Maximum Consecutive Floors Without Special Floors

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ArraysGreedy Algorithms

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the maximum values for `top`, `bottom`, and the elements within the `special` array? Are we dealing with integers?
Can the `special` array be empty? If so, what should I return?
Is the `special` array guaranteed to contain only unique values, or could there be duplicates?
Is the `special` array guaranteed to contain only values within the range [bottom, top]?
Could `bottom` and `top` be equal? If so, and `special` is empty, what is the expected output?

Brute Force Solution

Approach

The brute force way to find the maximum gap between special floors involves checking every possible range of floors. We will exhaustively consider each possible set of consecutive floors to see how many regular floors are included in each.

Here's how the algorithm would work step-by-step:

Think of each pair of floors as defining a range of floors we want to check.
Consider every possible pair of floors between the bottom and top floor: the first and second floor, first and third floor, and so on, until we check between the first and top floor.
Then start at the second floor and repeat, checking the ranges: second and third, second and fourth, and so on.
Continue until you check every possible range by starting at each floor and going up to the top floor.
For each range of floors you've identified, count the number of regular floors within that range (floors that are not special floors).
Keep track of the largest number of consecutive regular floors found in any of the ranges you examined.
The largest number you kept track of is your answer.

Code Implementation

def max_consecutive_floors_brute_force(bottom, top, special):
    max_consecutive_non_special_floors = 0

    for start_floor in range(bottom, top + 1):
        for end_floor in range(start_floor, top + 1):
            # Consider each possible range of floors from start to end

            number_of_non_special_floors = 0

            for current_floor in range(start_floor, end_floor + 1):
                # Iterate to count the number of non-special floors

                if current_floor not in special:
                    number_of_non_special_floors += 1

            # Update the maximum if the current range has more floors
            if number_of_non_special_floors > max_consecutive_non_special_floors:
                max_consecutive_non_special_floors = number_of_non_special_floors

    return max_consecutive_non_special_floors

Big(O) Analysis

Time Complexity

O(n³) – The described brute force approach involves iterating through all possible pairs of floors. For each possible range defined by a pair of floors, the algorithm counts the number of non-special floors within that range. The outer loops iterate through all possible starting and ending floors, which takes O(n²) time, where n is the total number of floors. The inner operation of counting non-special floors in a range of floors could take O(n) time in the worst case. Thus, the total time complexity is O(n² * n) which simplifies to O(n³).

Space Complexity

O(1) – The brute force approach, as described, doesn't use any auxiliary data structures that scale with the input size, specifically the number of special floors or the range of floors. It only requires a few constant space variables to keep track of the current range of floors being considered, the count of consecutive regular floors within that range, and the maximum count found so far. Therefore, the space complexity remains constant regardless of the input. This translates to an O(1) space complexity.

Optimal Solution

Approach

The key to efficiently finding the maximum consecutive floors without special floors is to realize that the special floors split the range into subranges. We can then examine these subranges to find the largest one. This allows us to avoid checking every single floor individually.

Here's how the algorithm would work step-by-step:

First, sort the list of special floors in ascending order.
Next, calculate the difference between the bottom floor and the lowest special floor. This represents the number of consecutive safe floors at the bottom of the building.
Also, calculate the difference between the highest special floor and the top floor. This represents the number of consecutive safe floors at the top of the building.
Then, iterate through the sorted list of special floors, calculating the difference between each adjacent pair of special floors and subtract one (since the special floors themselves are not safe). This represents the number of consecutive safe floors between those two special floors.
Finally, compare all three of the calculated lengths (bottom safe floors, top safe floors, and the maximum gap between special floors) and return the largest of the three. This is the maximum number of consecutive safe floors.

Code Implementation

def max_consecutive(bottom, top, special):
    special.sort()

    # Calculate consecutive floors at the bottom.
    bottom_gap = special[0] - bottom

    # Calculate consecutive floors at the top.
    top_gap = top - special[-1]

    max_gap = 0
    # Find the maximum gap between special floors
    for i in range(len(special) - 1):
        gap = special[i+1] - special[i] - 1

        #Update the max gap if needed
        if gap > max_gap:
            max_gap = gap

    return max(bottom_gap, top_gap, max_gap)

Big(O) Analysis

Time Complexity

O(n log n) – The most significant factor determining the time complexity is sorting the special floors list, which contains n elements. Common sorting algorithms like merge sort or quicksort have a time complexity of O(n log n). The subsequent calculations of differences between adjacent special floors involve iterating through the sorted list once, which takes O(n) time. Since O(n log n) dominates O(n), the overall time complexity is O(n log n).

Space Complexity

O(1) – The algorithm sorts the special floors in place, so no extra space is used for sorting if the sorting algorithm used is in-place. Besides sorting, the algorithm only uses a few variables to store the bottom floor difference, top floor difference, maximum gap, and potentially a loop counter. These variables consume a constant amount of space regardless of the number of special floors N, so the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
bottom == top, implying zero floors in total	Return 0 as there are no floors to consider.
special is null or empty	Return top - bottom as all floors are available.
special contains duplicate floor numbers	Sorting and iterating will handle the duplicates without issue as only adjacent differences matter after sorting.
special contains floor numbers outside the range [bottom, top]	Filter the special array to only include valid floors within the range before processing.
special is very large, approaching system memory limits	The sorting operation should be done in place and the diff should be done iteratively, avoiding excessive memory allocation.
bottom and top are very large integers, causing potential integer overflow during subtraction	Use long data type for storing the difference between top and bottom to avoid overflow.
special contains bottom or top	The algorithm implicitly handles these by considering the difference between bottom and the first special floor and the top and last special floor.
bottom > top	Return 0 as this signifies an invalid floor range.

bottom == top, implying zero floors in total

How to Handle:

Return 0 as there are no floors to consider.

special is null or empty

How to Handle:

Return top - bottom as all floors are available.

special contains duplicate floor numbers

How to Handle:

Sorting and iterating will handle the duplicates without issue as only adjacent differences matter after sorting.

special contains floor numbers outside the range [bottom, top]

How to Handle:

Filter the special array to only include valid floors within the range before processing.

special is very large, approaching system memory limits

How to Handle:

The sorting operation should be done in place and the diff should be done iteratively, avoiding excessive memory allocation.

bottom and top are very large integers, causing potential integer overflow during subtraction

How to Handle:

Use long data type for storing the difference between top and bottom to avoid overflow.

special contains bottom or top

How to Handle:

The algorithm implicitly handles these by considering the difference between bottom and the first special floor and the top and last special floor.

bottom > top

How to Handle:

Return 0 as this signifies an invalid floor range.