最大按位或子集的数量
Count Number of Maximum Bitwise-OR Subsets
题目详情
问题：最大按位或子集的数量
考察：位运算、递归
来源：Citadel
链接：https://www.jointaro.com/interviews/questions/count-number-of-maximum-bitwise-or-subsets/
Given an integer array nums, find the maximum possible bitwise OR of a subset of nums and return the number of different non-empty subsets with the maximum bitwise OR.
An array a is a subset of an array b if a can be obtained from b by deleting some (possibly zero) elements of b. Two subsets are considered different if the indices of the elements chosen are different.
The bitwise OR of an array a is equal to a[0] OR a[1] OR ... OR a[a.length - 1] (0-indexed).
Example 1:
Input: nums = [3,1] Output: 2 Explanation: The maximum possible bitwise OR of a subset is 3. There are 2 subsets with a bitwise OR of 3: - [3] - [3,1]
Example 2:
Input: nums = [2,2,2] Output: 7 Explanation: All non-empty subsets of [2,2,2] have a bitwise OR of 2. There are 23 - 1 = 7 total subsets.
Example 3:
Input: nums = [3,2,1,5] Output: 6 Explanation: The maximum possible bitwise OR of a subset is 7. There are 6 subsets with a bitwise OR of 7: - [3,5] - [3,1,5] - [3,2,5] - [3,2,1,5] - [2,5] - [2,1,5]
Constraints:
1 <= nums.length <= 161 <= nums[i] <= 105
解析
思路：先求所有元素按位或得到最大可能 OR。然后用回溯或 DP 统计子集 OR 值，最终计数 OR 等于最大值的子集。
复杂度：回溯 O(2^n)，也可用 OR 状态 DP，空间为不同 OR 状态数。