Leetcode 1863: Sum of All Subset XOR Totals

Input: The input is an array `nums` consisting of integers.

Example: nums = [1, 3]

Constraints:

• 1 <= nums.length <= 12

• 1 <= nums[i] <= 20

Output: The output is an integer representing the sum of all XOR totals for every subset of the array `nums`.

Example: Output = 6

Constraints:

• The output is the sum of XOR totals for every subset of the array.

Goal: The goal is to calculate the XOR total for every subset and return their sum.

Steps:

• Step 1: Generate all subsets of the array.

• Step 2: For each subset, calculate the XOR total by performing a bitwise XOR of all elements in the subset.

• Step 3: Add the XOR totals of all subsets and return the sum.

Goal: The constraints ensure the array is small enough for generating all subsets and calculating XOR totals efficiently.

Steps:

• The array contains between 1 and 12 elements.

• Each integer in the array is between 1 and 20.

Assumptions:

• The input array contains at least one element.

• Input: Input: [1, 3]

• Explanation: There are 4 subsets: the empty subset, [1], [3], and [1,3]. The XOR totals for each subset are 0, 1, 3, and 2, respectively. The sum of these totals is 6.

• Input: Input: [4, 1, 7]

• Explanation: The eight subsets of [4, 1, 7] and their corresponding XOR totals are: 0, 4, 1, 7, 5, 3, 6, and 2. The sum of these totals is 35.

Approach: To solve this problem, we generate all possible subsets of the array and compute the XOR total for each subset. Then, we sum all these XOR totals.

Observations:

• The problem requires generating all subsets and computing XOR totals for each subset.

• Given that the number of elements in the array is small (at most 12), generating all subsets (2^n subsets) is feasible.

Steps:

• Step 1: Loop through all numbers from 1 to 2^n - 1, where n is the size of the array. Each number represents a subset using its binary representation.

• Step 2: For each subset, calculate the XOR total by checking which elements are included (using the binary number) and performing XOR.

• Step 3: Sum all the XOR totals and return the result.

Empty Inputs:

• The input array is never empty as per the constraints.

Large Inputs:

• The maximum length of the array is 12, so generating subsets and calculating XOR totals will work efficiently within the time limits.

Special Values:

• Consider cases where all elements are the same, which will result in subsets with identical XOR totals.

Constraints:

• The array contains between 1 and 12 elements, each between 1 and 20.

int subsetXORSum(vector<int>& nums) {
    int res = 0;
    for (auto i = 1; i < (1 << nums.size()); ++i) {
        int total = 0;
        for (auto j = 0; j < nums.size(); ++j)
            if (i & (1 << j))
                total ^= nums[j];
        res += total;
    }
    return res;
}

1 : Function Definition

int subsetXORSum(vector<int>& nums) {

Define the function `subsetXORSum`, which takes a reference to a vector `nums` containing integers and returns an integer.

2 : Variable Initialization

    int res = 0;

Initialize a variable `res` to store the cumulative XOR sum of all subsets.

3 : Outer Loop (Subset Generation)

    for (auto i = 1; i < (1 << nums.size()); ++i) {

Start a loop to iterate through all possible subsets of `nums` by generating numbers from 1 to `2^n - 1` using bitwise operations.

4 : Variable Initialization (Subset XOR)

        int total = 0;

Initialize a variable `total` to store the XOR of the current subset.

5 : Inner Loop (Element Check)

        for (auto j = 0; j < nums.size(); ++j)

Start a loop to iterate through each element of the `nums` array to check if the current element belongs to the current subset.

6 : Bitwise Check

            if (i & (1 << j))

Check if the `j`-th element is included in the current subset by using a bitwise AND operation. If the `j`-th bit of `i` is set, the element is included.

7 : Subset XOR Calculation

                total ^= nums[j];

XOR the value of the `j`-th element of `nums` with the current `total` to update the XOR for the current subset.

8 : Update Result

        res += total;

Add the XOR result of the current subset (`total`) to the running total `res`.

9 : Return Result

    return res;

Return the final result `res`, which contains the sum of XORs of all subsets.

Best Case: O(2^n * n)

Average Case: O(2^n * n)

Worst Case: O(2^n * n)

Description: The time complexity is O(2^n * n), where n is the length of the input array. There are 2^n subsets and for each subset, we need to compute the XOR total which takes O(n) time.

Best Case: O(2^n)

Worst Case: O(2^n)

Description: The space complexity is O(2^n) due to the space needed for storing the subsets and the XOR totals.

Leetcode 1863: Sum of All Subset XOR Totals

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