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

Approach 01:

class Solution {
public:
    int countMaxOrSubsets(vector<int>& nums) {
        const int maxOrValue = accumulate(nums.begin(), nums.end(), 0, bit_or<>());
        int subsetCount = 0;
        dfs(nums, 0, 0, maxOrValue, subsetCount);
        return subsetCount;
    }

private:
    void dfs(const vector<int>& nums, int index, int currentOrValue, const int& maxOrValue, int& subsetCount) {
        if (index == nums.size()) {
            if (currentOrValue == maxOrValue)
                ++subsetCount;
            return;
        }

        dfs(nums, index + 1, currentOrValue, maxOrValue, subsetCount);
        dfs(nums, index + 1, currentOrValue | nums[index], maxOrValue, subsetCount);
    }
};

class Solution:
    def countMaxOrSubsets(self, nums: list[int]) -> int:
        maxOrValue = 0
        for num in nums:
            maxOrValue |= num  # Calculate the maximum OR value
        
        subsetCount = [0]  # Use a list to allow modification within dfs
        self.dfs(nums, 0, 0, maxOrValue, subsetCount)
        return subsetCount[0]  # Return the count stored in the list

    def dfs(self, nums: list[int], index: int, currentOrValue: int, maxOrValue: int, subsetCount: list[int]) -> None:
        if index == len(nums):
            if currentOrValue == maxOrValue:
                subsetCount[0] += 1  # Increment the count
            return

        # Exclude the current number
        self.dfs(nums, index + 1, currentOrValue, maxOrValue, subsetCount)

        # Include the current number
        self.dfs(nums, index + 1, currentOrValue | nums[index], maxOrValue, subsetCount)

Time Complexity

• Subset Generation:

  The function uses a depth-first search (DFS) to explore all subsets of the input vector nums. For a list of size n, there are a total of 2^n possible subsets. Hence, the DFS traversal results in a time complexity of \(O(2^n)\).

• Overall Time Complexity:

  As a result, the overall time complexity is \(O(2^n)\).

Space Complexity

• Call Stack:

  The maximum depth of the recursion stack is n, where n is the size of the input vector. Therefore, the space complexity due to the recursion call stack is \(O(n)\).

• Overall Space Complexity:

  The overall space complexity is \(O(n)\).