# Solving Leetcode Interviews in Seconds with AI: Minimum Number of Chairs in a Waiting Room


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3168" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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.  

	# Explanation
	Here's the solution to determine the minimum number of chairs needed in the waiting room:

*   **Keep Track of Occupancy:** Maintain a counter representing the current number of people in the waiting room.
*   **Update Occupancy:** Increment the counter for each 'E' (enter) and decrement it for each 'L' (leave).
*   **Track Maximum:** Keep track of the maximum value the occupancy counter reaches. This maximum value represents the minimum number of chairs needed.

*   **Runtime Complexity:** O(n), where n is the length of the string s.
*   **Storage Complexity:** O(1)

	
	# Code
	```python
	def min_chairs(s: str) -> int:
    """
    Calculates the minimum number of chairs needed in a waiting room given a sequence of entry and exit events.

    Args:
        s: A string representing the sequence of events, where 'E' denotes entry and 'L' denotes exit.

    Returns:
        The minimum number of chairs needed.
    """

    occupancy = 0
    max_occupancy = 0

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

        max_occupancy = max(max_occupancy, occupancy)

    return max_occupancy
	```