Can you solve this real interview question? Count Number of Maximum Bitwise-OR Subsets - 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]
class Solution:
def backtrack(self, nums, index, currentOR, maxOR, count):
if currentOR == maxOR:
count[0] += 1
for i in range(index, len(nums)):
self.backtrack(nums, i + 1, currentOR | nums[i], maxOR, count)
def countMaxOrSubsets(self, nums: List[int]) -> int:
maxOR = 0
# Step 1: Compute the maximum OR
for num in nums:
maxOR |= num
count = [0]
# Step 2: Backtrack to count the subsets
self.backtrack(nums, 0, 0, maxOR, count)
return count[0]