LeetCode #3168 — EASY

Minimum Number of Chairs in a Waiting Room

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

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The Problem

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. Therefore, a minimum of 7 chairs is needed.

Example 2:

Input: s = "ELELEEL"

Output: 2

Explanation:

Let's consider that there are 2 chairs in the waiting room. The table below shows the state of the waiting room at each second.

Second Event People in the Waiting Room Available Chairs
0 Enter 1 1
1 Leave 0 2
2 Enter 1 1
3 Leave 0 2
4 Enter 1 1
5 Enter 2 0
6 Leave 1 1

Example 3:

Input: s = "ELEELEELLL"

Output: 3

Explanation:

Let's consider that there are 3 chairs in the waiting room. The table below shows the state of the waiting room at each second.

Second Event People in the Waiting Room Available Chairs
0 Enter 1 2
1 Leave 0 3
2 Enter 1 2
3 Enter 2 1
4 Leave 1 2
5 Enter 2 1
6 Enter 3 0
7 Leave 2 1
8 Leave 1 2
9 Leave 0 3

Constraints:

  • 1 <= s.length <= 50
  • s consists only of the letters 'E' and 'L'.
  • s represents a valid sequence of entries and exits.

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

"EEEEEEE"

Example 2

"ELELEEL"

Example 3

"ELEELEELLL"

Related Problems

  • Consecutive Characters (consecutive-characters)
Step 02

Core Insight

What unlocks the optimal approach

  • Iterate from left to right over the string and keep track of the number of people in the waiting room using a variable that you will increment on every occurrence of ‘E’ and decrement on every occurrence of ‘L’.
  • The answer is the maximum number of people in the waiting room at any instance.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #3168: Minimum Number of Chairs in a Waiting Room
class Solution {
    public int minimumChairs(String s) {
        int cnt = 0, left = 0;
        for (int i = 0; i < s.length(); ++i) {
            if (s.charAt(i) == 'E') {
                if (left > 0) {
                    --left;
                } else {
                    ++cnt;
                }
            } else {
                ++left;
            }
        }
        return cnt;
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(1)

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.