Sum of All Subset XOR Totals

Easy

Asked by:

50 views

Topics:

ArraysBit Manipulation

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum possible length of the input array `nums`?
Can the integers in `nums` be negative?
Are there any constraints on the range of values within the `nums` array?
If `nums` is empty, should I return 0?
Are duplicate numbers allowed in the input array `nums`?

Brute Force Solution

Approach

The brute force approach in this problem involves finding every possible combination of numbers from the given set. We will then calculate a special value (XOR total) for each combination. Finally, we add up all these special values to get the final answer.

Here's how the algorithm would work step-by-step:

Consider each number in the set; each number is either 'in' or 'out' of a subset.
For every possible subset, calculate its XOR total: This means doing a bitwise XOR operation between all numbers that are 'in' that subset.
Once you have the XOR total for a subset, add that total to a running sum.
Repeat the previous two steps for every single possible subset.
After considering all possible subsets, the final running sum is the answer.

Code Implementation

def sum_subset_xor_totals_brute_force(numbers):
    total_xor_sum = 0
    number_of_elements = len(numbers)

    # Iterate through all possible subsets
    for i in range(1 << number_of_elements):
        current_xor_total = 0

        # Iterate through each element to construct the subset
        for j in range(number_of_elements):
            # Check if the j-th element is present in the current subset.
            if (i >> j) & 1:

                # XOR the element if it's in the current subset
                current_xor_total ^= numbers[j]

        # Add the XOR total of this subset to the overall sum.
        total_xor_sum += current_xor_total

    return total_xor_sum

Big(O) Analysis

Time Complexity

O(n * 2^n) – The algorithm iterates through all possible subsets of the input array nums of size n. There are 2^n possible subsets. For each subset, it computes the XOR sum, which takes O(n) time in the worst case (when the subset contains all n elements). Therefore, the overall time complexity is O(n * 2^n).

Space Complexity

O(1) – The provided description outlines a brute force approach that iterates through all possible subsets and calculates XOR sums. It does not mention creation of any auxiliary data structures like lists or hash maps to store intermediate subsets or results. The algorithm only seems to use a running sum variable to accumulate the XOR totals, which consumes constant extra space irrespective of the input array size N. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The problem asks to calculate the sum of XOR results from all possible subsets of a given set of numbers. Instead of generating all subsets and calculating the XOR sum for each, we leverage the properties of the XOR operation and bitwise operations to find the solution much faster.

Here's how the algorithm would work step-by-step:

Understand that a number's presence or absence in a subset is independent of other numbers.
Recognize that each bit in the input numbers contributes independently to the final XOR sum.
For each bit position (e.g., the rightmost bit, the next bit, and so on), determine if that bit will appear an odd or even number of times across all subset XOR totals.
A bit position contributes to the final sum only if it appears in an odd number of subset XOR totals.
The crucial insight is that if any number in the input set has a '1' at a specific bit position, then that bit position will be '1' in exactly half of all possible subsets.
Therefore, for each number in the input, check each of its bits. If any of the bits are '1', add the corresponding power of 2 to a running total which represents the sum of all subsets XOR totals multiplied by half the possibilities.
Multiply this running total by half the number of all possible subsets, which is computed as 2 to the power of number of input numbers minus 1.
The final result is this product.

Code Implementation

def sum_subset_xor_totals(numbers):
    subset_xor_sum = 0
    number_of_elements = len(numbers)

    # Iterate through each number to identify set bits.
    for number in numbers:
        subset_xor_sum |= number

    #Every bit set will appear in exactly half of the subsets.
    return subset_xor_sum * (1 << (number_of_elements - 1))

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each number in the input array nums of size n. For each number, it checks each bit position to see if it is set to '1'. Since the number of bits in each number is constant (determined by the integer size, e.g., 32 bits), this bitwise check takes constant time. Thus, the overall time complexity is dominated by the loop through the n numbers, resulting in O(n) time complexity. The power of 2 calculation and final multiplication are constant time operations and do not affect the overall complexity.

Space Complexity

O(1) – The algorithm calculates the sum of XOR totals using a running total and iterates through the input numbers, checking each bit. It uses a few variables to store intermediate calculations like the running total and powers of 2. The memory used by these variables remains constant regardless of the number of input numbers (N). Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 as the sum of XOR totals of subsets of an empty set is 0.
Input array with a single element	Return the value of that single element, as it's the only possible subset XOR total.
Input array with all elements being zero	Return 0, as the XOR total of any subset will be 0.
Input array with a large number of elements (scalability)	Ensure the solution uses an efficient algorithm (e.g., bit manipulation) to avoid exponential time complexity and potential timeouts.
Input array with very large integer values (potential overflow)	Use a data type capable of holding large values (e.g., long in Java/C++) to prevent integer overflow during XOR calculations.
Input array contains only positive numbers	The algorithm should work correctly without special handling.
Input array with all elements being identical non-zero values	The XOR total will depend on the parity (even/odd) of the subset size, and the solution needs to correctly account for this.
Input array where the XOR of all elements is zero	The solution should correctly calculate the sum of XOR totals even when the complete set's XOR is zero.

Null or empty input array

How to Handle:

Return 0 as the sum of XOR totals of subsets of an empty set is 0.

Input array with a single element

How to Handle:

Return the value of that single element, as it's the only possible subset XOR total.

Input array with all elements being zero

How to Handle:

Return 0, as the XOR total of any subset will be 0.

Input array with a large number of elements (scalability)

How to Handle:

Ensure the solution uses an efficient algorithm (e.g., bit manipulation) to avoid exponential time complexity and potential timeouts.

Input array with very large integer values (potential overflow)

How to Handle:

Use a data type capable of holding large values (e.g., long in Java/C++) to prevent integer overflow during XOR calculations.

Input array contains only positive numbers

How to Handle:

The algorithm should work correctly without special handling.

Input array with all elements being identical non-zero values

How to Handle:

The XOR total will depend on the parity (even/odd) of the subset size, and the solution needs to correctly account for this.

Input array where the XOR of all elements is zero

How to Handle:

The solution should correctly calculate the sum of XOR totals even when the complete set's XOR is zero.