Introduction

In this blog post, we will explore how to solve the LeetCode problem "3097" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an array nums of non-negative integers and an integer k. An array is called special if the bitwise OR of all of its elements is at least k. Return the length of the shortest special non-empty subarray of nums, or return -1 if no special subarray exists. Example 1: Input: nums = [1,2,3], k = 2 Output: 1 Explanation: The subarray [3] has OR value of 3. Hence, we return 1. Example 2: Input: nums = [2,1,8], k = 10 Output: 3 Explanation: The subarray [2,1,8] has OR value of 11. Hence, we return 3. Example 3: Input: nums = [1,2], k = 0 Output: 1 Explanation: The subarray [1] has OR value of 1. Hence, we return 1. Constraints: 1 <= nums.length <= 2 * 105 0 <= nums[i] <= 109 0 <= k <= 109

Explanation

• Sliding Window: Use a sliding window approach to efficiently explore all possible subarrays. Expand the window until the bitwise OR of elements within the window is greater than or equal to k.
  • Minimize Window Size: Once a valid subarray is found, contract the window from the left to find the smallest possible valid subarray.
  • Bitwise OR Accumulation: Maintain a running bitwise OR of the elements within the window to avoid redundant calculations.

• Time Complexity: O(n), Space Complexity: O(1)

Code

    def shortest_subarray(nums, k):
    n = len(nums)
    min_len = float('inf')
    left = 0
    current_or = 0

    for right in range(n):
        current_or |= nums[right]

        while current_or >= k:
            min_len = min(min_len, right - left + 1)
            current_or ^= nums[left]  # Remove nums[left] from current_or
            left += 1
            if left > right and left < n :
              current_or=0 # reset the or
              for i in range(left,right+1):
                 current_or |= nums[i]
              break



    if min_len == float('inf'):
        return -1
    else:
        return min_len