2044. Count Number of Maximum Bitwise-OR Subsets

Problem

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

Solution (Java)

class Solution {
    private int count = 0;

    public int countMaxOrSubsets(int[] nums) {
        int lookfor = 0;
        for (int i : nums) {
            lookfor = lookfor | i;
        }
        countsub(nums, 0, lookfor, 0);
        return count;
    }

    private void countsub(int[] nums, int index, int lookfor, int sofar) {
        if (lookfor == sofar) {
            count++;
        }
        if (index >= nums.length) {
            return;
        }

        for (int start = index; start < nums.length; start++) {
            countsub(nums, start + 1, lookfor, sofar | nums[start]);
        }
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).