What is the time complexity of Maximum Consecutive Floors Without Special Floors?

The optimal solution for Maximum Consecutive Floors Without Special Floors runs in O(n log n) time and uses O(1) space. The time complexity is O(n log n) due to sorting the special floors. The space complexity is O(1) as we only use a few variables for counting.

What is the brute force solution for Maximum Consecutive Floors Without Special Floors?

The brute force approach for Maximum Consecutive Floors Without Special Floors is Brute Force, with O(n) time complexity and O(n) space complexity. The brute force approach checks each floor from bottom to top and counts the number of consecutive non-special floors. This is straightforward but inefficient for large ranges. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Consecutive Floors Without Special Floors?

Maximum Consecutive Floors Without Special Floors uses the following concepts: Array, Sorting. The recommended patterns to study are: Sorting, Array.

What are the interview tips for Maximum Consecutive Floors Without Special Floors?

Always consider edge cases, like all floors being special. Explain your thought process clearly while coding. Practice identifying patterns in problems, such as sorting or using sets.

What are common mistakes when solving Maximum Consecutive Floors Without Special Floors?

Not sorting the special floors before calculating gaps. Overlooking the gaps before the first and after the last special floor.

#2274

Maximum Consecutive Floors Without Special Floors

Medium

ArraySortingSortingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(n log n)
Space	O(n)	O(1)

💡

Intuition

Time O(n log n)Space O(1)

By sorting the special floors and calculating gaps between them, we can efficiently find the maximum number of consecutive non-special floors. This avoids unnecessary checks for every floor.

⚙️

Algorithm

5 steps

1Step 1: Sort the special array.
2Step 2: Initialize max_count with the gap before the first special floor and after the last special floor.
3Step 3: Iterate through the sorted special floors, calculating the gaps between consecutive special floors.
4Step 4: Update max_count with the maximum gap found.
5Step 5: Return max_count.

solution.py7 lines

1def maxConsecutive(bottom, top, special):
2    special.sort()
3    max_count = special[0] - bottom
4    for i in range(1, len(special)):
5        max_count = max(max_count, special[i] - special[i - 1] - 1)
6    max_count = max(max_count, top - special[-1])
7    return max_count

ℹ

Complexity note: The time complexity is O(n log n) due to sorting the special floors. The space complexity is O(1) as we only use a few variables for counting.

1Sorting helps in efficiently calculating gaps between special floors.
2The maximum gap can occur before the first special floor or after the last special floor.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.