What is the time complexity of The Number of the Smallest Unoccupied Chair?

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve The Number of the Smallest Unoccupied Chair?

This problem uses Array, Hash Table, Heap (Priority Queue). Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving The Number of the Smallest Unoccupied Chair?

Insight 1: Sorting the friends by arrival time is crucial to simulate the real-time arrival process. Insight 2: Maintaining a data structure (like a min-heap or priority queue) to track available and occupied chairs allows efficient retrieval of the smallest available chair and management of chair availability upon departure.

Which companies ask LeetCode 1942. The Number of the Smallest Unoccupied Chair in interviews?

This problem is frequently asked at Otter.ai.

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The Number of the Smallest Unoccupied Chair - LeetCode 1942 Solution

The Number of the Smallest Unoccupied Chair - Complete Solution Guide

The Number of the Smallest Unoccupied Chair is LeetCode problem 1942, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There is a party where n friends numbered from 0 to n - 1 are attending. There is an infinite number of chairs in this party that are numbered from 0 to infinity . When a friend arrives at the party, they sit on the unoccupied chair with the smallest number . For example, if chairs 0 , 1 , and 5 are occupied when a friend comes, they will sit on chair number 2 . When a friend leaves the party, their chair becomes unoccupied at the moment they leave. If another friend arrives at that same moment,

Detailed Explanation

The problem simulates a party seating arrangement where friends arrive and leave at different times. The goal is to determine which chair number a specific friend (identified by `targetFriend`) will occupy. Chairs are numbered from 0 to infinity, and arriving friends always take the smallest available chair. When a friend leaves, their chair becomes immediately available for others. We are given an array `times` where each element `times[i] = [arrival_time, leaving_time]` represents the arrival and leaving times of the i-th friend. All arrival times are distinct.

Solution Approach

The solution uses a combination of sorting and priority queues to efficiently simulate the party seating process. First, the friends are sorted by their arrival times. Two priority queues are used: one to store the available chair numbers (`available_chairs`) and another to store the occupied chairs along with their leaving times (`occupied_chairs`). As we iterate through the sorted friends, we first free up any chairs that have become available (leaving time is less than or equal to the current friend's arrival time). Then, we assign the smallest available chair to the current friend. If no chairs are available, we assign the next available chair number. Finally, we store the chair number and the leaving time of the current friend in the `occupied_chairs` queue. The algorithm returns the chair number assigned to the `targetFriend`.

Step-by-Step Algorithm

Step 1: Create a list of tuples `(arrival_time, leaving_time, friend_id)` and sort it based on arrival time.
Step 2: Initialize two min-heaps (priority queues): `available_chairs` (initially empty) and `occupied_chairs` (initially empty). Also initialize `next_chair = 0` to track the next available chair number.
Step 3: Iterate through the sorted list of friends.
Step 4: For each friend, check if any chairs in `occupied_chairs` have become available (leaving time <= current friend's arrival time). If so, move those chairs to `available_chairs`.
Step 5: Assign a chair to the current friend. If `available_chairs` is not empty, assign the smallest chair from it. Otherwise, assign `next_chair` and increment `next_chair`.
Step 6: If the current friend is the `targetFriend`, return the assigned chair number.
Step 7: Add the assigned chair and the friend's leaving time to the `occupied_chairs`.
Step 8: After processing all friends, return the result.

Key Insights

Insight 1: Sorting the friends by arrival time is crucial to simulate the real-time arrival process.
Insight 2: Maintaining a data structure (like a min-heap or priority queue) to track available and occupied chairs allows efficient retrieval of the smallest available chair and management of chair availability upon departure.
Insight 3: We need to keep track of currently occupied chairs with their leaving times so we can free them up at the appropriate arrival times of other guests.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Heap (Priority Queue).

Companies

Asked at: Otter.ai.

The Number of the Smallest Unoccupied Chair | LeetCode 1942

Medium

Array Hash Table Heap (Priority Queue)

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

Solution Code

Available in:

import java.util.Arrays;
import java.util.PriorityQueue;

class Solution {
    public int smallestChair(int[][] times, int targetFriend) {
        int n = times.length;

        int[][] sorted_friends = new int[n][3];
        for (int i = 0; i < n; i++) {
            sorted_friends[i][0] = times[i][0];
            sorted_friends[i][1] = times[i][1];
            sorted_friends[i][2] = i;
        }
        Arrays.sort(sorted_friends, (a, b) -> Integer.compare(a[0], b[0]));
        
        PriorityQueue<Integer> available_chairs = new PriorityQueue<>();
        PriorityQueue<int[]> occupied_chairs = new PriorityQueue<>((a, b) -> Integer.compare(a[0], b[0]));

        int next_chair = 0;

        for (int i = 0; i < n; i++) {
            int arrival = sorted_friends[i][0];
            int leaving = sorted_friends[i][1];
            int friend_id = sorted_friends[i][2];

            while (!occupied_chairs.isEmpty() && occupied_chairs.peek()[0] <= arrival) {
                int[] chair_info = occupied_chairs.poll();
                int chair = chair_info[1];
                available_chairs.offer(chair);
            }

            int assigned_chair;
            if (!available_chairs.isEmpty()) {
                assigned_chair = available_chairs.poll();
            } else {
                assigned_chair = next_chair;
                next_chair++;
            }

            if (friend_id == targetFriend) {
                return assigned_chair;
            }

            occupied_chairs.offer(new int[]{leaving, assigned_chair});
        }
        return -1; 
    }
}

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Problem Info

Description

Problem 1942: The Number of the Smallest Unoccupied Chair - Detailed explanation and solution approaches.

Topics

Array Hash Table Heap (Priority Queue)

Companies

Otter.ai

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1942

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Create a list of tuples `(arrival_time, leaving_time, friend_id)` and sort it based on arrival time.
2
Step 2: Initialize two min-heaps (priority queues): `available_chairs` (initially empty) and `occupied_chairs` (initially empty). Also initialize `next_chair = 0` to track the next available chair number.
3
Step 3: Iterate through the sorted list of friends.
4
Step 4: For each friend, check if any chairs in `occupied_chairs` have become available (leaving time <= current friend's arrival time). If so, move those chairs to `available_chairs`.
5
Step 5: Assign a chair to the current friend. If `available_chairs` is not empty, assign the smallest chair from it. Otherwise, assign `next_chair` and increment `next_chair`.
6
Step 6: If the current friend is the `targetFriend`, return the assigned chair number.
7
Step 7: Add the assigned chair and the friend's leaving time to the `occupied_chairs`.
8
Step 8: After processing all friends, return the result.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by sorting the friends (O(n log n)). The priority queue operations (insertion and extraction) take O(log n) time each, and these operations are performed at most n times. Therefore, the overall time complexity is O(n log n). The space complexity is O(n) because we store at most n chair numbers and leaving times in the priority queues.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using a Set for Available Chairs

Instead of a priority queue for available chairs, you could use a set. Sets provide O(log n) time for insertion, deletion, and finding the minimum element.

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

Trade-offs: The performance difference is negligible in practice. Sets might be slightly easier to implement in some languages, but priority queues clearly express the need for smallest element retrieval.

Brute Force Approach

Iterate through all chair numbers starting from 0 and for each chair, check if it is occupied at the arrival time of the target friend. This approach is very inefficient.

Time: O(n*M), where M is the chair numberSpace: O(n)

Trade-offs: Avoid this approach as it is not optimal, especially for larger input sizes. The value of M can grow to be extremely large, making the runtime unacceptable. This approach is more intuitive but lacks efficiency.

✨

Key Insights

1Insight 1: Sorting the friends by arrival time is crucial to simulate the real-time arrival process.
2Insight 2: Maintaining a data structure (like a min-heap or priority queue) to track available and occupied chairs allows efficient retrieval of the smallest available chair and management of chair availability upon departure.
3Insight 3: We need to keep track of currently occupied chairs with their leaving times so we can free them up at the appropriate arrival times of other guests.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: `targetFriend` is the first to arrive. The algorithm correctly assigns chair 0 to them.
Edge case 2: All chairs are occupied when a friend arrives. The algorithm correctly assigns a new chair (incrementing `next_chair`) to them.
Edge case 3: `targetFriend` is the last to arrive. The algorithm correctly assigns a chair after processing all other friends.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to free up chairs that have become available before assigning a new chair. This can lead to incorrect chair assignments.

✗Common mistake 2: Not sorting the friends by their arrival times. This will result in an incorrect simulation of the seating arrangement.

✗Common mistake 3: Using an inefficient data structure for managing available chairs. Linear search for the smallest available chair will increase the time complexity to O(n^2).

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation in a system where resources are dynamically assigned and released. For example, allocating server instances to incoming requests based on availability.

🏦Real-world scenario 2: Managing parking spaces in a parking lot. Arriving cars are assigned the smallest available parking spot, and when a car leaves, the spot becomes available for others.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 21, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "The Number of the Smallest Unoccupied Chair"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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