LeetCode Problem 3168

Difficulty: 🟢 Easy
Topics: String, Simulation

Solution

LeetCode 3168, Minimum Number of Chairs in a Waiting Room

Problem Understanding

The problem gives us a string s that represents events happening in a waiting room over time. Each character corresponds to one second.

If the character is 'E', one person enters the room and occupies a chair. If the character is 'L', one person leaves the room and frees a chair.

The waiting room starts empty, meaning there are initially zero people inside. Our task is to determine the minimum number of chairs required so that every person who enters always has a chair available.

Another way to think about the problem is that we need to track how many people are simultaneously inside the waiting room at any moment. The answer is simply the maximum number of people present at the same time.

For example, if the sequence is "ELELEEL":

• The first person enters, so 1 chair is occupied.
• That person leaves, so occupancy becomes 0.
• Another person enters, occupancy becomes 1 again.
• Eventually, two people are inside at the same time, so we need at least 2 chairs.

The constraints are very small:

• 1 <= s.length <= 50
• The string contains only 'E' and 'L'
• The sequence is guaranteed to be valid

The validity guarantee is important. It means we never encounter a situation where someone leaves an already empty room. Therefore, the number of people inside never becomes negative.

Even though the constraints are tiny, the goal is still to design the cleanest and most efficient simulation possible.

Some important edge cases include:

• A string containing only 'E', where occupancy continuously increases.
• Alternating enter and leave events such as "ELELEL", where only one chair is needed.
• Cases where the maximum occupancy occurs near the middle or end of the sequence.
• Cases where many people accumulate before any leave events occur.

Approaches

Brute Force Approach

A brute force solution could simulate the waiting room while explicitly tracking every occupied chair individually.

For example, we could maintain a list of chair states, assigning a new chair whenever someone enters and freeing one whenever someone leaves. At every step, we would count how many chairs are currently occupied and track the maximum.

This approach works because it directly models the real-world process described in the problem. However, it introduces unnecessary complexity because we do not actually care which specific chair a person occupies. We only care about how many people are inside at any moment.

Although the constraints are small enough that this would still run quickly, it wastes memory and performs more operations than necessary.

Optimal Approach

The key observation is that the exact chair assignments do not matter.

We only need two values:

• current_people, the number of people currently in the room
• max_people, the largest number of people seen at any point

As we iterate through the string:

• 'E' increases current_people
• 'L' decreases current_people

After each update, we compare current_people with max_people.

The maximum occupancy encountered during the simulation is exactly the minimum number of chairs required.

This works because every person currently inside requires one chair, and the peak occupancy determines the smallest capacity that can support the entire sequence.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n)	O(n)	Explicitly tracks chair assignments
Optimal	O(n)	O(1)	Tracks only current and maximum occupancy

Algorithm Walkthrough

1. Initialize two integer variables:

• current_people = 0
• max_people = 0

The first variable tracks how many people are currently inside the waiting room. The second tracks the maximum occupancy observed so far. 2. Iterate through each character in the string s. 3. If the current character is 'E', increment current_people by 1.

This simulates a person entering and occupying a chair. 4. If the current character is 'L', decrement current_people by 1.

This simulates a person leaving and freeing a chair. 5. After processing the event, update max_people.

Set:

max_people = max(max_people, current_people)

This ensures we remember the highest occupancy reached during the simulation. 6. After the loop finishes, return max_people.

This value represents the minimum number of chairs required.

Why it works

At every moment, current_people exactly represents the number of chairs currently needed. Since every person occupies one chair, the minimum total number of chairs must be at least the maximum occupancy encountered during the simulation.

The algorithm maintains this invariant throughout the traversal, so the final max_people value is guaranteed to be correct.

Python Solution

class Solution:
    def minimumChairs(self, s: str) -> int:
        current_people = 0
        max_people = 0

        for event in s:
            if event == 'E':
                current_people += 1
            else:
                current_people -= 1

            max_people = max(max_people, current_people)

        return max_people

The implementation follows the simulation directly.

The variable current_people keeps track of how many people are inside the room at the current moment. Whenever we encounter 'E', we increment it because someone enters. Whenever we encounter 'L', we decrement it because someone leaves.

After processing each event, we update max_people to record the highest occupancy seen so far.

Finally, we return max_people, which represents the minimum number of chairs necessary to handle the busiest moment in the sequence.

The implementation is concise because the problem only requires tracking counts, not individual chair assignments.

Go Solution

func minimumChairs(s string) int {
    currentPeople := 0
    maxPeople := 0

    for _, event := range s {
        if event == 'E' {
            currentPeople++
        } else {
            currentPeople--
        }

        if currentPeople > maxPeople {
            maxPeople = currentPeople
        }
    }

    return maxPeople
}

The Go implementation mirrors the Python logic closely.

Go iterates through the string using a range loop, where each character is returned as a rune. Since the input contains only ASCII characters 'E' and 'L', direct rune comparison works naturally.

No special handling for overflow is needed because the maximum string length is only 50, so all counts easily fit within Go's standard integer type.

Worked Examples

Example 1

Input:

s = "EEEEEEE"

Step	Event	Current People	Maximum So Far
1	E	1	1
2	E	2	2
3	E	3	3
4	E	4	4
5	E	5	5
6	E	6	6
7	E	7	7

Final answer:

7

Since nobody leaves, occupancy continuously increases.

Example 2

Input:

s = "ELELEEL"

Step	Event	Current People	Maximum So Far
1	E	1	1
2	L	0	1
3	E	1	1
4	L	0	1
5	E	1	1
6	E	2	2
7	L	1	2

Final answer:

2

The largest occupancy occurs after the sixth event, when two people are simultaneously inside.

Example 3

Input:

s = "ELEELEELLL"

Step	Event	Current People	Maximum So Far
1	E	1	1
2	L	0	1
3	E	1	1
4	E	2	2
5	L	1	2
6	E	2	2
7	E	3	3
8	L	2	3
9	L	1	3
10	L	0	3

Final answer:

3

The peak occupancy is 3, so at least 3 chairs are required.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	We process each character exactly once
Space	O(1)	Only two integer variables are used

The algorithm performs a single pass through the input string. Each character requires only constant time work, so the total runtime is linear in the length of the string.

The memory usage is constant because we store only two counters regardless of input size.

Test Cases

# Provided examples
assert Solution().minimumChairs("EEEEEEE") == 7  # all enter events
assert Solution().minimumChairs("ELELEEL") == 2  # alternating with brief overlap
assert Solution().minimumChairs("ELEELEELLL") == 3  # larger peak occupancy

# Boundary cases
assert Solution().minimumChairs("E") == 1  # minimum length input
assert Solution().minimumChairs("EL") == 1  # one enter and one leave

# Alternating pattern
assert Solution().minimumChairs("ELELELEL") == 1  # occupancy never exceeds 1

# Increasing occupancy before leaving
assert Solution().minimumChairs("EEELLL") == 3  # all enters first

# Peak occurs in middle
assert Solution().minimumChairs("EELLEEELLL") == 3  # middle section creates max

# Occupancy repeatedly rises and falls
assert Solution().minimumChairs("EELEELELL") == 3  # multiple fluctuations

# Longest possible occupancy buildup
assert Solution().minimumChairs("E" * 50) == 50  # maximum constraint size

Test	Why
"EEEEEEE"	Validates continuous occupancy growth
"ELELEEL"	Tests alternating events with overlap
"ELEELEELLL"	Tests multiple occupancy fluctuations
"E"	Smallest possible input
"EL"	Simple enter then leave
"ELELELEL"	Ensures occupancy never exceeds 1
"EEELLL"	Tests maximum buildup before exits
"EELLEEELLL"	Verifies middle-position peak occupancy
"EELEELELL"	Tests repeated occupancy changes
"E" * 50	Validates upper constraint boundary

Edge Cases

One important edge case is when the string contains only enter events, such as "EEEE". A buggy implementation might fail to keep updating the maximum occupancy after each step. In this case, occupancy continuously increases, so the correct answer equals the entire string length. The implementation handles this naturally by updating max_people after every event.

Another important case is an alternating pattern like "ELELEL". Here, occupancy repeatedly rises to 1 and falls back to 0. A flawed solution might incorrectly count total enters instead of simultaneous occupancy. The current implementation correctly tracks only the current number of people inside, so it returns 1.

A third important edge case is when the maximum occupancy occurs late in the sequence rather than at the beginning. For example, "ELEELEE" gradually builds up occupancy after several earlier fluctuations. Some incorrect implementations might stop tracking once occupancy decreases. This solution continuously updates the maximum throughout the full traversal, ensuring later peaks are captured correctly.

A final subtle edge case is ensuring occupancy never becomes negative. The problem guarantees the sequence is valid, meaning there is never a leave event before a matching enter event. Because of this guarantee, the implementation can safely decrement without additional validation checks.