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, 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

    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