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Leetcode 1863. Sum of All Subset XOR Totals

Problem Statement

Given an integer array nums, find the sum of all XOR totals for every subset of nums. The XOR total of a subset is defined as the bitwise XOR of all its elements if the subset is not empty.

Clarifying Questions

  1. Input Constraints: What is the range of values for the elements in the nums array?
    • Elements in nums are non-negative integers and the length of nums is between 1 and 12.
  2. Output Constraints: Are there any specific format requirements for the output?
    • The output should be a single integer which is the sum of XOR values of all subsets.
  3. Edge Cases:
    • What should be returned if the input array contains only one element?
      • The XOR value would be the element itself.

Strategy

  1. Generate All Subsets: Use backtracking to generate all possible subsets of nums.
  2. Calculate XOR Total: For each subset, calculate its XOR total (skip empty subsets).
  3. Sum XOR Totals: Sum up the XOR totals of all subsets.
  4. Optimization Insight: Since each element appears in exactly half of the generated subsets given any input of length n, we can simplify the computation.

Code

import java.util.ArrayList;
import java.util.List;

public class Solution {
    public int subsetXORSum(int[] nums) {
        List<Integer> currentSubset = new ArrayList<>();
        return backtrack(nums, currentSubset, 0);
    }

    private int backtrack(int[] nums, List<Integer> currentSubset, int index) {
        if (index == nums.length) {
            // Calculate the XOR total for the current subset (excluding empty subset)
            if (currentSubset.isEmpty()) return 0;
            int xorTotal = 0;
            for (int num : currentSubset) {
                xorTotal ^= num;
            }
            return xorTotal;
        }
        // Include the current element
        currentSubset.add(nums[index]);
        int include = backtrack(nums, currentSubset, index + 1);
        currentSubset.remove(currentSubset.size() - 1);
        
        // Exclude the current element
        int exclude = backtrack(nums, currentSubset, index + 1);
        
        return include + exclude;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        int[] nums = {1, 3};
        System.out.println(sol.subsetXORSum(nums));  // Output: 6
    }
}

Time Complexity

The time complexity for this backtracking solution is exponential, O(2^n), where n is the number of elements in the nums array. This is because we generate all possible subsets (which is 2^n subsets). In practice, given nums array length constraints (1 ≤ n ≤ 12), this exponential time complexity is manageable.

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