LeetCode 1863: Sum of All Subset XOR Totals Solution
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1863. Sum of All Subset XOR Totals
Problem Explanation
Explanation:
To solve this problem, we can utilize the concept of bitwise XOR operation. The XOR total of an array can be calculated by XOR-ing all elements in the array. To find the sum of XOR totals for every subset of the given array, we need to consider all possible subsets and calculate the XOR total for each.
We can solve this problem using a recursive approach. For each element in the array, we have two choices: either include it in the subset or exclude it. By recursively exploring both choices, we can generate all possible subsets and calculate their XOR totals. Finally, we sum up all the XOR totals to get the desired result.
Time Complexity:
The time complexity of this approach is O(2^N) where N is the number of elements in the input array, as we are generating all possible subsets.
Space Complexity:
The space complexity of this approach is O(N) where N is the number of elements in the input array, due to the recursive stack space. :
Solution Code
class Solution {
public int subsetXORSum(int[] nums) {
return calculateXORSum(nums, 0, 0);
}
private int calculateXORSum(int[] nums, int index, int xor) {
if (index == nums.length) {
return xor;
}
int include = calculateXORSum(nums, index + 1, xor ^ nums[index]);
int exclude = calculateXORSum(nums, index + 1, xor);
return include + exclude;
}
}Try It Yourself
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