Problem Description

Alice has rented multiple consecutive floors in a building, from floor bottom to floor top (inclusive). Among these floors, she has designated some as "special floors" that are used only for relaxation.

You are given:

• bottom: the lowest floor number Alice has rented
• top: the highest floor number Alice has rented
• special: an array containing the floor numbers that are designated as special floors

Your task is to find the maximum number of consecutive floors that are NOT special floors.

For example, if Alice rented floors 2 through 9 and floors 4 and 6 are special, then:

• Floors 2, 3 are consecutive non-special (2 floors)
• Floor 5 is non-special (1 floor)
• Floors 7, 8, 9 are consecutive non-special (3 floors)

The answer would be 3, as that's the maximum consecutive stretch of non-special floors.

The solution works by sorting the special floors array, then checking:

1. The gap between bottom and the first special floor
2. The gaps between each pair of adjacent special floors
3. The gap between the last special floor and top

The maximum gap among all these represents the longest consecutive sequence of non-special floors.

Intuition

To find the maximum consecutive non-special floors, we need to think about where these consecutive sequences can occur. Since special floors act as "barriers" that break up consecutive sequences, the non-special floors must exist in the gaps between special floors.

If we visualize the floors from bottom to top as a line segment with special floors marked as points on this line, we're essentially looking for the largest gap between these points.

The key insight is that once we sort the special floors, we can easily identify all possible gaps:

• There's a gap before the first special floor (from bottom to the first special floor)
• There are gaps between consecutive special floors
• There's a gap after the last special floor (from the last special floor to top)

For the gaps between consecutive special floors, if we have special floors at positions x and y, the number of non-special floors between them is y - x - 1 (we subtract 1 because we don't count the special floors themselves).

For the boundary cases:

• The gap at the beginning is special[0] - bottom floors
• The gap at the end is top - special[-1] floors

By sorting the special floors first, we can iterate through them in order and calculate all these gaps efficiently. The maximum among all these gaps is our answer.

This approach ensures we check every possible consecutive sequence of non-special floors with a time complexity of O(n log n) due to sorting, where n is the number of special floors.

Learn more about Sorting patterns.

Solution Approach

The solution follows a straightforward sorting-based approach to find the maximum gap between special floors.

Step 1: Sort the special floors

special.sort()

We sort the special array in ascending order. This allows us to process the floors sequentially and identify all gaps between consecutive special floors.

Step 2: Calculate boundary gaps

ans = max(special[0] - bottom, top - special[-1])

We initialize our answer with the maximum of two boundary gaps:

• special[0] - bottom: The number of non-special floors from the bottom floor up to (but not including) the first special floor
• top - special[-1]: The number of non-special floors from after the last special floor up to and including the top floor

Step 3: Calculate gaps between consecutive special floors

for x, y in pairwise(special):
    ans = max(ans, y - x - 1)

We iterate through pairs of consecutive special floors using pairwise(). For each pair (x, y):

• The gap between them is y - x - 1 (we subtract 1 because both x and y are special floors and shouldn't be counted)
• We update ans to keep track of the maximum gap found so far

Step 4: Return the result

return ans

After checking all possible gaps, we return the maximum value found.

Time Complexity: O(n log n) where n is the length of the special array, dominated by the sorting operation.

Space Complexity: O(1) if we don't count the space used by the sorting algorithm (which is typically O(log n) for the recursion stack in languages like Python).

Example Walkthrough

Let's walk through a concrete example to understand how the solution works.

Example:

• bottom = 2
• top = 9
• special = [4, 6]

Step 1: Sort the special floors The array [4, 6] is already sorted, so it remains [4, 6].

Step 2: Calculate boundary gaps

• Gap from bottom to first special floor: special[0] - bottom = 4 - 2 = 2
  • This represents floors 2 and 3 (both non-special)
• Gap from last special floor to top: top - special[-1] = 9 - 6 = 3
  • This represents floors 7, 8, and 9 (all non-special)
• Initial answer: ans = max(2, 3) = 3

Step 3: Calculate gaps between consecutive special floors We have one pair of consecutive special floors: (4, 6)

• Gap between them: 6 - 4 - 1 = 1
  • This represents floor 5 (the only non-special floor between 4 and 6)
• Update answer: ans = max(3, 1) = 3

Step 4: Return the result The maximum consecutive non-special floors is 3.

Visual representation:

Floor:  2  3  4  5  6  7  8  9
Type:   N  N  S  N  S  N  N  N
Gap:    [--2--]  [1]  [---3---]

Where N = Non-special, S = Special

The three gaps are:

• Floors 2-3: 2 consecutive non-special floors
• Floor 5: 1 non-special floor
• Floors 7-9: 3 consecutive non-special floors

The maximum is 3, which corresponds to floors 7, 8, and 9.

Solution Implementation

Time and Space Complexity

The time complexity is O(n × log n), where n is the length of the array special. This is dominated by the sorting operation special.sort(), which takes O(n × log n) time. After sorting, the code iterates through the array once using pairwise(special), which takes O(n) time. The operations to find the maximum value and calculate differences are all O(1). Therefore, the overall time complexity is O(n × log n) + O(n) = O(n × log n).

The space complexity is O(log n). The sorting algorithm (typically Timsort in Python) requires O(log n) space for its recursive call stack in the average case. The pairwise function creates an iterator that only maintains references to consecutive pairs without creating a new array, so it uses O(1) additional space. The variable ans and loop variables x, y also use O(1) space. Therefore, the overall space complexity is O(log n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Off-by-One Errors in Gap Calculations

One of the most common mistakes is incorrectly calculating the gaps between special floors or boundaries.

Incorrect Implementation:

# Wrong: Forgetting to subtract 1 between special floors
for current_floor, next_floor in pairwise(special):
    gap = next_floor - current_floor  # Missing -1!
    max_consecutive = max(max_consecutive, gap)

# Wrong: Including the special floor in boundary calculations
max_consecutive = max(special[0] - bottom + 1, top - special[-1] + 1)

Why it's wrong:

• Between two special floors at positions 4 and 7, the non-special floors are 5 and 6 (2 floors), which is 7 - 4 - 1 = 2, not 7 - 4 = 3
• From bottom=2 to first special floor=4, non-special floors are 2 and 3 (2 floors), which is 4 - 2 = 2, not 4 - 2 + 1 = 3

2. Not Handling Edge Cases

Missing validation when special array has only one element:

# Problematic if not careful with boundaries
def maxConsecutive(self, bottom: int, top: int, special: List[int]) -> int:
    special.sort()
    # If special has only 1 element, pairwise() won't iterate
    # Must ensure boundary gaps are properly calculated

Solution: The current implementation handles this correctly by always calculating boundary gaps first, which works even with a single special floor.

3. Forgetting to Sort the Special Array

Incorrect Implementation:

def maxConsecutive(self, bottom: int, top: int, special: List[int]) -> int:
    # Forgot to sort! This will give wrong results
    max_consecutive = max(special[0] - bottom, top - special[-1])
  
    for current_floor, next_floor in pairwise(special):
        gap = next_floor - current_floor - 1
        max_consecutive = max(max_consecutive, gap)
  
    return max_consecutive

Why it fails: If special = [6, 4, 8], without sorting, we'd check gaps between 6→4 (negative!) and 4→8, missing the actual consecutive arrangement.

4. Integer Overflow in Other Languages

While not an issue in Python, languages like C++ or Java might face integer overflow:

// In C++, this could overflow if top and bottom are near INT_MAX
int gap = next_floor - current_floor - 1;

Solution for other languages: Use appropriate data types (long long in C++, long in Java) when dealing with potentially large floor numbers.

5. Misunderstanding Problem Boundaries

Common misconception: Thinking that bottom and top might be special floors themselves.

# Wrong assumption leading to unnecessary checks
if bottom in special or top in special:
    # Special handling that's actually not needed

Clarification: The problem guarantees that all special floors are within the range [bottom, top], but bottom and top themselves are not necessarily special floors. The current solution handles this correctly by treating them as boundaries.