A non-recursive trick for subsets generation II

Many years ago I wrote about this simple trick to generate all subsets of a set: A non-recursive trick for subsets generation (anothercasualcoder.blogspot.com). The problem in this post requires the same approach. Generate all subsets, and keep track of the number of subsets satisfying a certain criteria. The code is down below, cheers, ACC.

Count Number of Maximum Bitwise-OR Subsets - LeetCode

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

Accepted

6,027

Submissions

8,064

public int CountMaxOrSubsets(int[] nums)
{
    SortedList sortedSets = new SortedList();

    for (int i = 0; i < Math.Pow(2, nums.Length); i++)
    {
        int index = nums.Length - 1;
        int temp = i;
        int countOr = 0;
        while (index >= 0)
        {
            if (temp % 2 == 1)
            {
                countOr |= nums[index];
            }
            index--;
            temp /= 2;
        }

        if (!sortedSets.ContainsKey(countOr)) sortedSets.Add(countOr, 0);
        sortedSets[countOr]++;
    }

    return sortedSets.Last().Value;
}