How do you solve Minimum Number of Chairs in a Waiting Room on LeetCode?

Minimum Number of Chairs in a Waiting Room (LeetCode #3168) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The maximum number of people in the waiting room at any time is equal to the number of chairs needed.

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

The optimal solution for Minimum Number of Chairs in a Waiting Room runs in O(n) time and uses O(1) space. The complexity is O(n) because we only make a single pass through the string, processing each character once.

What is the brute force solution for Minimum Number of Chairs in a Waiting Room?

The brute force approach for Minimum Number of Chairs in a Waiting Room is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach simulates each event in the string and counts the number of people in the waiting room. For each 'E', we add a person, and for each 'L', we remove one. We track the maximum number of people present at any time to determine the number of chairs needed. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Number of Chairs in a Waiting Room?

Minimum Number of Chairs in a Waiting Room uses the following concepts: String, Simulation. The recommended patterns to study are: Simulation, Counting.

What are the interview tips for Minimum Number of Chairs in a Waiting Room?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, such as all entries or all exits. Explain your thought process while coding to show your understanding.

What are common mistakes when solving Minimum Number of Chairs in a Waiting Room?

Failing to account for the initial state of the waiting room (starting from zero). Not updating the maximum correctly after each event.

#3168

Minimum Number of Chairs in a Waiting Room

Easy

StringSimulationSimulationCounting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution efficiently counts the number of people in the waiting room using a single pass through the string. We maintain a count of current people and track the maximum number of people present at any time, which directly gives us the number of chairs needed.

⚙️

Algorithm

3 steps

1Step 1: Initialize a variable to track the current number of people in the waiting room.
2Step 2: Initialize a variable to track the maximum number of people at any time.
3Step 3: Iterate through each character in the string: increment for 'E' and decrement for 'L'. Update the maximum if the current number exceeds it.

solution.py11 lines

1def min_chairs(s):
2    current_people = 0
3    max_people = 0
4    for event in s:
5        if event == 'E':
6            current_people += 1
7        else:
8            current_people -= 1
9        max_people = max(max_people, current_people)
10    return max_people
11

ℹ

Complexity note: The complexity is O(n) because we only make a single pass through the string, processing each character once.

1The maximum number of people in the waiting room at any time is equal to the number of chairs needed.
2Each 'E' increases the count while each 'L' decreases it.

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