How do you solve The Number of the Smallest Unoccupied Chair on LeetCode?

The Number of the Smallest Unoccupied Chair (LeetCode #1942) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Use a priority queue to efficiently manage available chairs.

What is the brute force solution for The Number of the Smallest Unoccupied Chair?

The brute force approach for The Number of the Smallest Unoccupied Chair is Brute Force, with O(n²) time complexity and O(1) space complexity. We can simulate the arrival and departure of each friend, checking for the smallest unoccupied chair each time a friend arrives. This is straightforward but inefficient as it checks all chairs repeatedly. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve The Number of the Smallest Unoccupied Chair?

The Number of the Smallest Unoccupied Chair uses the following concepts: Array, Hash Table, Heap (Priority Queue). The recommended patterns to study are: Hash Map, Array.

What are the interview tips for The Number of the Smallest Unoccupied Chair?

Always clarify the problem and constraints before jumping into coding. Discuss your thought process and approach with the interviewer. Consider edge cases, such as all friends arriving at the same time or leaving immediately.

What are common mistakes when solving The Number of the Smallest Unoccupied Chair?

Not managing the leaving times correctly, leading to incorrect chair assignments. Forgetting to check if a chair is available before assigning it.

#1942

The Number of the Smallest Unoccupied Chair

Medium

ArrayHash TableHeap (Priority Queue)Hash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

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

We can use a priority queue (min-heap) to efficiently manage the chairs. By keeping track of the next available chair and the leaving times, we can quickly determine which chair is free when a friend arrives.

⚙️

Algorithm

3 steps

1Step 1: Sort the times array by arrival time.
2Step 2: Initialize a priority queue to manage available chairs and a list to track occupied chairs.
3Step 3: For each friend, check the priority queue for available chairs and update it based on leaving times.

solution.py18 lines

1# Full working Python code
2import heapq
3
4def smallestChair(times, targetFriend):
5    times.sort()  # Sort by arrival time
6    available_chairs = []  # Min-heap for available chairs
7    occupied = []  # List to keep track of occupied chairs
8    for i in range(len(times)):
9        arrival, leaving = times[i]
10        # Free chairs that have been left
11        while occupied and occupied[0][0] <= arrival:
12            heapq.heappush(available_chairs, occupied.pop(0)[1])
13        # Get the smallest available chair
14        chair_num = heapq.heappop(available_chairs) if available_chairs else len(occupied)
15        occupied.append((leaving, chair_num))  # Mark chair as occupied
16        if i == targetFriend:
17            return chair_num
18    return -1

ℹ

Complexity note: The time complexity is O(n log n) due to sorting the arrival times and managing the priority queue, while the space complexity is O(n) for storing occupied chairs.

1Use a priority queue to efficiently manage available chairs.
2Sorting the arrival times is crucial for correct chair allocation.

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