How do you solve Count Number of Maximum Bitwise-OR Subsets on LeetCode?

Count Number of Maximum Bitwise-OR Subsets (LeetCode #2044) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The maximum bitwise OR can be obtained by ORing all elements together.

What is the brute force solution for Count Number of Maximum Bitwise-OR Subsets?

The brute force approach for Count Number of Maximum Bitwise-OR Subsets is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible subsets of the array and calculate their bitwise OR. This will allow us to find the maximum bitwise OR and count how many subsets achieve it. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Count Number of Maximum Bitwise-OR Subsets?

Always clarify if subsets should be non-empty. Consider edge cases, such as arrays with duplicate elements. Practice generating subsets and counting them to understand the problem better.

What are common mistakes when solving Count Number of Maximum Bitwise-OR Subsets?

Forgetting to consider all elements when calculating the maximum OR. Not correctly counting subsets formed by unique contributing elements.

#2044

Count Number of Maximum Bitwise-OR Subsets

Medium

ArrayBacktrackingBit ManipulationEnumerationBit ManipulationEnumeration

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

Instead of generating all subsets, we can find the maximum bitwise OR directly by ORing all elements together. Then, we count how many subsets can achieve this maximum OR using the unique elements contributing to it.

⚙️

Algorithm

3 steps

1Step 1: Calculate the maximum bitwise OR by ORing all elements in the array.
2Step 2: Identify which elements contribute to this maximum OR.
3Step 3: Count the number of subsets formed by these elements using the formula 2^k - 1, where k is the count of unique elements contributing to the maximum OR.

solution.py13 lines

1# Full working Python code
2def countMaxBitwiseOR(nums):
3    max_or = 0
4    for num in nums:
5        max_or |= num
6    count = 0
7    for num in nums:
8        if (max_or & num) == num:
9            count += 1
10    return (1 << count) - 1
11
12# Example usage
13print(countMaxBitwiseOR([3, 1]))  # Output: 2

ℹ

Complexity note: The time complexity is O(n) because we traverse the array twice: once to calculate the maximum OR and once to count the contributing elements. The space complexity is O(1) as we only use a few variables.

1The maximum bitwise OR can be obtained by ORing all elements together.
2Counting subsets can be simplified by focusing on the unique elements that contribute to the maximum OR.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.