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Leetcode 1863. Sum of All Subset XOR Totals

Problem Statement

You are given an integer array nums. 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.

Return the sum of all XOR totals for every subset of nums.

A subset of an array is a selection of elements (possibly empty) of the array.

Clarifying Questions

1. Range of Input Size: How many numbers can the array nums contain?
   • The array length can be up to 12 (1 <= nums.length <= 12).
2. Range of Element Values: What is the range of the values in nums?
   • The elements in the array can range from 0 to 1000 (0 <= nums[i] <= 1000).

Strategy

We need to compute the sum of XOR totals for all possible subsets. Given that the array size is relatively small (max 12), an exhaustive approach is feasible. Each subset can be generated and evaluated using bitwise XOR.

Steps:

1. Generate All Subsets: Utilize a bitmask approach to generate all possible subsets of the given array.
2. Compute XOR for Each Subset: For each subset, calculate the XOR of the elements.
3. Sum the XOR Values: Accumulate the XOR values to get the final result.

Code

Let’s translate the strategy into C++ code.

#include <vector>
#include <cmath>

class Solution {
public:
    int subsetXORSum(std::vector<int>& nums) {
        int n = nums.size();
        int totalSubsets = 1 << n; // 2^n subsets
        int sumOfXORs = 0;

        // Generate each subset
        for (int subsetMask = 0; subsetMask < totalSubsets; ++subsetMask) {
            int currentXOR = 0;
            // Calculate XOR for the current subset
            for (int i = 0; i < n; ++i) {
                if (subsetMask & (1 << i)) {
                    currentXOR ^= nums[i];
                }
            }
            sumOfXORs += currentXOR;
        }

        return sumOfXORs;
    }
};

Time Complexity

• Generation of Subsets: There are (2^n) subsets for an array of size (n).
• XOR Calculation: For each subset, we iterate through (n) elements to compute the XOR.

Hence, the time complexity is: [ O(n \cdot 2^n) ]

This is manageable given the constraints ((n \leq 12)).

Space Complexity

• The space complexity is (O(1)), as we are only using a few additional variables for counting and accumulating the XOR values, regardless of the input size.

This approach efficiently addresses the problem within the given constraints.

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