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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
    }
}

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