Problem Description

You are given a string s that represents events happening at each second in a waiting room. Each character in the string represents what happens at that second:

• If s[i] == 'E': A person enters the waiting room and takes one of the chairs
• If s[i] == 'L': A person leaves the waiting room and frees up a chair

The waiting room starts empty with no chairs initially. Your task is to find the minimum number of chairs needed so that every person who enters the waiting room can have a chair to sit on.

For example, if s = "EELLL":

• At second 0: Person enters, need 1 chair (total chairs needed: 1)
• At second 1: Another person enters, need 1 more chair (total chairs needed: 2)
• At second 2: Person leaves, 1 chair becomes free
• At second 3: Person leaves, 2 chairs are now free
• At second 4: Person leaves (but only 2 people entered, so this represents an already empty chair)

The key insight is that we need to track the maximum number of people in the room at any given time, which determines the minimum chairs required.

Intuition

The key observation is that we need to track two things: the total number of chairs we've acquired so far, and how many of those chairs are currently empty (available for reuse).

Think of it this way - when someone enters ('E'), they need a chair. If there's an empty chair available from someone who left earlier, we can reuse it. If not, we need to get a new chair. Once we get a chair, it stays in the waiting room forever - we never remove chairs, we just track whether they're occupied or empty.

When someone leaves ('L'), they free up a chair that can be reused by the next person who enters.

This leads us to a greedy approach:

• Keep a counter cnt for the total number of chairs we've acquired
• Keep a counter left for the number of currently empty chairs
• When we see 'E':
  • If left > 0, someone can use an empty chair (decrease left by 1)
  • If left == 0, all chairs are occupied, so we need a new chair (increase cnt by 1)
• When we see 'L':
  • A chair becomes empty (increase left by 1)

The beauty of this approach is that we only add new chairs when absolutely necessary - when someone enters and all existing chairs are occupied. This naturally gives us the minimum number of chairs needed.

Solution Approach

We implement a simulation approach that processes each event in the string sequentially:

Variables Used:

• cnt: Tracks the total number of chairs we've acquired (this is our answer)
• left: Tracks the number of currently empty/available chairs

Algorithm Steps:

1. Initialize both cnt and left to 0, since the waiting room starts empty with no chairs.

2. Iterate through each character c in the string s:

   When c == 'E' (person enters):

   • Check if there are any empty chairs available (left > 0)
   • If yes: Use one empty chair by decrementing left by 1
   • If no: All existing chairs are occupied, so we need to add a new chair by incrementing cnt by 1

   When c == 'L' (person leaves):

   • The person frees up their chair, so increment left by 1

3. After processing all events, return cnt as the minimum number of chairs needed.

Example Walkthrough:

For s = "EELEE":

• i=0, 'E': left=0, so add new chair → cnt=1, left=0
• i=1, 'E': left=0, so add new chair → cnt=2, left=0
• i=2, 'L': person leaves → cnt=2, left=1
• i=3, 'E': left=1, use empty chair → cnt=2, left=0
• i=4, 'E': left=0, so add new chair → cnt=3, left=0

Final answer: 3 chairs needed

The time complexity is O(n) where n is the length of the string, and space complexity is O(1) as we only use two variables.

Example Walkthrough

Let's walk through the example s = "EELEELL" to understand how our solution works:

Initial State:

• cnt = 0 (no chairs acquired yet)
• left = 0 (no empty chairs available)

Processing each second:

Second 0: 'E' (person enters)

• Check: Are there empty chairs? left = 0, so NO
• Action: Need a new chair, so cnt = 1
• State: cnt = 1, left = 0

Second 1: 'E' (person enters)

• Check: Are there empty chairs? left = 0, so NO
• Action: Need another new chair, so cnt = 2
• State: cnt = 2, left = 0

Second 2: 'L' (person leaves)

• Action: A chair becomes empty, so left = 1
• State: cnt = 2, left = 1

Second 3: 'E' (person enters)

• Check: Are there empty chairs? left = 1, so YES
• Action: Use the empty chair, so left = 0
• State: cnt = 2, left = 0

Second 4: 'E' (person enters)

• Check: Are there empty chairs? left = 0, so NO
• Action: Need a new chair, so cnt = 3
• State: cnt = 3, left = 0

Second 5: 'L' (person leaves)

• Action: A chair becomes empty, so left = 1
• State: cnt = 3, left = 1

Second 6: 'L' (person leaves)

• Action: Another chair becomes empty, so left = 2
• State: cnt = 3, left = 2

Final Answer: 3 chairs

The key insight: We only added new chairs (increased cnt) when someone entered and all existing chairs were occupied. When chairs became available from people leaving, we reused them instead of adding more chairs. This greedy approach ensures we use the minimum number of chairs possible.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the string s. The algorithm iterates through the string exactly once, performing constant-time operations (O(1)) for each character - checking if the character is "E" or "L", and updating the counters accordingly.

The space complexity is O(1). The algorithm only uses a fixed amount of extra space regardless of the input size, storing just two integer variables: cnt (to track the minimum chairs needed) and left (to track available chairs after people leave).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Tracking Current Occupancy Instead of Total Chairs

The Mistake: Many people initially try to solve this by tracking the current number of people in the room and returning the maximum occupancy:

def minimumChairs(self, s: str) -> int:
    current_people = 0
    max_people = 0
  
    for event in s:
        if event == "E":
            current_people += 1
            max_people = max(max_people, current_people)
        else:  # event == "L"
            current_people -= 1
  
    return max_people

Why It's Wrong: This approach assumes we can dynamically add and remove chairs as needed. However, once we acquire a chair, we keep it for the entire duration. The problem asks for the minimum number of chairs we need to acquire in total, not the maximum occupancy at any point.

The Fix: Track two separate values:

• Total chairs acquired (which never decreases)
• Available empty chairs (which increases when people leave)

Pitfall 2: Not Handling the Initial Empty State Correctly

The Mistake: Starting with available_chairs = 1 or some non-zero value, thinking the room has chairs initially:

def minimumChairs(self, s: str) -> int:
    chairs_needed = 1  # Wrong: assuming we start with a chair
    available_chairs = 1  # Wrong: the room starts empty
    # ... rest of code

Why It's Wrong: The problem clearly states "The waiting room starts empty with no chairs initially." We only add chairs when people enter and no empty chairs are available.

The Fix: Initialize both chairs_needed and available_chairs to 0.

Pitfall 3: Misunderstanding What "L" Means

The Mistake: Assuming that "L" always corresponds to a person who previously entered:

def minimumChairs(self, s: str) -> int:
    # Trying to validate that leaves match enters
    if s.count('L') > s.count('E'):
        return -1  # Invalid input

Why It's Wrong: While logically there shouldn't be more leaves than enters in a real scenario, the problem doesn't require us to validate this. We should process the events as given and trust that the input is valid according to the problem constraints.

The Fix: Simply process each 'L' as incrementing available chairs without validation. The problem guarantees valid input sequences.