README.md

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

Solutions (Rust)

1. Solution

use std::collections::HashMap;

impl Solution {
    pub fn count_max_or_subsets(nums: Vec<i32>) -> i32 {
        let mut count0 = HashMap::from([(0, 1)]);
        let mut count1 = HashMap::new();

        for &num in &nums {
            count1.clear();

            for (&x, &y) in count0.iter() {
                count1.entry(x | num).and_modify(|c| *c += y).or_insert(y);
            }

            for (&x, &y) in count1.iter() {
                count0.entry(x).and_modify(|c| *c += y).or_insert(y);
            }
        }

        count0[&nums.iter().fold(0, |a, b| a | b)]
    }
}