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Given an integer array nums, return the sum of all XOR totals for every subset of nums. The XOR total of an array is defined as the bitwise XOR of all its elements, and the XOR total of the empty subset is 0.

Clarifying Questions

  1. Constraints on Input Size:
    • What is the maximum length of the array nums?
    • What is the range of values for each element in nums?
  2. Return Value:
    • Should the solution handle large sums, or are there any constraints on the sum value?

Strategy

We need to consider all subsets of the array nums and calculate the XOR for each subset. The sum of these XOR values will be our answer.

  1. Generate Subsets: We’ll use the power set approach to generate all subsets of nums. Each element can either be in a subset or not, leading to 2^n subsets for an array of length n.

  2. XOR Calculation: For each subset, calculate the XOR of its elements.

  3. Sum the XOR Values: Sum all the XOR values obtained for each subset.

To efficiently generate subsets and handle XOR calculations, we can use backtracking.

Code

def subsetXORSum(nums):
    def backtrack(start, current_xor):
        nonlocal total_sum
        total_sum += current_xor
        
        for i in range(start, len(nums)):
            backtrack(i + 1, current_xor ^ nums[i])

    total_sum = 0
    backtrack(0, 0)
    
    return total_sum

# Example usage:
nums = [1, 3]
print(subsetXORSum(nums))  # Output: 6

Explanation

  1. Backtrack Function (Recursive):
    • start: Index to start from for generating subsets.
    • current_xor: Current XOR value of the subset being generated.
    • For each function call, add the current_xor to the total_sum.
  2. Recursive Call:
    • Iterate from the current start index to the length of nums.
    • For each index, calculate the new XOR by including nums[i] and recursively call backtrack.
  3. Initial Call:
    • Initialize total_sum to 0.
    • Start the backtracking process from index 0 with current_xor as 0.

Time Complexity

The algorithm’s complexity can be broken down as follows:

So, the worst-case time complexity is O(n * 2^n).

This approach is efficient for small to moderately sized arrays but may become infeasible for large n due to the exponential growth of subsets.

Conclusion

This solution efficiently generates all possible subsets and calculates the XOR sum needed while maintaining manageable complexity for reasonable input sizes.

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