Problem Description

You are given a string s representing a sequence of events. Each character in the string can either be E or L:

• E: A person enters the waiting room and occupies a chair.
• L: A person leaves the waiting room, freeing up a chair.

The task is to determine the minimum number of chairs needed so that, at any given time, there is a chair available for every person who enters the waiting room. The waiting room is initially empty.

Intuition

To solve this problem, we simulate the sequence of events described by the string s while keeping track of the number of chairs in use and the number of chairs available.

We use two variables:

• cnt: This keeps track of the minimum number of chairs currently required.
• left: This accounts for empty chairs that are available after someone leaves.

As we traverse through each character in the string:

• If the character is E, we check if there are any empty chairs (left). If there are, we simply decrement the number of available chairs (left). If not, it means we need an additional chair, so we increment cnt.
• If the character is L, we increase the count of available chairs (left) because a person has left, freeing up a chair.

By the end of the traversal, cnt will represent the minimum number of chairs needed to accommodate all incoming people without delay.

Solution Approach

We utilize a simple simulation approach to determine the minimum number of chairs needed in the waiting room:

1. Initialize two variables:

   • cnt to count the minimum number of chairs required.
   • left to track the number of available empty chairs at any time.

2. Traverse the string s character by character:

   • If the current character is 'E', check if there is an available empty chair (left > 0):
     • If there is an empty chair, decrement left since an incoming person will occupy it.
     • If no empty chairs are available, increase cnt to indicate a need for an additional chair.
   • If the current character is 'L', increment left as a person leaving frees up a chair.

3. The final value of cnt after processing the entire string represents the minimum number of chairs needed.

On a high-level view, this solution uses a greedy approach with two counters to dynamically manage the allocation and deallocation of chairs efficiently and ensure optimal usage at any time given the sequence of events.

Here’s the code implementing this solution:

class Solution:
    def minimumChairs(self, s: str) -> int:
        cnt = left = 0
        for c in s:
            if c == "E":
                if left:
                    left -= 1
                else:
                    cnt += 1
            else:
                left += 1
        return cnt

Example Walkthrough

Let's walk through the solution approach with a small example string s = "EELLE":

1. Initialize Variables:

   • cnt = 0 (This tracks the minimum number of chairs needed).
   • left = 0 (This tracks available empty chairs).

2. Process Each Character:

   • Character 1 ('E'):

     • Since there are no empty chairs (left = 0), increment cnt to 1.
     • Now, cnt = 1, left = 0.

   • Character 2 ('E'):

     • Again, no empty chairs are available (left = 0), so increment cnt to 2.
     • Now, cnt = 2, left = 0.

   • Character 3 ('L'):

     • A person leaves, which frees up a chair, so increment left by 1.
     • Now, cnt = 2, left = 1.

   • Character 4 ('L'):

     • Another person leaves, freeing up another chair, so increment left to 2.
     • Now, cnt = 2, left = 2.

   • Character 5 ('E'):

     • An entering person uses one of the empty chairs, so decrement left by 1.
     • Now, cnt = 2, left = 1.

3. Final Result:

   • After processing the entire string, cnt = 2, which represents the minimum number of chairs needed throughout the events to accommodate all entering persons without delay.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n), where n is the length of the string s. This is because the code iterates over each character in the string exactly once, performing constant-time operations for each character. The space complexity is O(1), as it uses a fixed amount of extra space (variables cnt and left) irrespective of the input size.

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