What is the time complexity of Minimum Number of Chairs in a Waiting Room?

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

What data structures are used to solve Minimum Number of Chairs in a Waiting Room?

This problem uses String, Simulation. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Minimum Number of Chairs in a Waiting Room?

Insight 1: The problem can be solved by tracking the number of people currently in the waiting room. The maximum number of people in the room at any given time determines the minimum number of chairs needed. Insight 2: A simple counter is sufficient to track the number of occupied chairs; no complex data structures are needed.

Which companies ask LeetCode 3168. Minimum Number of Chairs in a Waiting Room in interviews?

This problem is frequently asked at Expedia.

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Minimum Number of Chairs in a Waiting Room - LeetCode 3168 Solution

Minimum Number of Chairs in a Waiting Room - Complete Solution Guide

Minimum Number of Chairs in a Waiting Room is LeetCode problem 3168, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a string s . Simulate events at each second i : If s[i] == 'E' , a person enters the waiting room and takes one of the chairs in it. If s[i] == 'L' , a person leaves the waiting room, freeing up a chair. Return the minimum number of chairs needed so that a chair is available for every person who enters the waiting room given that it is initially empty . Example 1: Input: s = "EEEEEEE" Output: 7 Explanation: After each second, a person enters the waiting room and no person leaves it

Detailed Explanation

The problem asks you to determine the minimum number of chairs needed in a waiting room to accommodate a sequence of arrivals and departures. The input is a string `s` where 'E' represents a person entering the room and 'L' represents a person leaving. The room starts empty. The output is a single integer representing the minimum number of chairs required such that there's always a chair available for anyone entering the room. The constraints specify that the input string length is between 1 and 50, contains only 'E' and 'L' characters, and represents a valid sequence (no one leaves before entering).

Solution Approach

The solution uses a single integer variable `chairs` to keep track of the number of people currently in the waiting room. It iterates through the input string `s`. If it encounters an 'E', it increments `chairs`. If it encounters an 'L', it decrements `chairs`. A second variable `max_chairs` (or `chairsNeeded` in some implementations) keeps track of the maximum value of `chairs` encountered so far. After iterating through the entire string, `max_chairs` holds the minimum number of chairs needed.

Step-by-Step Algorithm

Step 1: Initialize two integer variables: `chairs` to 0 (representing the number of occupied chairs) and `max_chairs` to 0 (representing the maximum number of occupied chairs encountered so far).
Step 2: Iterate through each character `c` in the input string `s`.
Step 3: If `c` is 'E', increment `chairs` and update `max_chairs` to be the maximum of `max_chairs` and `chairs`.
Step 4: If `c` is 'L', decrement `chairs`.
Step 5: After iterating through the entire string, return `max_chairs`.

Key Insights

Insight 1: The problem can be solved by tracking the number of people currently in the waiting room. The maximum number of people in the room at any given time determines the minimum number of chairs needed.
Insight 2: A simple counter is sufficient to track the number of occupied chairs; no complex data structures are needed.
Insight 3: The problem is about finding the peak number of occupied chairs, not the total number of entries or exits. The algorithm iterates through the string, incrementing the counter for 'E' and decrementing for 'L', and keeping track of the maximum value.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Simulation.

Companies

Asked at: Expedia.

Minimum Number of Chairs in a Waiting Room | LeetCode 3168

Easy

String Simulation

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minimumChairs(String s) {
        int chairsNeeded = 0;
        int occupiedChairs = 0;
        for (char c : s.toCharArray()) {
            if (c == 'E') {
                occupiedChairs++;
                chairsNeeded = Math.max(chairsNeeded, occupiedChairs);
            } else {
                occupiedChairs--;
            }
        }
        return chairsNeeded;
    }
}

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Problem Info

Description

Problem 3168: Minimum Number of Chairs in a Waiting Room - Detailed explanation and solution approaches.

Topics

String Simulation

Companies

Expedia

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3168

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

5 steps

1
Step 1: Initialize two integer variables: `chairs` to 0 (representing the number of occupied chairs) and `max_chairs` to 0 (representing the maximum number of occupied chairs encountered so far).
2
Step 2: Iterate through each character `c` in the input string `s`.
3
Step 3: If `c` is 'E', increment `chairs` and update `max_chairs` to be the maximum of `max_chairs` and `chairs`.
4
Step 4: If `c` is 'L', decrement `chairs`.
5
Step 5: After iterating through the entire string, return `max_chairs`.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because each character in the input string is processed exactly once. The space complexity is constant because the number of variables used does not depend on the input size.

📖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

✨

Key Insights

1Insight 1: The problem can be solved by tracking the number of people currently in the waiting room. The maximum number of people in the room at any given time determines the minimum number of chairs needed.
2Insight 2: A simple counter is sufficient to track the number of occupied chairs; no complex data structures are needed.
3Insight 3: The problem is about finding the peak number of occupied chairs, not the total number of entries or exits. The algorithm iterates through the string, incrementing the counter for 'E' and decrementing for 'L', and keeping track of the maximum value.

📚

Key Terms Explained

For Students

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty string. The code handles this correctly; the result will be 0.
Edge case 2: String with only 'E' or only 'L' characters. The code correctly handles this, resulting in the count of 'E's for a string of only 'E's and 0 for a string of only 'L's.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to update `max_chairs` every time `chairs` increases. This would lead to an incorrect result.

✗Common mistake 2: Not handling the case of an empty input string correctly, leading to errors or unexpected results.

✅

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 systems. This algorithm can be used to determine the minimum number of resources (e.g., servers, threads) needed to handle a stream of requests (entries) and releases (exits).

🏦Real-world scenario 2: Managing a queue or buffer. It can model situations where items enter and leave a queue, determining the minimum buffer size to prevent overflow.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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: September 17, 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

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

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Minimum Number of Chairs in a Waiting Room"?

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

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