Introduction

In this blog post, we will explore how to solve the LeetCode problem "2401" 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 consisting of positive integers. We call a subarray of nums nice if the bitwise AND of every pair of elements that are in different positions in the subarray is equal to 0. Return the length of the longest nice subarray. A subarray is a contiguous part of an array. Note that subarrays of length 1 are always considered nice. Example 1: Input: nums = [1,3,8,48,10] Output: 3 Explanation: The longest nice subarray is [3,8,48]. This subarray satisfies the conditions: - 3 AND 8 = 0. - 3 AND 48 = 0. - 8 AND 48 = 0. It can be proven that no longer nice subarray can be obtained, so we return 3. Example 2: Input: nums = [3,1,5,11,13] Output: 1 Explanation: The length of the longest nice subarray is 1. Any subarray of length 1 can be chosen. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 109

Explanation

• Sliding Window: Use a sliding window to maintain a candidate subarray.
  • Bitwise AND Check: Efficiently check if the bitwise AND of all pairs within the window is zero using a running bitwise OR.
  • Window Adjustment: Expand the window to the right, and shrink from the left if the condition is violated.

• Runtime Complexity: O(n), Storage Complexity: O(1)

Code

    def longestNiceSubarray(nums):
    """
    Finds the length of the longest nice subarray.

    Args:
        nums: A list of positive integers.

    Returns:
        The length of the longest nice subarray.
    """

    left = 0
    current_or = 0
    max_len = 0

    for right in range(len(nums)):
        while (current_or & nums[right]) != 0:
            current_or ^= nums[left]
            left += 1

        current_or |= nums[right]
        max_len = max(max_len, right - left + 1)

    return max_len