# Solving Leetcode Interviews in Seconds with AI: Maximum Consecutive Floors Without Special Floors


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2274" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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
	> Alice manages a company and has rented some floors of a building as office space. Alice has decided some of these floors should be special floors, used for relaxation only. You are given two integers bottom and top, which denote that Alice has rented all the floors from bottom to top (inclusive). You are also given the integer array special, where special[i] denotes a special floor that Alice has designated for relaxation. Return the maximum number of consecutive floors without a special floor.   Example 1:  Input: bottom = 2, top = 9, special = [4,6] Output: 3 Explanation: The following are the ranges (inclusive) of consecutive floors without a special floor: - (2, 3) with a total amount of 2 floors. - (5, 5) with a total amount of 1 floor. - (7, 9) with a total amount of 3 floors. Therefore, we return the maximum number which is 3 floors.  Example 2:  Input: bottom = 6, top = 8, special = [7,6,8] Output: 0 Explanation: Every floor rented is a special floor, so we return 0.    Constraints:  1 <= special.length <= 105 1 <= bottom <= special[i] <= top <= 109 All the values of special are unique.  

	# Explanation
	Here's a breakdown of the solution:

*   **Sort Special Floors:** Sort the `special` array to easily find consecutive ranges.
*   **Calculate Differences:** Calculate the differences between consecutive special floors, and the differences between the bottom floor and the first special floor, and the last special floor and the top floor.
*   **Find Maximum:** Return the maximum of these differences.

*   **Runtime Complexity:** O(n log n), where n is the length of the `special` array (due to sorting). **Storage Complexity:** O(1) (excluding the input array).

	
	# Code
	```python
	def maxConsecutive(bottom: int, top: int, special: list[int]) -> int:
    """
    Calculates the maximum number of consecutive floors without a special floor.

    Args:
        bottom: The bottom floor of the rented range.
        top: The top floor of the rented range.
        special: A list of special floor numbers.

    Returns:
        The maximum number of consecutive floors without a special floor.
    """

    special.sort()
    max_consecutive = max(special[0] - bottom, top - special[-1])

    for i in range(len(special) - 1):
        max_consecutive = max(max_consecutive, special[i+1] - special[i] - 1)

    return max_consecutive
	```