2170 Count Number Of Maximum Bitwise Or Subsets

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

impl Solution {
    pub fn count_max_or_subsets(nums: Vec<i32>) -> i32 {
        // bitwise or is max if done for all elements ?
        //length is 16 
        // so there can be 2^16 subsets , which is quite lrge 
        // but maybe still under pow(10,5) ? IDK 
        // idea is if there is a 1 at a ceratin place 
        // we could exclude the other 1 from other integeres
        // say we have an array of 16
        // we could all 1s at each place,
        // indexes with just 1 1s are mandatory
        // rest are optional
        // final answer would be (optional + 1)*(mandatory count)

        // 010 101

        let mut final_result = 0;
        let mut result = 0;
        for k in &nums {
            final_result |= k;
        }
        println!("{}", final_result);

        fn subset( arr : &Vec<i32>, total : i32, final_result : i32, mut result : &mut i32, index : i32) {
            //println!("{} {} {} {}", total, final_result, result, index);
            if total == final_result && index == arr.len() as i32  {
                *result = *result + 1;
            }
            if index >= arr.len() as i32 {
                return 
            }
            subset(arr, total | arr[index as usize] , final_result, result, index + 1 );
            subset(arr, total , final_result, result, index + 1 );
            

        }
        subset(&nums, 0, final_result, &mut result, 0);
        return result;
    }
}