LeetCode #2274 — MEDIUM

Maximum Consecutive Floors Without Special Floors

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

2
9
[4,6]

Example 2

6
8
[7,6,8]

Core Insight

What unlocks the optimal approach

Say we have a pair of special floors (x, y) with no other special floors in between. There are x - y - 1 consecutive floors in between them without a special floor.
Say there are n special floors. After sorting special, we have answer = max(answer, special[i] – special[i – 1] – 1) for all 0 < i < n.
However, there are two special cases left to consider: the floors before special[0] and after special[n-1].
To consider these cases, we have answer = max(answer, special[0] – bottom, top – special[n-1]).

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2274: Maximum Consecutive Floors Without Special Floors
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;
    }
}

// Accepted solution for LeetCode #2274: Maximum Consecutive Floors Without Special Floors
func maxConsecutive(bottom int, top int, special []int) int {
	sort.Ints(special)
	ans := max(special[0]-bottom, top-special[len(special)-1])
	for i, x := range special[1:] {
		ans = max(ans, x-special[i]-1)
	}
	return ans
}

# Accepted solution for LeetCode #2274: Maximum Consecutive Floors Without Special Floors
class Solution:
    def maxConsecutive(self, bottom: int, top: int, special: List[int]) -> int:
        special.sort()
        ans = max(special[0] - bottom, top - special[-1])
        for x, y in pairwise(special):
            ans = max(ans, y - x - 1)
        return ans

// Accepted solution for LeetCode #2274: Maximum Consecutive Floors Without Special Floors
/**
 * [2274] Maximum Consecutive Floors Without Special Floors
 *
 * 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.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/maximum-consecutive-floors-without-special-floors/
// discuss: https://leetcode.com/problems/maximum-consecutive-floors-without-special-floors/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn max_consecutive(bottom: i32, top: i32, special: Vec<i32>) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2274_example_1() {
        let bottom = 2;
        let top = 9;
        let special = vec![4, 6];

        let result = 3;

        assert_eq!(Solution::max_consecutive(bottom, top, special), result);
    }

    #[test]
    #[ignore]
    fn test_2274_example_2() {
        let bottom = 6;
        let top = 8;
        let special = vec![7, 6, 8];

        let result = 0;

        assert_eq!(Solution::max_consecutive(bottom, top, special), result);
    }
}

// Accepted solution for LeetCode #2274: Maximum Consecutive Floors Without Special Floors
function maxConsecutive(bottom: number, top: number, special: number[]): number {
    special.sort((a, b) => a - b);
    const n = special.length;
    let ans = Math.max(special[0] - bottom, top - special[n - 1]);
    for (let i = 1; i < n; ++i) {
        ans = Math.max(ans, special[i] - special[i - 1] - 1);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(log n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.