Formatted question description: https://leetcode.ca/all/2044.html

# 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] <= 105

Similar Questions:

1. Compute goal which is the maximum possible bitwise OR of a subset of nums, i.e. the bitwise OR of all the numbers in nums.
2. Enumerate all non-empty subsets of nums using bitmask and compute the bitwise OR of each of them. Increment answer if the subset’s bitwise OR is the same as goal.
// OJ: https://leetcode.com/problems/count-number-of-maximum-bitwise-or-subsets/
// 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;
}
};


We can use DP to reduce the time complexity at the cost of space complexity

// OJ: https://leetcode.com/problems/count-number-of-maximum-bitwise-or-subsets/
// Time: O(2^N)
// Space: O(2^N)
class Solution {
public:
int countMaxOrSubsets(vector<int>& A) {
int goal = 0, N = A.size(), ans = 0;
vector<int> dp(1 << N);
for (int n : A) goal |= n;
for (int m = 1; m < 1 << N; ++m) {
int lowbit = m & -m;
dp[m] = dp[m - lowbit] | A[__builtin_ctz(lowbit)];
if (dp[m] == goal) ++ans;
}
return ans;
}
};