Formatted question description: https://leetcode.ca/all/2044.html

2044. Count Number of Maximum Bitwise-OR Subsets (Medium)

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

Similar Questions:

Subsets (Medium)

Solution 1. Bitmask

Compute goal which is the maximum possible bitwise OR of a subset of nums, i.e. the bitwise OR of all the numbers in nums.
Enumerate all non-empty subsets of nums using bitmask and compute the bitwise OR of each of them. Increment answer if the subset’s bitwise OR is the same as goal.

// OJ: https://leetcode.com/problems/count-number-of-maximum-bitwise-or-subsets/
// Time: O(2^N * N)
// Space: O(1)
class Solution {
public:
    int countMaxOrSubsets(vector<int>& A) {
        int goal = 0, N = A.size(), ans = 0;
        for (int n : A) goal |= n;
        for (int m = 1; m < 1 << N; ++m) {
            int x = 0;
            for (int i = 0; i < N; ++i) {
                if (m >> i & 1) x |= A[i];
            }
            if (x == goal) ++ans;
        }
        return ans;
    }
};

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

############

# 2044. Count Number of Maximum Bitwise-OR Subsets
# https://leetcode.com/problems/count-number-of-maximum-bitwise-or-subsets/

class Solution:
    def countMaxOrSubsets(self, nums: List[int]) -> int:
        n = len(nums)
        mp = collections.defaultdict(int)
        mmax = float('-inf')
        
        for mask in range(1, 1 << n):
            s = 0
            for j in range(n):
                if (mask >> j) & 1:
                    s |= nums[j]

            mp[s] += 1
            mmax = max(mmax, s)

        return mp[mmax]

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]);
    }
}

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
    }
}

We can use DP to reduce the time complexity at the cost of space complexity

// OJ: https://leetcode.com/problems/count-number-of-maximum-bitwise-or-subsets/
// Time: O(2^N)
// Space: O(2^N)
class Solution {
public:
    int countMaxOrSubsets(vector<int>& A) {
        int goal = 0, N = A.size(), ans = 0;
        vector<int> dp(1 << N);
        for (int n : A) goal |= n;
        for (int m = 1; m < 1 << N; ++m) {
            int lowbit = m & -m;
            dp[m] = dp[m - lowbit] | A[__builtin_ctz(lowbit)];
            if (dp[m] == goal) ++ans;
        }
        return ans;
    }
};

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

############

# 2044. Count Number of Maximum Bitwise-OR Subsets
# https://leetcode.com/problems/count-number-of-maximum-bitwise-or-subsets/

class Solution:
    def countMaxOrSubsets(self, nums: List[int]) -> int:
        n = len(nums)
        mp = collections.defaultdict(int)
        mmax = float('-inf')
        
        for mask in range(1, 1 << n):
            s = 0
            for j in range(n):
                if (mask >> j) & 1:
                    s |= nums[j]

            mp[s] += 1
            mmax = max(mmax, s)

        return mp[mmax]

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]);
    }
}

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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https://leetcode.com/problems/count-number-of-maximum-bitwise-or-subsets/discuss/1525216/C%2B%2B-Bitmask-%2B-DP

2044 - Count Number of Maximum Bitwise-OR Subsets

2044. Count Number of Maximum Bitwise-OR Subsets (Medium)

Solution 1. Bitmask

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2044. Count Number of Maximum Bitwise-OR Subsets (Medium)

Solution 1. Bitmask

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