Sum Of All Subset Xor Totals - Leetcode Solution

Problem Statement: Given an integer array nums, return the sum of all XOR totals for every subset of nums.

A subset is a group of elements that can be selected from an array, without necessarily selecting all the elements in the array. For example, a subset of the array [1, 2, 3] can be [1, 3] or [2], but not [1, 4].

The XOR total of a set is the bitwise XOR (exclusive or) of all the elements in that set. For example, the XOR total of the set [1, 3, 6] is equal to 1 XOR 3 XOR 6 = 4.

Solution: A brute force method to solve this problem can be to generate all the possible subsets for an array and calculate the XOR total for each of them, which would have a time complexity of O(2^n*n), where n is the length of the array.

But this approach is highly inefficient as the time complexity could go up to 2^20*20 which is quite a large number. So, we need a more efficient solution.

One efficient solution is to use the properties of bitwise XOR. The bitwise XOR of two numbers will be 1 in the ith position if the ith bit of the two numbers is different, otherwise, the ith bit will be 0. So, the problem can be solved by counting the number of subsets in which the ith element is included and the number of subsets in which it is not included. The XOR total of a subset with the ith element will be equal to the XOR total of a subset without the ith element XOR with the ith element. Calculating for all the elements and adding all XOR totals of all these subsets is the required answer.

A recursive approach can be used to solve this problem, where the recursion will be done on the index of the array, and each time we will have two choices, whether to include the current element in the subset or not. This approach will have a time complexity of O(2^n).

Algorithm:

1. Define a function named 'subsetXORSum' which takes an array of integers 'nums' as input and returns an integer.
2. Define a variable 'result' and initialize it with 0.
3. Define a recursive function named 'helper' which takes three parameters, 'i' (index), 'xorVal' (current XOR value) and 'nums' (array).
4. Inside the 'helper' function, check if the 'i' >= length of 'nums', if yes, then return the 'xorVal'.
5. Otherwise, return the result of two recursive calls to the 'helper' function, one by passing 'i+1' and 'xorVal' as it is, and another by passing 'i+1' and 'xorVal^nums[i]' (current XOR value XOR current element).
6. Call the 'helper' function by passing two parameters, 0 and 0.
7. Add the result obtained from the helper function to 'result' and return it.

Code:

Below is the implementation of the above algorithm in Python:

def subsetXORSum(nums: List[int]) -> int: # Step 2 result = 0

# Step 3
def helper(i, xorVal, nums):
    # Step 4
    if i >= len(nums):
        return xorVal
    
    # Step 5
    return helper(i+1, xorVal, nums) + helper(i+1, xorVal^nums[i], nums)

# Step 6
result += helper(0, 0, nums)

# Step 7
return result

The time complexity of this solution is O(2^n), where n is the length of the input array, and the space complexity is O(n) due to the recursion stack.

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