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Formatted question description: https://leetcode.ca/all/1863.html

1863. Sum of All Subset XOR Totals

Level

Easy

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

Solution

Enumerate all subsets of nums. For each subset, calculate the XOR total. Finally, calculate the sum of all subsets’ XOR totals and return.

  • class Solution {
    public int subsetXORSum(int[] nums) {
        int sum = 0;
        int length = nums.length;
        int count = 1 << length;
        for (int i = 0; i < count; i++) {
            int xor = getXor(nums, i);
            sum += xor;
        }
        return sum;
    }

    public int getXor(int[] nums, int curNum) {
        int xor = 0;
        int index = 0;
        while (curNum > 0) {
            int remainder = curNum % 2;
            if (remainder == 1)
                xor ^= nums[index];
            curNum >>= 1;
            index++;
        }
        return xor;
    }
}

############

class Solution {
    private int res;

    public int subsetXORSum(int[] nums) {
        dfs(nums, 0, 0);
        return res;
    }

    private void dfs(int[] nums, int depth, int prev) {
        res += prev;
        for (int i = depth; i < nums.length; ++i) {
            prev ^= nums[i];
            dfs(nums, ++depth, prev);
            prev ^= nums[i];
        }
    }
}

  • // OJ: https://leetcode.com/problems/sum-of-all-subset-xor-totals/
// 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;
    }
};

  • class Solution:
    def subsetXORSum(self, nums: List[int]) -> int:
        def dfs(nums, depth, prev):
            self.res += prev
            for num in nums[depth:]:
                prev ^= num
                depth += 1
                dfs(nums, depth, prev)
                prev ^= num

        self.res = 0
        dfs(nums, 0, 0)
        return self.res

############

# 1863. Sum of All Subset XOR Totals
# https://leetcode.com/problems/sum-of-all-subset-xor-totals/

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

            res += t
        
        return res


  • /**
 * @param {number[]} nums
 * @return {number}
 */
var subsetXORSum = function (nums) {
    let res = [];
    let prev = 0;
    dfs(nums, 0, prev, res);
    return res.reduce((a, c) => a + c, 0);
};

function dfs(nums, depth, prev, res) {
    res.push(prev);
    for (let i = depth; i < nums.length; i++) {
        prev ^= nums[i];
        depth++;
        dfs(nums, depth, prev, res);
        // bracktrack
        prev ^= nums[i];
    }
}


  • func subsetXORSum(nums []int) (ans int) {
	n := len(nums)
	var dfs func(int, int)
	dfs = func(i, s int) {
		if i >= n {
			ans += s
			return
		}
		dfs(i+1, s)
		dfs(i+1, s^nums[i])
	}
	dfs(0, 0)
	return
}

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

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