totals

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] 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.

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).

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