2044.count-number-of-maximum-bitwise-or-subsets

Statement

``````输入：nums = [3,1]

- [3]
- [3,1]
``````

``````输入：nums = [2,2,2]

``````

``````输入：nums = [3,2,1,5]

- [3,5]
- [3,1,5]
- [3,2,5]
- [3,2,1,5]
- [2,5]
- [2,1,5]``````

• `1 <= nums.length <= 16`
• `1 <= nums[i] <= 105`

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 <= 16`
• `1 <= nums[i] <= 105`

Solution

``````from typing import List

class Solution:
def countMaxOrSubsets(self, nums: List[int]) -> int:
n = len(nums)
max_res = 0
num = 0
for S in range(1 << n):
cur = 0
for i in range(n):
if (S >> i) & 1:
cur |= nums[i]

if cur > max_res:
max_res = cur
num = 1
elif cur == max_res:
num += 1

return num
``````