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

Solution 1: DFS

The maximum bitwise OR value \(\textit{mx}\) in the array \(\textit{nums}\) can be obtained by performing bitwise OR on all elements in the array.

Then we can use depth-first search to enumerate all subsets and count the number of subsets whose bitwise OR equals \(\textit{mx}\). We design a function \(\text{dfs(i, t)}\), which represents the number of subsets starting from index \(\textit{i}\) with the current bitwise OR value being \(\textit{t}\). Initially, \(\textit{i} = 0\) and \(\textit{t} = 0\).

In the function \(\text{dfs(i, t)}\), if \(\textit{i}\) equals the array length, it means we have enumerated all elements. At this point, if \(\textit{t}\) equals \(\textit{mx}\), we increment the answer by one. Otherwise, we can choose to either exclude the current element \(\textit{nums[i]}\) or include the current element \(\textit{nums[i]}\), so we can recursively call \(\text{dfs(i + 1, t)}\) and \(\text{dfs(i + 1, t | nums[i])}\).

Finally, we return the answer.

The time complexity is \(O(2^n)\), and the space complexity is \(O(n)\), where \(n\) is the length of the array \(\textit{nums}\).

Python3JavaC++GoTypeScriptRust

class Solution:
    def countMaxOrSubsets(self, nums: List[int]) -> int:
        def dfs(i, t):
            nonlocal ans, mx
            if i == len(nums):
                if t == mx:
                    ans += 1
                return
            dfs(i + 1, t)
            dfs(i + 1, t | nums[i])

        ans = 0
        mx = reduce(lambda x, y: x | y, nums)
        dfs(0, 0)
        return ans

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 countMaxOrSubsets(vector<int>& nums) {
        int ans = 0;
        int mx = accumulate(nums.begin(), nums.end(), 0, bit_or<int>());
        auto dfs = [&](this auto&& dfs, int i, int t) {
            if (i == nums.size()) {
                if (t == mx) {
                    ans++;
                }
                return;
            }
            dfs(i + 1, t);
            dfs(i + 1, t | nums[i]);
        };
        dfs(0, 0);
        return ans;
    }
};

func countMaxOrSubsets(nums []int) (ans int) {
    mx := 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
}

function countMaxOrSubsets(nums: number[]): number {
    let ans = 0;
    const mx = nums.reduce((x, y) => x | y, 0);

    const dfs = (i: number, t: number) => {
        if (i === nums.length) {
            if (t === mx) {
                ans++;
            }
            return;
        }
        dfs(i + 1, t);
        dfs(i + 1, t | nums[i]);
    };

    dfs(0, 0);
    return ans;
}

impl Solution {
    pub fn count_max_or_subsets(nums: Vec<i32>) -> i32 {
        let mut ans = 0;
        let mx = nums.iter().fold(0, |x, &y| x | y);

        fn dfs(i: usize, t: i32, nums: &Vec<i32>, mx: i32, ans: &mut i32) {
            if i == nums.len() {
                if t == mx {
                    *ans += 1;
                }
                return;
            }
            dfs(i + 1, t, nums, mx, ans);
            dfs(i + 1, t | nums[i], nums, mx, ans);
        }

        dfs(0, 0, &nums, mx, &mut ans);
        ans
    }
}

Solution 2: Binary Enumeration

We can use binary enumeration to count the bitwise OR results of all subsets. For an array \(\textit{nums}\) of length \(n\), we can use an integer \(\textit{mask}\) to represent a subset, where the \(i\)-th bit of \(\textit{mask}\) being 1 means including element \(\textit{nums[i]}\), and 0 means not including it.

We can iterate through all possible \(\textit{mask}\) values from \(0\) to \(2^n - 1\). For each \(\textit{mask}\), we can calculate the bitwise OR result of the corresponding subset and update the maximum value \(\textit{mx}\) and answer \(\textit{ans}\).

The time complexity is \(O(2^n \cdot n)\), where \(n\) is the length of the array \(\textit{nums}\). The space complexity is \(O(1)\).