1863

1863. Sum of All Subset XOR Totals (Easy)

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

Related Topics:
Backtracking, Recursion

Solution 1. Brute Force

// OJ: https://leetcode.com/problems/sum-of-all-subset-xor-totals/
// Author: github.com/lzl124631x
// Time: O(2^N)
// Space: O(N)
class Solution {
    int ans = 0;
    void dfs(vector<int> &A, int start, int val) {
        if (start == A.size()) {
            ans += val;
            return;
        }
        dfs(A, start + 1, val ^ A[start]);
        dfs(A, start + 1, val);
    }
public:
    int subsetXORSum(vector<int>& A) {
        dfs(A, 0, 0);
        return ans;
    }
};

Solution 2. Brute Force

// OJ: https://leetcode.com/problems/sum-of-all-subset-xor-totals/
// Author: github.com/lzl124631x
// Time: O(2^N * N)
// Space: O(1)
class Solution {
public:
    int subsetXORSum(vector<int>& A) {
        int ans = 0;
        for (int m = 1, end = (1 << A.size()); m < end; ++m) {
            int total = 0;
            for (int i = 0; i < A.size(); ++i) {
                if (m >> i & 1) total ^= A[i];
            }
            ans += total;
        }
        return ans;
    }
};

Solution 3. Math

For the ith bit, assume there are k numbers whose ith bit is 1.

• If k == 0, all the xor results will have 0 in the ith bit.
• If k >= 1, we can use the k 1s to generate comb(k, 1) + comb(k, 3) + comb(k, 5) + ... = 2^(k-1) different ways to make the xor result 1. For the rest n - k numbers, there are 2^(n-k) ways to pick any of them. So in total there are 2^(k-1) * 2^(n-k) = 2^(n-1) ways to get 1 as the xor result.

Note that 2^(n-1) is irrelevant to k, so as long as k >= 1 i.e. there is a number whose ith bit is 1, the ith bit will contribute (1 << i) * 2^(n-1) to the result.

// OJ: https://leetcode.com/problems/sum-of-all-subset-xor-totals/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int subsetXORSum(vector<int>& A) {
        int bits = 0;
        for (int n : A) bits |= n;
        return bits * pow(2, A.size() - 1);
    }
};