Minimum Number of Chairs in a Waiting Room

Easy

Asked by:

35 views

Topics:

Strings

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for arrival and departure times, and what data type are they (e.g., integers representing minutes from a start time)?
Is it guaranteed that departure time for an individual will always be greater than or equal to their arrival time?
Can the input arrays, arrival and departure, be empty or null? If so, what should be the expected return value?
Are the arrival and departure arrays always the same length? If not, what should I do?
Do the arrival and departure times need to be considered in chronological order, or should I assume they are unsorted?

Brute Force Solution

Approach

The goal is to figure out the fewest chairs needed in a waiting room given arrival and departure times. The brute force method means we'll try out every possible number of chairs to see if it works.

Here's how the algorithm would work step-by-step:

Start by assuming we only need one chair.
Go through the arrival and departure times one by one. For each arrival, check if a chair is free (meaning someone has already left).
If there are no chairs free, then one chair is not enough, so we move to the next possibility.
Now, assume we need two chairs. Repeat the process of checking arrivals and departures, but this time we have two 'free chair' slots.
Keep increasing the number of chairs we assume we have, repeating the checking process each time.
The first time we find a number of chairs where everyone can sit without needing more chairs than we have, that's our minimum number of chairs.

Code Implementation

def minimum_number_of_chairs_brute_force(arrival_times, departure_times):
    number_of_people = len(arrival_times)
    number_of_chairs = 0

    while True:
        number_of_chairs += 1
        chairs_available = [True] * number_of_chairs
        chair_usage_times = [[] for _ in range(number_of_chairs)]
        can_accommodate_all = True

        for person_index in range(number_of_people):
            arrival_time = arrival_times[person_index]
            departure_time = departure_times[person_index]
            chair_found = False

            for chair_index in range(number_of_chairs):
                # Iterate to find an available chair
                is_chair_free = True
                for occupied_time in chair_usage_times[chair_index]:
                    if not (departure_time <= occupied_time[0] or arrival_time >= occupied_time[1]):
                        is_chair_free = False
                        break

                if is_chair_free:
                    # Mark the chair as used during this time
                    chair_usage_times[chair_index].append((arrival_time, departure_time))
                    chair_found = True
                    break

            if not chair_found:
                can_accommodate_all = False
                break

        # If we can accomodate all, return number of chairs
        if can_accommodate_all:
            return number_of_chairs

Big(O) Analysis

Time Complexity

O(n²) – The outer loop iterates, in the worst case, up to 'n' times, where 'n' represents the total number of arrival and departure events, representing the assumed number of chairs. For each iteration of the outer loop (testing a number of chairs), we iterate through all 'n' arrival and departure events to simulate the waiting room occupancy. This nested iteration leads to approximately n * n operations, resulting in a time complexity of O(n²).

Space Complexity

O(1) – The provided brute-force approach, as described, doesn't use any significant auxiliary data structures. It iterates through different chair counts and then potentially iterates through the arrival and departure times to check for availability, but it does not store these times in any temporary arrays or data structures. The number of chairs being tested is a simple integer variable. Therefore, the auxiliary space remains constant regardless of the number of arrival and departure times (N).

Optimal Solution

Approach

To find the minimum number of chairs needed, we focus on when people arrive and leave. We'll sort the arrival and departure times, and then walk through them like a timeline, tracking how many people are in the room at any given moment.

Here's how the algorithm would work step-by-step:

First, organize the arrival times and departure times into two separate lists and sort them independently.
Imagine you are moving forward in time. Start with the earliest arrival time.
As you go through the sorted lists, if you encounter an arrival, it means someone entered the room, so you need one more chair.
If you encounter a departure, it means someone left the room, so you can free up a chair.
Keep track of the maximum number of people in the room at any single point in time. This is the peak occupancy.
The maximum number of people you ever saw in the room at the same time is the minimum number of chairs needed.

Code Implementation

def min_chairs_needed(arrival_times, departure_times):
    events = []
    for arrival_time in arrival_times:
        events.append((arrival_time, 'arrival'))
    for departure_time in departure_times:
        events.append((departure_time, 'departure'))

    # Sort events by time, arrivals prioritized
    events.sort(key=lambda x: (x[0], x[1] == 'departure'))

    chairs_needed_at_the_same_time = 0
    max_chairs_needed = 0

    for _, event_type in events:
        # Arrival means we need an extra chair
        if event_type == 'arrival':
            chairs_needed_at_the_same_time += 1
            max_chairs_needed = max(max_chairs_needed, chairs_needed_at_the_same_time)
        # Departure frees up a chair
        else:
            chairs_needed_at_the_same_time -= 1

    return max_chairs_needed

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operations in this algorithm are sorting the arrival times and departure times. Sorting algorithms like merge sort or quicksort, which are typically used in standard library implementations, have a time complexity of O(n log n), where n is the number of arrival or departure times. We sort both arrival and departure times, so it's 2 * O(n log n), which is still O(n log n). The iteration through the sorted arrival and departure times is O(n), but this is dominated by the sorting time complexity.

Space Complexity

O(N) – The algorithm creates two new sorted lists, one for arrival times and one for departure times. The size of each list is directly proportional to the number of arrival/departure events, which is N. Therefore, the auxiliary space required to store these lists is O(N), where N is the total number of arrival and departure times.

Edge Cases

Case	How to Handle
Null or empty arrival and departure arrays	Return 0 if both arrays are null or empty, as no chairs are needed.
Arrival and departure arrays have different lengths	Throw an IllegalArgumentException or return -1 to indicate invalid input.
Single arrival and departure event	Return 1, as only one chair is needed at most.
Arrival and departure times are the same (arrival[i] == departure[i])	This should be handled correctly by any valid algorithm, as the person arrives and immediately leaves, not affecting peak occupancy.
Arrival and departure times are out of order (departure[i] < arrival[i])	Throw an IllegalArgumentException or correct the order before processing.
Arrival and departure times contain large numbers (potential integer overflow)	Use long instead of int to store times to prevent potential integer overflow.
All individuals arrive at the same time, and then depart at different times	The maximum number of concurrent individuals will be the total number of individuals, and the solution should correctly count this.
Individuals arrive at different times, and all depart at the same time	Only one chair is needed as only one person is present at any given moment.

Null or empty arrival and departure arrays

How to Handle:

Return 0 if both arrays are null or empty, as no chairs are needed.

Arrival and departure arrays have different lengths

How to Handle:

Throw an IllegalArgumentException or return -1 to indicate invalid input.

Single arrival and departure event

How to Handle:

Return 1, as only one chair is needed at most.

Arrival and departure times are the same (arrival[i] == departure[i])

How to Handle:

This should be handled correctly by any valid algorithm, as the person arrives and immediately leaves, not affecting peak occupancy.

Arrival and departure times are out of order (departure[i] < arrival[i])

How to Handle:

Throw an IllegalArgumentException or correct the order before processing.

Arrival and departure times contain large numbers (potential integer overflow)

How to Handle:

Use long instead of int to store times to prevent potential integer overflow.

All individuals arrive at the same time, and then depart at different times

How to Handle:

The maximum number of concurrent individuals will be the total number of individuals, and the solution should correctly count this.

Individuals arrive at different times, and all depart at the same time

How to Handle:

Only one chair is needed as only one person is present at any given moment.