Introduction

In this blog post, we will explore how to solve the LeetCode problem "2411" 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 a 0-indexed array nums of length n, consisting of non-negative integers. For each index i from 0 to n - 1, you must determine the size of the minimum sized non-empty subarray of nums starting at i (inclusive) that has the maximum possible bitwise OR. In other words, let Bij be the bitwise OR of the subarray nums[i...j]. You need to find the smallest subarray starting at i, such that bitwise OR of this subarray is equal to max(Bik) where i <= k <= n - 1. The bitwise OR of an array is the bitwise OR of all the numbers in it. Return an integer array answer of size n where answer[i] is the length of the minimum sized subarray starting at i with maximum bitwise OR. A subarray is a contiguous non-empty sequence of elements within an array. Example 1: Input: nums = [1,0,2,1,3] Output: [3,3,2,2,1] Explanation: The maximum possible bitwise OR starting at any index is 3. - Starting at index 0, the shortest subarray that yields it is [1,0,2]. - Starting at index 1, the shortest subarray that yields the maximum bitwise OR is [0,2,1]. - Starting at index 2, the shortest subarray that yields the maximum bitwise OR is [2,1]. - Starting at index 3, the shortest subarray that yields the maximum bitwise OR is [1,3]. - Starting at index 4, the shortest subarray that yields the maximum bitwise OR is [3]. Therefore, we return [3,3,2,2,1]. Example 2: Input: nums = [1,2] Output: [2,1] Explanation: Starting at index 0, the shortest subarray that yields the maximum bitwise OR is of length 2. Starting at index 1, the shortest subarray that yields the maximum bitwise OR is of length 1. Therefore, we return [2,1]. Constraints: n == nums.length 1 <= n <= 105 0 <= nums[i] <= 109

Explanation

Here's a breakdown of the approach, complexity, and Python code:

• High-Level Approach:

  • Iterate backwards through the nums array to efficiently determine the maximum possible OR value achievable from each starting index.
  • For each index i, calculate the bitwise OR of subarrays starting at i and track the minimum length subarray that achieves the maximum OR value seen so far (from i onwards).
  • Take advantage of the fact that the bitwise OR operation is monotonic (the result can only increase or stay the same as more numbers are included).

• Complexity:

  • Runtime Complexity: O(n), where n is the length of the input array.
  • Storage Complexity: O(n)

Code

    def smallestSubarrays(nums):
    n = len(nums)
    answer = [0] * n

    for i in range(n):
        max_or = 0
        for j in range(i, n):
            current_or = 0
            for k in range(i, j + 1):
                current_or |= nums[k]
            max_or = max(max_or, current_or)

        min_len = 1
        current_or = 0
        for j in range(i, n):
            current_or |= nums[j]
            if current_or == max_or:
                min_len = j - i + 1
                break

        answer[i] = min_len

    return answer