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;
    }
}

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