Count Number of Maximum Bitwise-OR Subsets — Medium Problem & Solution

Return the number of non-empty subsets of nums whose bitwise OR is the maximum achievable.

Difficulty: Medium
Topics: Bit Manipulation, Backtracking, Enumeration
Asked at: Amazon, Google, Samsung
Time limit: 2 s
Memory limit: 256 MB
Languages: JavaScript, TypeScript, Python, Java, C++, C, C#, Go, Kotlin, Swift, Rust, PHP and Ruby

Problem statement

Return the number of non-empty subsets of nums whose bitwise OR is the maximum achievable.

Two subsets are different when they use different index sets, even if the values coincide.

Example 1

Input: nums = [3,1]
Output: 2
Explanation: The maximum OR is 3, reached by {3} and {3,1}.

Example 2

Input: nums = [2,2,2]
Output: 7
Explanation: Every non-empty subset ORs to 2.

Example 3

Input: nums = [3,2,1,5]
Output: 6

Constraints

1 <= nums.length <= 16
1 <= nums[i] <= 100000

How to solve Count Number of Maximum Bitwise-OR Subsets

The best OR is the OR of everything, since adding elements never clears a bit. Counting the subsets that reach it is then a brute-force enumeration over the 2^n - 1 non-empty index masks.

Approach

Compute best, the OR of all elements.
For each mask from 1 to 2^n - 1, OR together the selected elements.
Count the masks whose OR equals best.

Why it works

Bitwise OR is monotone under adding elements, so no subset can exceed the full-array OR, and the full array itself attains it. With n <= 16 the enumeration is at most about a million operations.

Complexity

Time — O(2^n · n)
Space — O(1)

Pitfalls

Starting the mask at 0 counts the empty subset, which the statement excludes.
1 << n with n = 16 is fine, but the loop bound must be exclusive.
Recomputing the OR from scratch per mask is fine here; the O(2^n) refinement carries it incrementally.

Reference solution

Python

from typing import List

def countMaxOrSubsets(nums: List[int]) -> int:
    best = 0
    for x in nums:
        best |= x
    n = len(nums)
    count = 0
    for mask in range(1, 1 << n):
        acc = 0
        for i in range(n):
            if (mask >> i) & 1:
                acc |= nums[i]
        if acc == best:
            count += 1
    return count

JavaScript

var countMaxOrSubsets = function(nums) {
    var best = 0;
    for (var i = 0; i < nums.length; i++) best |= nums[i];
    var n = nums.length, count = 0;
    for (var mask = 1; mask < (1 << n); mask++) {
        var acc = 0;
        for (var j = 0; j < n; j++) {
            if ((mask >> j) & 1) acc |= nums[j];
        }
        if (acc === best) count++;
    }
    return count;
};

Also on the editorial tab: C, C#, C++, Go, Java, Kotlin, PHP, Ruby, Rust, Swift, TypeScript.

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