2044. 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 <= 16
• 1 <= nums[i] <= 105

C++

// Time: O(2^N * N)
// Space: O(1)
class Solution {
public:
    int countMaxOrSubsets(vector<int>& A) {
        int goal = 0, N = A.size(), ans = 0;
        for (int n : A) goal |= n;
        for (int m = 1; m < 1 << N; ++m) {
            int x = 0;
            for (int i = 0; i < N; ++i) {
                if (m >> i & 1) x |= A[i];
            }
            if (x == goal) ++ans;
        }
        return ans;
    }
};

JAVA

• class Solution {
      private int mx;
      private int ans;
      private int[] nums;
  
      public int countMaxOrSubsets(int[] nums) {
          mx = 0;
          for (int x : nums) {
              mx |= x;
          }
          this.nums = nums;
          dfs(0, 0);
          return ans;
      }
  
      private void dfs(int i, int t) {
          if (i == nums.length) {
              if (t == mx) {
                  ++ans;
              }
              return;
          }
          dfs(i + 1, t);
          dfs(i + 1, t | nums[i]);
      }
  }

PYTHON

• class Solution:
      def countMaxOrSubsets(self, A):
          goal = 0
          N = len(A)
          ans = 0
          for n in A:
              goal |= n
          for m in range(1, 1 << N):
              x = 0
              for i in range(N):
                  if m >> i & 1:
                      x |= A[i]
              if x == goal:
                  ans += 1
          return ans