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2274. Maximum Consecutive Floors Without Special Floors
Description
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 <= 10^{5}
1 <= bottom <= special[i] <= top <= 10^{9}
 All the values of
special
are unique.
Solutions

class Solution { public int maxConsecutive(int bottom, int top, int[] special) { Arrays.sort(special); int n = special.length; int ans = Math.max(special[0]  bottom, top  special[n  1]); for (int i = 1; i < n; ++i) { ans = Math.max(ans, special[i]  special[i  1]  1); } return ans; } }

class Solution: def maxConsecutive(self, bottom: int, top: int, special: List[int]) > int: special.sort() ans = max(special[0]  bottom, top  special[1]) for i in range(1, len(special)): ans = max(ans, special[i]  special[i  1]  1) return ans

function maxConsecutive(bottom: number, top: number, special: number[]): number { let nums = special.slice().sort((a, b) => a  b); nums.unshift(bottom  1); nums.push(top + 1); let ans = 0; const n = nums.length; for (let i = 1; i < n; i++) { ans = Math.max(ans, nums[i]  nums[i  1]  1); } return ans; }