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

Shortest Subarray With OR at Least K II (LeetCode #3097) 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: Bitwise OR accumulates bits, never unsetting them.

What is the time complexity of Shortest Subarray With OR at Least K II?

The optimal solution for Shortest Subarray With OR at Least K II runs in O(n) time and uses O(n) space. This complexity is achieved because we only traverse the array once, and the deque operations are amortized constant time.

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

The brute force approach for Shortest Subarray With OR at Least K II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible subarray to see if its bitwise OR meets or exceeds the value of k. This is straightforward but inefficient, especially 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 II?

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

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

Always clarify the problem constraints and edge cases. Discuss your thought process and approach before jumping into coding. Optimize your solution step-by-step, explaining why each change improves efficiency.

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

Not handling edge cases where no valid subarray exists. Incorrectly calculating the OR value by not resetting it properly.

#3097

Shortest Subarray With OR at Least K II

Medium

ArrayBit ManipulationSliding WindowSliding WindowDeque

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 maintain a dynamic subarray and efficiently compute the bitwise OR. This reduces the time complexity significantly by avoiding redundant calculations.

⚙️

Algorithm

5 steps

1Step 1: Initialize a deque to store indices and a variable for the current OR value.
2Step 2: Iterate through the array, updating the current OR value with each element.
3Step 3: While the current OR is at least k, update the minimum length and remove elements from the front of the deque.
4Step 4: Add the current index to the deque.
5Step 5: Return the minimum length found or -1 if no valid subarray exists.

solution.py15 lines

1# Full working Python code
2from collections import deque
3
4def shortestSubarray(nums, k):
5    n = len(nums)
6    min_length = float('inf')
7    current_or = 0
8    dq = deque()
9    for i in range(n):
10        current_or |= nums[i]
11        dq.append(i)
12        while current_or >= k:
13            min_length = min(min_length, dq[-1] - dq[0] + 1)
14            current_or ^= nums[dq.popleft()]  # Remove the leftmost element
15    return min_length if min_length != float('inf') else -1

ℹ

Complexity note: This complexity is achieved because we only traverse the array once, and the deque operations are amortized constant time.

1Bitwise OR accumulates bits, never unsetting them.
2Using a sliding window allows for efficient tracking of valid subarrays.

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