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:2Explanation: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:7Explanation:All non-empty subsets of [2,2,2] have a bitwise OR of 2. There are 2^{3}- 1 = 7 total subsets.

**Example 3:**

Input:nums = [3,2,1,5]Output:6Explanation: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}

**Similar Questions**:

## Solution 1. Bitmask

- 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`

. - 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;
}
};
```

## Discuss

https://leetcode.com/problems/count-number-of-maximum-bitwise-or-subsets/discuss/1525216/C%2B%2B-Bitmask-%2B-DP