2411. Smallest Subarrays With Maximum Bitwise OR
Description
You are given a 0-indexed array nums
of length n
, consisting of non-negative integers. For each index i
from 0
to n - 1
, you must determine the size of the minimum sized non-empty subarray of nums
starting at i
(inclusive) that has the maximum possible bitwise OR.
- In other words, let
Bij
be the bitwise OR of the subarraynums[i...j]
. You need to find the smallest subarray starting ati
, such that bitwise OR of this subarray is equal tomax(Bik)
wherei <= k <= n - 1
.
The bitwise OR of an array is the bitwise OR of all the numbers in it.
Return an integer array answer
of size n
where answer[i]
is the length of the minimum sized subarray starting at i
with maximum bitwise OR.
A subarray is a contiguous non-empty sequence of elements within an array.
Example 1:
Input: nums = [1,0,2,1,3] Output: [3,3,2,2,1] Explanation: The maximum possible bitwise OR starting at any index is 3. - Starting at index 0, the shortest subarray that yields it is [1,0,2]. - Starting at index 1, the shortest subarray that yields the maximum bitwise OR is [0,2,1]. - Starting at index 2, the shortest subarray that yields the maximum bitwise OR is [2,1]. - Starting at index 3, the shortest subarray that yields the maximum bitwise OR is [1,3]. - Starting at index 4, the shortest subarray that yields the maximum bitwise OR is [3]. Therefore, we return [3,3,2,2,1].
Example 2:
Input: nums = [1,2] Output: [2,1] Explanation: Starting at index 0, the shortest subarray that yields the maximum bitwise OR is of length 2. Starting at index 1, the shortest subarray that yields the maximum bitwise OR is of length 1. Therefore, we return [2,1].
Constraints:
n == nums.length
1 <= n <= 105
0 <= nums[i] <= 109
Solution
smallest-subarrays-with-maximum-bitwise-or.py
class Solution:
def smallestSubarrays(self, nums: List[int]) -> List[int]:
N = len(nums)
bits = [0] * 32
res = [0] * N
j = N - 1
curr = target = 0
for i in range(N - 1, -1, -1):
target |= nums[i]
curr |= nums[i]
for k in range(31, -1, -1):
if (nums[i] & (1 << k)) > 0:
bits[k] += 1
while curr == target and i <= j:
for k in range(31, -1, -1):
if (nums[j] & (1 << k)) > 0:
bits[k] -= 1
if bits[k] == 0:
curr ^= (1 << k)
j -= 1
res[i] = j - i + 2
return res