Problem 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] 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.
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.
Solution Code
class Solution {
public:
int subsetXORSum(vector& a) {
int n = a.size();
int bitSize = 1 << n;
int res = 0;
for (int i = 0; i < bitSize; i++) {
int xorSum = 0;
for (int j = 0; j < n; j++) {
if (i & (1 << j)) {
xorSum ^= a[j];
}
}
res += xorSum;
}
return res;
}
};
Solution Explanation
The problem involves calculating the sum of XOR totals for every subset of a given array. The solution follows these steps:
- Generate Subsets: For an array of size
n, there are2^nsubsets. We use a bitmask approach to generate all possible subsets. Each bit in the bitmask represents whether a corresponding element in the array is included in the subset or not. - Calculate XOR Total: For each subset represented by a bitmask, calculate the XOR total by iterating through the elements and applying XOR operation based on the bitmask.
- Sum XOR Totals: Accumulate the XOR totals of all subsets to get the final result.