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 <= 105
1 <= bottom <= special[i] <= top <= 109
- All the values of
special
are unique.
Solutions
Solution 1: Sorting
We can sort the special floors in ascending order, then calculate the number of floors between each pair of adjacent special floors. Finally, we calculate the number of floors between the first special floor and $\textit{bottom}$, as well as the number of floors between the last special floor and $\textit{top}$. The maximum of these floor counts is the answer.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array $\textit{special}$.
1 2 3 4 5 6 7 |
|
1 2 3 4 5 6 7 8 9 10 11 |
|
1 2 3 4 5 6 7 8 9 10 11 |
|
1 2 3 4 5 6 7 8 |
|
1 2 3 4 5 6 7 8 9 |
|