How do you solve Sum of All Subset XOR Totals on LeetCode?

Sum of All Subset XOR Totals (LeetCode #1863) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Each number contributes equally to the XOR of all subsets.

What is the brute force solution for Sum of All Subset XOR Totals?

The brute force approach for Sum of All Subset XOR Totals is Brute Force, with O(n * 2^n) time complexity and O(1) space complexity. The brute force approach involves generating all possible subsets of the input array and calculating the XOR for each subset. This is straightforward but can be inefficient for larger arrays due to the exponential growth of subsets. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Sum of All Subset XOR Totals?

Clarify the problem requirements and constraints before diving into coding. Think about the properties of operations like XOR to simplify your approach. Practice generating subsets and their properties to build intuition.

What are common mistakes when solving Sum of All Subset XOR Totals?

Not considering the contribution of each number directly. Forgetting that the empty subset has an XOR of 0.

#1863

Sum of All Subset XOR Totals

Easy

ArrayMathBacktrackingBit ManipulationCombinatoricsEnumerationBit ManipulationCombinatorics

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n * 2^n)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal approach leverages the properties of XOR and the fact that each number contributes to multiple subsets. Specifically, each number appears in half of the subsets, allowing us to calculate the contribution of each number directly without generating all subsets.

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Algorithm

3 steps

1Step 1: Calculate the number of subsets that include each number, which is 2^(n-1) where n is the length of the array.
2Step 2: For each number in the array, compute its contribution to the total XOR sum as num * (2^(n-1)).
3Step 3: Sum all contributions to get the final result.

solution.py6 lines

1def subsetXORSum(nums):
2    n = len(nums)
3    total_sum = 0
4    for num in nums:
5        total_sum += num * (1 << (n - 1))  # 2^(n-1)
6    return total_sum

ℹ

Complexity note: The time complexity is O(n) because we iterate through the array once. The space complexity is O(1) since we use a constant amount of space.

1Each number contributes equally to the XOR of all subsets.
2The number of subsets including a number is 2^(n-1).

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.