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
Solution: Brute Force
Try all possible subsets
Time complexity: O(n*2n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: int countMaxOrSubsets(vector<int>& nums) { const int n = nums.size(); int max_or = 0; int count = 0; for (int s = 0; s < 1 << n; ++s) { int cur_or = 0; for (int i = 0; i < n; ++i) if (s >> i & 1) cur_or |= nums[i]; if (cur_or > max_or) { max_or = cur_or; count = 1; } else if (cur_or == max_or) { ++count; } } return count; } }; |
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