Leetcode #2411: Smallest Subarrays With Maximum Bitwise OR
In this guide, we solve Leetcode #2411 Smallest Subarrays With Maximum Bitwise OR in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Bit Manipulation, Array, Binary Search, Sliding Window
Intuition
We are looking for a contiguous region that satisfies a constraint, which is a classic sliding-window signal.
Expanding and shrinking the window lets us maintain validity without restarting the scan.
Approach
Grow the window with a right pointer, and shrink from the left only when the constraint is violated.
Track the best window as you go to keep the solution linear.
Steps:
- Expand the right end of the window.
- While invalid, move the left end to restore constraints.
- Update the best window found.
Example
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].
Python Solution
class Solution:
def smallestSubarrays(self, nums: List[int]) -> List[int]:
n = len(nums)
ans = [1] * n
f = [-1] * 32
for i in range(n - 1, -1, -1):
t = 1
for j in range(32):
if (nums[i] >> j) & 1:
f[j] = i
elif f[j] != -1:
t = max(t, f[j] - i + 1)
ans[i] = t
return ans
Complexity
The time complexity is , where is the length of the array , and is the maximum value in the array . The space complexity is O(1) to O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.