Is Count Number of Maximum Bitwise-OR Subsets easy or hard?

The problem is categorized as Medium on LeetCode. The implementation is straightforward once you recognize that the maximum OR equals the OR of all elements. The main challenge is efficiently enumerating subsets using backtracking or bitmasking.

Count Number of Maximum Bitwise-OR Subsets Python/Java solution

In Python or Java, compute the target OR by OR-ing all elements in the array. Then generate subsets using recursion or iterate through bitmasks from 0 to (1 << n) - 1. For each subset, calculate the OR and increase the counter if it equals the maximum. Both implementations run in O(2^n) time.

How to solve Count Number of Maximum Bitwise-OR Subsets in O(n)?

An O(n) solution does not exist because the problem inherently requires examining subsets. The optimal strategy enumerates all 2^n subsets and checks their OR values. The typical implementation uses backtracking or bitmask enumeration, giving O(2^n) time complexity.

What is the best approach for Count Number of Maximum Bitwise-OR Subsets?

The most practical approach is subset enumeration using backtracking. First compute the maximum OR by OR-ing all numbers in the array. Then recursively generate all subsets while tracking the running OR value. Whenever the OR equals the target maximum, increment the count. The algorithm runs in O(2^n) time with O(n) recursion stack space.

Is Count Number of Maximum Bitwise-OR Subsets asked at Google/Amazon/Meta?

Bit manipulation and subset enumeration problems frequently appear in interviews at companies like Amazon, Google, and Meta. Variations of subset OR, AND, or XOR enumeration are common because they test understanding of recursion, bit operations, and combinatorial reasoning.

What data structure is used in Count Number of Maximum Bitwise-OR Subsets?

The solution mainly relies on arrays and bitwise operations rather than complex data structures. Backtracking uses the call stack to explore subsets, while the iterative method uses integer bitmasks to represent subsets.

What is the time complexity of Count Number of Maximum Bitwise-OR Subsets?

The time complexity is O(2^n) because every subset of the array must be evaluated. With recursive enumeration, each subset is processed once and only a constant-time OR operation is applied at each step. Space complexity is O(n) due to the recursion depth.

Description Hints 2 Solution Video 21:07

2044. Count Number of Maximum Bitwise-OR Subsets

Q: How to solve Count Number of Maximum Bitwise-OR Subsets in O(n)?

An O(n) solution does not exist because the problem inherently requires examining subsets. The optimal strategy enumerates all 2^n subsets and checks their OR values. The typical implementation uses backtracking or bitmask enumeration, giving O(2^n) time complexity.

Q: What is the best approach for Count Number of Maximum Bitwise-OR Subsets?

The most practical approach is subset enumeration using backtracking. First compute the maximum OR by OR-ing all numbers in the array. Then recursively generate all subsets while tracking the running OR value. Whenever the OR equals the target maximum, increment the count. The algorithm runs in O(2^n) time with O(n) recursion stack space.

Q: Is Count Number of Maximum Bitwise-OR Subsets asked at Google/Amazon/Meta?

Bit manipulation and subset enumeration problems frequently appear in interviews at companies like Amazon, Google, and Meta. Variations of subset OR, AND, or XOR enumeration are common because they test understanding of recursion, bit operations, and combinatorial reasoning.

Q: What data structure is used in Count Number of Maximum Bitwise-OR Subsets?

The solution mainly relies on arrays and bitwise operations rather than complex data structures. Backtracking uses the call stack to explore subsets, while the iterative method uses integer bitmasks to represent subsets.

Q: What is the time complexity of Count Number of Maximum Bitwise-OR Subsets?

The time complexity is O(2^n) because every subset of the array must be evaluated. With recursive enumeration, each subset is processed once and only a constant-time OR operation is applied at each step. Space complexity is O(n) due to the recursion depth.

Medium89.5% acceptance7 companies

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: