How do you solve Bitwise ORs of Subarrays on LeetCode?

Bitwise ORs of Subarrays (LeetCode #898) 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: The bitwise OR operation accumulates bits, meaning once a bit is set, it remains set for all subsequent ORs.

What is the brute force solution for Bitwise ORs of Subarrays?

The brute force approach for Bitwise ORs of Subarrays is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves generating all possible subarrays and calculating their bitwise ORs. This is straightforward but inefficient for larger arrays due to the nested loops required to explore all subarrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Bitwise ORs of Subarrays?

Bitwise ORs of Subarrays uses the following concepts: Array, Dynamic Programming, Bit Manipulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Bitwise ORs of Subarrays?

Always consider edge cases, such as arrays with all identical elements or maximum values. Explain your thought process clearly; this helps interviewers understand your approach. Practice writing code without an IDE to simulate real interview conditions.

What are common mistakes when solving Bitwise ORs of Subarrays?

Failing to recognize that the bitwise OR can only grow or stay the same, leading to unnecessary calculations. Not using a set to track unique results, which can lead to incorrect counts.

#898

Bitwise ORs of Subarrays

Medium

ArrayDynamic ProgrammingBit ManipulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal approach leverages the properties of the bitwise OR operation, where the result can only grow as we include more elements. By maintaining a running OR and using a set to track distinct results, we can efficiently compute the unique ORs without generating all subarrays.

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Algorithm

3 steps

1Step 1: Initialize a set to store unique bitwise OR results.
2Step 2: Use a variable to keep track of the current OR and an array to store the last seen ORs for each position.
3Step 3: Iterate through the array, updating the current OR and adding it to the set, while also updating the last seen ORs.

solution.py12 lines

1def subarrayBitwiseOR(arr):
2    unique_or = set()
3    current_or = 0
4    last_seen = []
5    for num in arr:
6        current_or = 0
7        for last in last_seen:
8            current_or |= last
9        current_or |= num
10        unique_or.add(current_or)
11        last_seen.append(current_or)
12    return len(unique_or)

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array once. The space complexity is O(n) due to the storage of unique OR results in a set.

1The bitwise OR operation accumulates bits, meaning once a bit is set, it remains set for all subsequent ORs.
2Distinct results can be efficiently tracked using a set, avoiding duplicates.

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