How do you solve Smallest Subarrays With Maximum Bitwise OR on LeetCode?

Smallest Subarrays With Maximum Bitwise OR (LeetCode #2411) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding bitwise operations is crucial for this problem.

What is the brute force solution for Smallest Subarrays With Maximum Bitwise OR?

The brute force approach for Smallest Subarrays With Maximum Bitwise OR is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible subarray starting from each index to find the maximum bitwise OR and its smallest length. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Smallest Subarrays With Maximum Bitwise OR?

Explain your thought process clearly while coding. Discuss edge cases like arrays with all zeros or single elements. Practice bit manipulation problems to build confidence.

What are common mistakes when solving Smallest Subarrays With Maximum Bitwise OR?

Not considering all bits when calculating maxOR. Overlooking the need to track the last seen index of bits.

#2411

Smallest Subarrays With Maximum Bitwise OR

Medium

ArrayBinary SearchBit ManipulationSliding WindowSliding WindowBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach uses a sliding window technique to efficiently find the smallest subarray with the maximum bitwise OR. By tracking the maximum OR and the last seen index of each bit, we can quickly determine the required lengths.

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Algorithm

3 steps

1Step 1: Initialize an array answer to store results and a variable maxOR to track the maximum bitwise OR.
2Step 2: Iterate through the array from the end to the beginning, updating maxOR with the current element's OR.
3Step 3: For each bit in maxOR, find the last index where that bit was set and calculate the length of the subarray from the current index to that last index.

solution.py13 lines

1def smallestSubarrays(nums):
2    n = len(nums)
3    answer = [0] * n
4    maxOR = 0
5    lastIndex = [-1] * 32
6    for i in range(n - 1, -1, -1):
7        maxOR |= nums[i]
8        answer[i] = n
9        for b in range(32):
10            if (maxOR >> b) & 1:
11                lastIndex[b] = i
12                answer[i] = min(answer[i], lastIndex[b] - i + 1)
13    return answer

ℹ

Complexity note: The time complexity is O(n) because we traverse the array once and check each bit position (constant time). The space complexity is O(n) due to the answer array.

1Understanding bitwise operations is crucial for this problem.
2Using a sliding window can significantly reduce the time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.