Introduction

In this blog post, we will explore how to solve the LeetCode problem "2147" 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

Along a long library corridor, there is a line of seats and decorative plants. You are given a 0-indexed string corridor of length n consisting of letters 'S' and 'P' where each 'S' represents a seat and each 'P' represents a plant. One room divider has already been installed to the left of index 0, and another to the right of index n - 1. Additional room dividers can be installed. For each position between indices i - 1 and i (1 <= i <= n - 1), at most one divider can be installed. Divide the corridor into non-overlapping sections, where each section has exactly two seats with any number of plants. There may be multiple ways to perform the division. Two ways are different if there is a position with a room divider installed in the first way but not in the second way. Return the number of ways to divide the corridor. Since the answer may be very large, return it modulo 109 + 7. If there is no way, return 0. Example 1: Input: corridor = "SSPPSPS" Output: 3 Explanation: There are 3 different ways to divide the corridor. The black bars in the above image indicate the two room dividers already installed. Note that in each of the ways, each section has exactly two seats. Example 2: Input: corridor = "PPSPSP" Output: 1 Explanation: There is only 1 way to divide the corridor, by not installing any additional dividers. Installing any would create some section that does not have exactly two seats. Example 3: Input: corridor = "S" Output: 0 Explanation: There is no way to divide the corridor because there will always be a section that does not have exactly two seats. Constraints: n == corridor.length 1 <= n <= 105 corridor[i] is either 'S' or 'P'.

Explanation

Here's a breakdown of the approach and the Python code:

• High-Level Approach:

  • Count the total number of seats. If the count is not even or is zero, there's no solution, return 0.
  • Iterate through the corridor, keeping track of the number of seats encountered so far. When we find the second seat in a section, count the number of plants until we find the next seat. The number of ways to place a divider between these plants is one more than the plant count. Multiply the number of ways in each valid section.
  • Apply modulo arithmetic to prevent integer overflow.

• Complexity:

  • Runtime Complexity: O(n)
  • Storage Complexity: O(1)

Code

    def numberOfWays(corridor: str) -> int:
    seats = 0
    for c in corridor:
        if c == 'S':
            seats += 1

    if seats % 2 != 0 or seats == 0:
        return 0

    ways = 1
    seats_found = 0
    plants = 0
    MOD = 10**9 + 7
    counting_plants = False

    for i in range(len(corridor)):
        if corridor[i] == 'S':
            seats_found += 1
            if seats_found > 2 and seats_found % 2 == 1:
                ways = (ways * (plants + 1)) % MOD
                plants = 0
                counting_plants = False
            elif seats_found % 2 == 0:
                counting_plants = True
        elif counting_plants:
            plants += 1

    return ways