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
}