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[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] <= 10^5

解題

1863 Sum of All Subset XOR Totals 的加強版。
func countMaxOrSubsets(nums []int) int {
max := nums[0]
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
}