How do you solve Find Subarray With Bitwise OR Closest to K on LeetCode?

Find Subarray With Bitwise OR Closest to K (LeetCode #3171) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The bitwise OR operation can only add bits, never remove them, which allows for efficient tracking of OR values.

What is the brute force solution for Find Subarray With Bitwise OR Closest to K?

The brute force approach for Find Subarray With Bitwise OR Closest to K is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible subarrays and calculating their bitwise OR. This is straightforward but inefficient for larger arrays, as it requires checking every combination. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Find Subarray With Bitwise OR Closest to K?

Always explain your thought process while coding; it shows your understanding. Consider edge cases, such as single-element arrays or all elements being the same. Practice optimizing your solutions, as interviewers often look for efficiency.

What are common mistakes when solving Find Subarray With Bitwise OR Closest to K?

Not considering the properties of bitwise OR, leading to unnecessary recalculations. Failing to optimize the search space with two pointers, resulting in a brute force approach.

#3171

Find Subarray With Bitwise OR Closest to K

Hard

ArrayBinary SearchBit ManipulationSegment TreeSliding WindowTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages the properties of bitwise OR and a sliding window approach to efficiently find the closest OR value to k without recalculating the OR for every subarray.

⚙️

Algorithm

6 steps

1Step 1: Initialize a variable to store the minimum absolute difference as infinity.
2Step 2: Use two pointers to represent the current subarray's start and end.
3Step 3: Expand the end pointer to include more elements and calculate the bitwise OR.
4Step 4: For each new OR value, calculate the absolute difference with k and update the minimum difference.
5Step 5: If the OR value exceeds k, move the start pointer to reduce the OR value.
6Step 6: Repeat until all elements are processed.

solution.py17 lines

1# Full working Python code
2
3def min_abs_diff_optimal(nums, k):
4    min_diff = float('inf')
5    n = len(nums)
6    current_or = 0
7    start = 0
8    for end in range(n):
9        current_or |= nums[end]
10        while start <= end and current_or > k:
11            current_or ^= nums[start]
12            start += 1
13        min_diff = min(min_diff, abs(current_or - k))
14    return min_diff
15
16# Example usage
17print(min_abs_diff_optimal([1, 2, 4, 5], 3))  # Output: 0

ℹ

Complexity note: The time complexity is O(n) because we traverse the array with two pointers, ensuring each element is processed a limited number of times. The space complexity is O(1) as we only use a few variables for tracking the current state.

1The bitwise OR operation can only add bits, never remove them, which allows for efficient tracking of OR values.
2Using a sliding window helps to dynamically adjust the subarray without recalculating from scratch.

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