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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] 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

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 $O(1)$.

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 $nums$.

  • class Solution {
    public int subsetXORSum(int[] nums) {
        int n = nums.length;
        int ans = 0;
        for (int i = 0; i < 1 << n; ++i) {
            int s = 0;
            for (int j = 0; j < n; ++j) {
                if ((i >> j & 1) == 1) {
                    s ^= nums[j];
                }
            }
            ans += s;
        }
        return ans;
    }
}

  • class Solution {
public:
    int subsetXORSum(vector<int>& nums) {
        int n = nums.size();
        int ans = 0;
        for (int i = 0; i < 1 << n; ++i) {
            int s = 0;
            for (int j = 0; j < n; ++j) {
                if (i >> j & 1) {
                    s ^= nums[j];
                }
            }
            ans += s;
        }
        return ans;
    }
};

  • class Solution:
    def subsetXORSum(self, nums: List[int]) -> int:
        ans, n = 0, len(nums)
        for i in range(1 << n):
            s = 0
            for j in range(n):
                if i >> j & 1:
                    s ^= nums[j]
            ans += s
        return ans


  • func subsetXORSum(nums []int) (ans int) {
	n := len(nums)
	for i := 0; i < 1<<n; i++ {
		s := 0
		for j, x := range nums {
			if i>>j&1 == 1 {
				s ^= x
			}
		}
		ans += s
	}
	return
}

  • function subsetXORSum(nums: number[]): number {
    let ans = 0;
    const n = nums.length;
    for (let i = 0; i < 1 << n; ++i) {
        let s = 0;
        for (let j = 0; j < n; ++j) {
            if ((i >> j) & 1) {
                s ^= nums[j];
            }
        }
        ans += s;
    }
    return ans;
}


  • /**
 * @param {number[]} nums
 * @return {number}
 */
var subsetXORSum = function (nums) {
    let ans = 0;
    const n = nums.length;
    for (let i = 0; i < 1 << n; ++i) {
        let s = 0;
        for (let j = 0; j < n; ++j) {
            if ((i >> j) & 1) {
                s ^= nums[j];
            }
        }
        ans += s;
    }
    return ans;
};

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