1863. Sum of All Subset XOR Totals
Description
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
Solutions
Solution 1: Binary Enumeration
We can use binary enumeration to enumerate all subsets, and then calculate the XOR sum of each subset.
Specifically, we enumerate $i$ in the range $[0, 2^n)$, where $n$ is the length of the array $nums$. If the $j$th bit of the binary representation of $i$ is $1$, it means that the $j$th element of $nums$ is in the current subset; if the $j$th bit is $0$, it means that the $j$th element of $nums$ is not in the current subset. We can get the XOR sum of the current subset according to the binary representation of $i$, and add it to the answer.
The time complexity is $O(n \times 2^n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.
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Solution 2: DFS (Depth-First Search)
We can also use depth-first search to enumerate all subsets, and then calculate the XOR sum of each subset.
We design a function $dfs(i, s)$, where $i$ represents the current search to the $i$th element of the array $nums$, and $s$ represents the XOR sum of the current subset. Initially, $i=0$, $s=0$. During the search, we have two choices each time:
- Add the $i$th element of $nums$ to the current subset, i.e., $dfs(i+1, s \oplus nums[i])$;
- Do not add the $i$th element of $nums$ to the current subset, i.e., $dfs(i+1, s)$.
When we have searched all elements of the array $nums$, i.e., $i=n$, the XOR sum of the current subset is $s$, and we can add it to the answer.
The time complexity is $O(2^n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.
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