# 2044. Count Number of Maximum Bitwise-OR Subsets

Medium
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 ``OR`` a ``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:
- 
- [3,1]
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] <= 10^5`

### 解題

1863 Sum of All Subset XOR Totals 的加強版。
func countMaxOrSubsets(nums []int) int {
max := nums
for i:=1; i<len(nums); i++ {
max |= nums[i]
}
ans := 0
var dfs func(int, int)
dfs = func(index int, xor int) {
if index >= len(nums) {
return
}
if xor | nums[index] == max {
ans++
}
dfs(index + 1, xor | nums[index]) //取
dfs(index + 1, xor ) //不取
return
}
dfs(0, 0)
return ans
}