# 2044. Count Number of Maximum Bitwise-OR Subsets

## Description

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

DFS.

• class Solution {
private int mx;
private int ans;
private int[] nums;

public int countMaxOrSubsets(int[] nums) {
mx = 0;
for (int x : nums) {
mx |= x;
}
this.nums = nums;
dfs(0, 0);
return ans;
}

private void dfs(int i, int t) {
if (i == nums.length) {
if (t == mx) {
++ans;
}
return;
}
dfs(i + 1, t);
dfs(i + 1, t | nums[i]);
}
}

• class Solution {
public:
int mx;
int ans;
vector<int> nums;

int countMaxOrSubsets(vector<int>& nums) {
this->nums = nums;
mx = 0;
ans = 0;
for (int x : nums) mx |= x;
dfs(0, 0);
return ans;
}

void dfs(int i, int t) {
if (i == nums.size()) {
if (t == mx) ++ans;
return;
}
dfs(i + 1, t);
dfs(i + 1, t | nums[i]);
}
};

• class Solution:
def countMaxOrSubsets(self, nums: List[int]) -> int:
mx = ans = 0
for x in nums:
mx |= x

def dfs(i, t):
nonlocal mx, ans
if i == len(nums):
if t == mx:
ans += 1
return
dfs(i + 1, t)
dfs(i + 1, t | nums[i])

dfs(0, 0)
return ans


• func countMaxOrSubsets(nums []int) int {
mx, ans := 0, 0
for _, x := range nums {
mx |= x
}

var dfs func(i, t int)
dfs = func(i, t int) {
if i == len(nums) {
if t == mx {
ans++
}
return
}
dfs(i+1, t)
dfs(i+1, t|nums[i])
}

dfs(0, 0)
return ans
}

• function countMaxOrSubsets(nums: number[]): number {
let n = nums.length;
let max = 0;
for (let i = 0; i < n; i++) {
max |= nums[i];
}
let ans = 0;
function dfs(pre: number, depth: number): void {
if (depth == n) {
if (pre == max) ++ans;
return;
}
dfs(pre, depth + 1);
dfs(pre | nums[depth], depth + 1);
}
dfs(0, 0);
return ans;
}


• impl Solution {
fn dfs(nums: &Vec<i32>, i: usize, sum: i32) -> (i32, i32) {
let n = nums.len();
let mut max = i32::MIN;
let mut res = 0;
for j in i..n {
let num = sum | nums[j];
if num >= max {
if num > max {
max = num;
res = 0;
}
res += 1;
}
let (r_max, r_res) = Self::dfs(nums, j + 1, num);
if r_max >= max {
if r_max > max {
max = r_max;
res = 0;
}
res += r_res;
}
}
(max, res)
}

pub fn count_max_or_subsets(nums: Vec<i32>) -> i32 {
Self::dfs(&nums, 0, 0).1
}
}