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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:

Solution 1. Bitmask

  1. 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.
  2. 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
        }
    }
    
    

Discuss

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

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