{ "data": { "question": { "questionId": "1993", "questionFrontendId": "1863", "categoryTitle": "Algorithms", "boundTopicId": 772194, "title": "Sum of All Subset XOR Totals", "titleSlug": "sum-of-all-subset-xor-totals", "content": "
The XOR total of an array is defined as the bitwise XOR
of all its elements, or 0
if the array is empty.
[2,5,6]
is 2 XOR 5 XOR 6 = 1
.Given an array nums
, return the sum of all XOR totals for every subset of nums
.
Note: Subsets with the same elements should be counted multiple times.
\n\nAn array a
is a subset of an array b
if a
can be obtained from b
by deleting some (possibly zero) elements of b
.
\n
Example 1:
\n\n\nInput: nums = [1,3]\nOutput: 6\nExplanation: The 4 subsets of [1,3] are:\n- The empty subset has an XOR total of 0.\n- [1] has an XOR total of 1.\n- [3] has an XOR total of 3.\n- [1,3] has an XOR total of 1 XOR 3 = 2.\n0 + 1 + 3 + 2 = 6\n\n\n
Example 2:
\n\n\nInput: nums = [5,1,6]\nOutput: 28\nExplanation: The 8 subsets of [5,1,6] are:\n- The empty subset has an XOR total of 0.\n- [5] has an XOR total of 5.\n- [1] has an XOR total of 1.\n- [6] has an XOR total of 6.\n- [5,1] has an XOR total of 5 XOR 1 = 4.\n- [5,6] has an XOR total of 5 XOR 6 = 3.\n- [1,6] has an XOR total of 1 XOR 6 = 7.\n- [5,1,6] has an XOR total of 5 XOR 1 XOR 6 = 2.\n0 + 5 + 1 + 6 + 4 + 3 + 7 + 2 = 28\n\n\n
Example 3:
\n\n\nInput: nums = [3,4,5,6,7,8]\nOutput: 480\nExplanation: The sum of all XOR totals for every subset is 480.\n\n\n
\n
Constraints:
\n\n1 <= nums.length <= 12
1 <= nums[i] <= 20
一个数组的 异或总和 定义为数组中所有元素按位 XOR
的结果;如果数组为 空 ,则异或总和为 0
。
[2,5,6]
的 异或总和 为 2 XOR 5 XOR 6 = 1
。给你一个数组 nums
,请你求出 nums
中每个 子集 的 异或总和 ,计算并返回这些值相加之 和 。
注意:在本题中,元素 相同 的不同子集应 多次 计数。
\n\n数组 a
是数组 b
的一个 子集 的前提条件是:从 b
删除几个(也可能不删除)元素能够得到 a
。
\n\n
示例 1:
\n\n输入:nums = [1,3]\n输出:6\n解释:[1,3] 共有 4 个子集:\n- 空子集的异或总和是 0 。\n- [1] 的异或总和为 1 。\n- [3] 的异或总和为 3 。\n- [1,3] 的异或总和为 1 XOR 3 = 2 。\n0 + 1 + 3 + 2 = 6\n\n\n
示例 2:
\n\n输入:nums = [5,1,6]\n输出:28\n解释:[5,1,6] 共有 8 个子集:\n- 空子集的异或总和是 0 。\n- [5] 的异或总和为 5 。\n- [1] 的异或总和为 1 。\n- [6] 的异或总和为 6 。\n- [5,1] 的异或总和为 5 XOR 1 = 4 。\n- [5,6] 的异或总和为 5 XOR 6 = 3 。\n- [1,6] 的异或总和为 1 XOR 6 = 7 。\n- [5,1,6] 的异或总和为 5 XOR 1 XOR 6 = 2 。\n0 + 5 + 1 + 6 + 4 + 3 + 7 + 2 = 28\n\n\n
示例 3:
\n\n输入:nums = [3,4,5,6,7,8]\n输出:480\n解释:每个子集的全部异或总和值之和为 480 。\n\n\n
\n\n
提示:
\n\n1 <= nums.length <= 12
1 <= nums[i] <= 20
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