How do you solve Shortest Subarray With OR at Least K I on LeetCode?

Shortest Subarray With OR at Least K I (LeetCode #3095) 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 accumulates bits, meaning once a bit is set to 1, it stays 1 for the rest of the OR operation.

What is the brute force solution for Shortest Subarray With OR at Least K I?

The brute force approach for Shortest Subarray With OR at Least K I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks all possible subarrays to see if their bitwise OR meets or exceeds the given value k. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Shortest Subarray With OR at Least K I?

Shortest Subarray With OR at Least K I uses the following concepts: Array, Bit Manipulation, Sliding Window. The recommended patterns to study are: Sliding Window, Bit Manipulation.

What are the interview tips for Shortest Subarray With OR at Least K I?

Always start with a brute force solution to understand the problem better before optimizing. Be clear about how bitwise operations work, especially OR, as they can be tricky. Practice explaining your thought process and code as you write it, as this will help in interviews.

What are common mistakes when solving Shortest Subarray With OR at Least K I?

Failing to reset the OR value correctly when adjusting the left pointer in the optimal solution. Not considering single-element subarrays as valid special subarrays.

#3095

Shortest Subarray With OR at Least K I

Easy

ArrayBit ManipulationSliding WindowSliding WindowBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal approach uses a sliding window technique to efficiently find the shortest subarray with a bitwise OR of at least k. This reduces the number of checks needed compared to the brute force method.

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Algorithm

3 steps

1Step 1: Initialize two pointers for the sliding window and a variable to keep track of the current OR.
2Step 2: Expand the right pointer to include elements until the OR is at least k.
3Step 3: Once the OR is at least k, try to shrink the left pointer to minimize the window size while maintaining the OR condition.

solution.py14 lines

1# Full working Python code
2
3def shortest_subarray_with_or(nums, k):
4    n = len(nums)
5    left = 0
6    current_or = 0
7    min_length = float('inf')
8    for right in range(n):
9        current_or |= nums[right]
10        while current_or >= k:
11            min_length = min(min_length, right - left + 1)
12            current_or ^= nums[left]
13            left += 1
14    return min_length if min_length != float('inf') else -1

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Complexity note: The time complexity is O(n) because each element is processed at most twice (once by the right pointer and once by the left pointer). The space complexity is O(1) since we only use a few extra variables.

1The bitwise OR operation accumulates bits, meaning once a bit is set to 1, it stays 1 for the rest of the OR operation.
2Shortening the subarray while maintaining the OR condition is crucial for optimizing the solution.

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